From 12e0f3ad209371bbe59e1ca0cde9806590e83023 Mon Sep 17 00:00:00 2001
From: sean-fitzpatrick
Date: Fri, 9 May 2025 12:25:20 -0600
Subject: [PATCH 2/9] Greg's edits - derivatives
---
ptx/sec_deriv_chainrule.ptx | 162 ++++++++++++++---------------
ptx/sec_deriv_implicit.ptx | 23 ++--
ptx/sec_deriv_interpret.ptx | 4 +-
ptx/sec_deriv_intro.ptx | 14 ++-
ptx/sec_deriv_inverse_function.ptx | 12 +--
5 files changed, 109 insertions(+), 106 deletions(-)
diff --git a/ptx/sec_deriv_chainrule.ptx b/ptx/sec_deriv_chainrule.ptx
index a7b3f39ab..1989cfc8e 100644
--- a/ptx/sec_deriv_chainrule.ptx
+++ b/ptx/sec_deriv_chainrule.ptx
@@ -166,7 +166,7 @@
The statement of takes care to ensure
- this problem does not arise, but our focus is more on the derivative result than
+ this problem does not arise. We will focus more on the derivative result than
on the domain/range conditions.
@@ -1034,90 +1034,86 @@
That is, the rate at which the u gear makes a revolution is twice as fast as the rate at which the x gear makes a revolution.
-
- Convention with s
+ Conventional notation
Using s(t) to represent position is a fairly common mathematical convention.
It is also common to use s to represent arc length.
@@ -1262,7 +1262,7 @@
- f(x) is line with slope 2 and x intercept . g(x) is a parabola with vertex at ( , )
+ f(x) is a line with slope 2 and x intercept . g(x) is a parabola with vertex at ( , )
The graph of f is a line with slope 2a and the x intercept h .
diff --git a/ptx/sec_deriv_intro.ptx b/ptx/sec_deriv_intro.ptx
index e295c9a28..ac3be7980 100644
--- a/ptx/sec_deriv_intro.ptx
+++ b/ptx/sec_deriv_intro.ptx
@@ -77,9 +77,9 @@
Units in Calculations
- In the above calculations,
- we left off the units until the end of the problem.
- You should always be sure that you label your answer with the correct units.
+ In our calculations of the difference quotients,
+ we did not label the units until giving the final answer.
+ Be sure to always label your answer with the correct units.
For example, if g(x) gave you the cost
(in $)
of producing x widgets,
@@ -1009,7 +1009,7 @@
\text{Product/sum limit rules}
\amp = \sin(x)\cdot 0 + \cos(x) \cdot 1 \amp \amp \text{Applied }
- \amp = \cos(x). \amp \amp \text{(Are you surprised?)}
+ \amp = \cos(x). \amp \amp
@@ -1511,6 +1511,12 @@
we conclude g is differentiable on its domain of [0,\infty) .
+
+ We state (without proof) that \gp(x) = 3\sqrt{x}/2 .
+ Note that \lim_{x\to 0^+}\gp(x) = 0 ;
+ again, this limit is easier to evluate than the limit of the difference quotient.
+
+
A graph of y=x^{1/2} and y=x^{3/2} in
diff --git a/ptx/sec_deriv_inverse_function.ptx b/ptx/sec_deriv_inverse_function.ptx
index 9d02bc7b8..0335e351b 100644
--- a/ptx/sec_deriv_inverse_function.ptx
+++ b/ptx/sec_deriv_inverse_function.ptx
@@ -180,8 +180,6 @@
-
-
Information about f |
Information about g=f^{-1} |
@@ -192,19 +190,19 @@
|
-
+
Slope of tangent line to f at x=1 is 3
-
+
|
-
+
Slope of tangent line to g at x=1.5 is 1/3
-
+
|
\fp(1) = 3 |
- g'(1.5) = 1/3 |
+ \gp(1.5) = 1/3 |
From 3662f964cdd7a64762a32b8864694f1970957ecd Mon Sep 17 00:00:00 2001
From: sean-fitzpatrick
Date: Fri, 9 May 2025 12:25:40 -0600
Subject: [PATCH 3/9] Greg's edits - differentials, concavity
---
ptx/sec_differentials.ptx | 31 +++++++++++++-----
ptx/sec_graph_concavity.ptx | 63 +++++++++++++++++++------------------
2 files changed, 55 insertions(+), 39 deletions(-)
diff --git a/ptx/sec_differentials.ptx b/ptx/sec_differentials.ptx
index 3296eaf01..e0699d4e9 100644
--- a/ptx/sec_differentials.ptx
+++ b/ptx/sec_differentials.ptx
@@ -164,15 +164,18 @@
f(c+\dx) \approx \ell(c+\dx)
,
since the tangent line to a function approximates well the values of that function near x=c .
- This tangent line approximation is used frequently enough in applications that we give it a name.
+ In fact, the tangent line is the graph of the linear function that best approximates the value of
+ f(x) for x near c .
+ Because of its value in applications, we give it a name.
- The function \ell(x) is often referred to as the linearization ,
+ Let f be differentiable on an open inverval I containing c .
+ The function \ell(x) = \fp(c)(x-c)+f(c) is called the linearization ,
or linear approximation of f at c .
- It is the linear function that best approximates the value of f(x) when x is close to c .
+
linearization
approximation tangent line
approximation linear
@@ -409,14 +412,26 @@
- PID controllers
+ Applications of differentials
- Another place differentials are used is in a PID controller,
- which stands for Proportional Integral Derivative .
- A PID controller uses concepts of both derivative and integral calculus to very accurately control a process
- (such as maintaining a stable temperature on an espresso machine).
+ Differentials are used within Proportional Integral Derivative (PID ) controllers,
+ which use both integral and differential calculus to accurately control a process.
+
+
+
+ For instance, consider the task of steering a self driving car.
+ If the vehicle drifts to the left, how much should the wheel be turned to the right to correct the path?
+ Too little and the problem may not be corrected at all;
+ too much and the car will overshoot its target and extreme corrective action will need to be taken.
+ PID controllers consider the rate at which the drift and corrective measures take effect.
+
+
+
+ Common, everyday applications of PID controllers are in automotive cruise control
+ and maintaining espresso machine temperature.
+
We use differentials once more to approximate the value of a function.
Even though calculators are very accessible,
diff --git a/ptx/sec_graph_concavity.ptx b/ptx/sec_graph_concavity.ptx
index bbd7fd02f..99e8ee825 100644
--- a/ptx/sec_graph_concavity.ptx
+++ b/ptx/sec_graph_concavity.ptx
@@ -56,6 +56,27 @@
+
+
+
+ Geometric concavity
+
+ Geometrically speaking, a function is concave up if its graph lies below its secant line segments,
+ and above its tangent lines; see and .
+
+
+
+ A function is concave down if its graph lies above its secant lines and below its tangent lines;
+ see and .
+
+
+
Geometrically, the condition in Equation
states that a graph is concave up if the midpoint of the secant line from (a,f(a))
@@ -129,15 +150,6 @@
-
-
-
Consider a function f such that f is continuous on [a,b] and differentiable on (a,b) .
Note that \frac{a+b}{2} is the midpoint of the interval [a,b] .
@@ -206,7 +218,7 @@
- As with ,
+
lets us conclude that the graph of a function is concave up (or down)
on a closed interval, assuming that the function is continuous on that interval.
Again, we follow the convention that when a problem asks us to give the intervals on which the graph is concave up or down,
@@ -464,14 +476,6 @@
-
- Geometric Concavity
-
- Geometrically speaking,
- a function is concave up if its graph lies above its tangent lines and below secant line segments.
- A function is concave down if its graph lies below its tangent lines and above secant line segments.
-
-
If knowing where a graph is concave up/down is important,
it makes sense that the places where the graph changes from one to the other is also important.
@@ -1208,19 +1212,16 @@
-
- Use Wisely
-
- The second derivative test can only be used on a function that is twice differentiable at c .
- For functions that are not twice differentiable at c ,
- you will need to use the .
- If you've already determined the sign diagram for \fp ,
- the is usually
- easier to apply, and it applies in cases when
- does not.
-
-
-
+
+ The second derivative test can only be used on a function that is twice differentiable at c .
+ For functions that are not twice differentiable at c ,
+ you will need to use the .
+ If you've already determined the sign diagram for \fp ,
+ the is usually
+ easier to apply, and it applies in cases when
+ does not.
+
+
We have been learning how the first and second derivatives of a function relate information about the graph of that function.
We have found intervals of increasing and decreasing,
From 8b0426e8eb2c6b58008cc87ee86974ab3ff97edf Mon Sep 17 00:00:00 2001
From: sean-fitzpatrick
Date: Fri, 9 May 2025 13:08:55 -0600
Subject: [PATCH 4/9] other graph edits
---
ptx/sec_graph_extreme_values.ptx | 34 +++----
ptx/sec_graph_incr_decr.ptx | 160 +++++++++++++++----------------
ptx/sec_graph_mvt.ptx | 4 +-
3 files changed, 99 insertions(+), 99 deletions(-)
diff --git a/ptx/sec_graph_extreme_values.ptx b/ptx/sec_graph_extreme_values.ptx
index c50288926..664b05dcb 100644
--- a/ptx/sec_graph_extreme_values.ptx
+++ b/ptx/sec_graph_extreme_values.ptx
@@ -306,6 +306,22 @@
+
+ Alternative Vocabulary
+
+ The terms local minimum
+ and local maximum
+ are often used as synonyms for relative minimum
+ and relative maximum .
+
+
+
+ As it makes intuitive sense that an absolute maximum is also a relative maximum,
+
+ allows a relative maximum to occur at an interval's endpoint.
+
+
+
Relative Minimum and Relative Maximum
@@ -344,7 +360,7 @@
-
+
The graph of a function is given below,
@@ -403,22 +419,6 @@
-
- Alternative Vocabulary
-
- The terms local minimum
- and local maximum
- are often used as synonyms for relative minimum
- and relative maximum .
-
-
-
- As it makes intuitive sense that an absolute maximum is also a relative maximum,
-
- allows a relative maximum to occur at an interval's endpoint.
-
-
-
We briefly practice using these definitions.
diff --git a/ptx/sec_graph_incr_decr.ptx b/ptx/sec_graph_incr_decr.ptx
index 43ebe2ae3..67d5e1099 100644
--- a/ptx/sec_graph_incr_decr.ptx
+++ b/ptx/sec_graph_incr_decr.ptx
@@ -85,7 +85,7 @@
- Caution: the definition we give in
+ Note: the definition we give in
is not the one you will find in formal mathematics textbooks.
Such texts define a function to be increasing on I if,
for every a\lt b in I , f(a)\leq f(b) .
@@ -1027,6 +1027,85 @@
breaking the number line into four subintervals as shown in .
+
+ Number line for f in
+
+
+
+ Number line showing critical points and intervals of increase and decrease
+
+
+ On a number line, three points are marked.
+ The first point is labeled below with -1 , and above with rel min ,
+ indicating that f has a relative minimum at the critical point x=-1 .
+
+
+
+ The next point is labeled below with 0 and above with rel max ,
+ indicating that f has a relative maximum at the critical point x=0 .
+
+
+
+ The last point is labeled below with 1 and above with rel min ,
+ indicating that f has a relative minimum at the critical point x=1 .
+
+
+
+ In between these points there is text indicating the sign of \fp(x) ,
+ and whether f is increasing or decreasing, as follows:
+
+
+
+ For x\lt -1 , f'\tt 0 and f is decreasing
+
+
+
+
+
+ For -1\lt x\lt 0 , \fp\gt 0 and f is increasing
+
+
+
+
+
+ For 0\lt x\lt 1 , \fp\lt 0 and f is decreasing
+
+
+
+
+
+ For x\gt 1 , \fp\gt 0 and f is increasing
+
+
+
+
+
+
+ \begin{tikzpicture}
+ \begin{axis}[
+ numberline,
+ xmin=-2,
+ xmax=2,
+ extra x ticks={-1, 0, 1},
+ extra x tick labels={$-1$,$0$, $1$},
+ ]
+ \addplot[guideline] coordinates {(-1,0) (-1,2)};
+ \addplot[guideline] coordinates {(0,0) (0,2)};
+ \addplot[guideline] coordinates {(1,0) (1,2)};
+ \addplot[mark=none] coordinates {(-1.5,1)} node {\parbox{3em}{\centering $\fp\lt0$\\$f$ decr }};
+ \addplot[mark=none] coordinates {(-0.5,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }};
+ \addplot[mark=none] coordinates {(0.5,1)} node {\parbox{3em}{\centering $\fp\lt0$\\$f$ decr }};
+ \addplot[mark=none] coordinates {(1.5,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }};
+ \addplot[mark=none] coordinates {(-1,2)} node[above] {\parbox{3em}{\centering rel\\min}};
+ \addplot[mark=none] coordinates {(0,2)} node[above] {\parbox{3em}{\centering rel\\max}};
+ \addplot[mark=none] coordinates {(1,2)} node[above] {\parbox{3em}{\centering rel\\min}};
+ \end{axis}
+ \end{tikzpicture}
+
+
+
+
+
@@ -1064,85 +1143,6 @@
.
-
- Number line for f in
-
-
-
- Number line showing critical points and intervals of increase and decrease
-
-
- On a number line, three points are marked.
- The first point is labeled below with -1 , and above with rel min ,
- indicating that f has a relative minimum at the critical point x=-1 .
-
-
-
- The next point is labeled below with 0 and above with rel max ,
- indicating that f has a relative maximum at the critical point x=0 .
-
-
-
- The last point is labeled below with 1 and above with rel min ,
- indicating that f has a relative minimum at the critical point x=1 .
-
-
-
- In between these points there is text indicating the sign of \fp(x) ,
- and whether f is increasing or decreasing, as follows:
-
-
-
- For x\lt -1 , f'\tt 0 and f is decreasing
-
-
-
-
-
- For -1\lt x\lt 0 , \fp\gt 0 and f is increasing
-
-
-
-
-
- For 0\lt x\lt 1 , \fp\lt 0 and f is decreasing
-
-
-
-
-
- For x\gt 1 , \fp\gt 0 and f is increasing
-
-
-
-
-
-
- \begin{tikzpicture}
- \begin{axis}[
- numberline,
- xmin=-2,
- xmax=2,
- extra x ticks={-1, 0, 1},
- extra x tick labels={$-1$,$0$, $1$},
- ]
- \addplot[guideline] coordinates {(-1,0) (-1,2)};
- \addplot[guideline] coordinates {(0,0) (0,2)};
- \addplot[guideline] coordinates {(1,0) (1,2)};
- \addplot[mark=none] coordinates {(-1.5,1)} node {\parbox{3em}{\centering $\fp\lt0$\\$f$ decr }};
- \addplot[mark=none] coordinates {(-0.5,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }};
- \addplot[mark=none] coordinates {(0.5,1)} node {\parbox{3em}{\centering $\fp\lt0$\\$f$ decr }};
- \addplot[mark=none] coordinates {(1.5,1)} node {\parbox{3em}{\centering $\fp>0$\\$f$ incr }};
- \addplot[mark=none] coordinates {(-1,2)} node[above] {\parbox{3em}{\centering rel\\min}};
- \addplot[mark=none] coordinates {(0,2)} node[above] {\parbox{3em}{\centering rel\\max}};
- \addplot[mark=none] coordinates {(1,2)} node[above] {\parbox{3em}{\centering rel\\min}};
- \end{axis}
- \end{tikzpicture}
-
-
-
-
-
We have two positive factors and one negative factor;
\fp(p)\lt 0 and so f is decreasing on (0,1) .
diff --git a/ptx/sec_graph_mvt.ptx b/ptx/sec_graph_mvt.ptx
index bb559278d..6927e8ce0 100644
--- a/ptx/sec_graph_mvt.ptx
+++ b/ptx/sec_graph_mvt.ptx
@@ -299,7 +299,7 @@
Let f be differentiable on (a,b) where f(a)=f(b) .
We consider two cases.
-
+ Case 1
Consider the case when f is constant on [a,b] ;
that is, f(x) = f(a) = f(b) for all x in [a,b] .
@@ -307,7 +307,7 @@
showing there is at least one value c in (a,b) where \fp(c)=0 .
-
+ Case 2
Now assume that f is not constant on [a,b] .
The Extreme Value Theorem guarantees that f has a maximal and minimal value on [a,b] ,
From 4fbce6cf6efc78782dc8b310d2e7f22c2c1df10e Mon Sep 17 00:00:00 2001
From: sean-fitzpatrick
Date: Fri, 9 May 2025 13:09:07 -0600
Subject: [PATCH 5/9] limit edits
---
ptx/sec_lhopitals_rule.ptx | 2 +-
ptx/sec_limit_analytically.ptx | 58 ++++++++++++++++------------------
ptx/sec_limit_def.ptx | 4 +--
3 files changed, 31 insertions(+), 33 deletions(-)
diff --git a/ptx/sec_lhopitals_rule.ptx b/ptx/sec_lhopitals_rule.ptx
index fe63bf38b..ac08917bc 100644
--- a/ptx/sec_lhopitals_rule.ptx
+++ b/ptx/sec_lhopitals_rule.ptx
@@ -50,7 +50,7 @@
- L'Hospital's Rule with indeterminate forms 0/0 and \infty/\infty
+ L'Hospital's Rule with Indeterminate Forms 0/0 and \infty/\infty
L'Hospital's Rule, Part 1
diff --git a/ptx/sec_limit_analytically.ptx b/ptx/sec_limit_analytically.ptx
index d5ad5b478..0c54ce582 100644
--- a/ptx/sec_limit_analytically.ptx
+++ b/ptx/sec_limit_analytically.ptx
@@ -199,7 +199,7 @@
\amp = 3\bigl(\lim_{x\to 2}x\bigr)^2-5\lim_{x\to 2}(x) +7
\amp = 3\cdot 2^2 - 5\cdot 2+7
\amp = 9
-
+ .
@@ -345,9 +345,6 @@
\lim_{x\to c}\sqrt[n]{x} = \sqrt[n]{c}
- ( follows from the
- and
- rules.)
@@ -478,7 +475,33 @@
That is what the Squeeze Theorem states.
This is illustrated in .
-
+
+
+ Squeeze Theorem
+
+ Let f ,
+ g and h be functions on an open interval I
+ containing c such that for all x in I ,
+
+ f(x)\leq g(x) \leq h(x)
+ .
+
+ limit Squeeze Theorem
+ Squeeze Theorem
+
+ If
+
+ \lim_{x\to c} f(x) = L = \lim_{x\to c} h(x)
+ ,
+ then
+
+ \lim_{x\to c} g(x) = L
+ .
+
+
+
+
+
An illustration of the Squeeze Theorem
@@ -520,31 +543,6 @@
-
- Squeeze Theorem
-
- Let f ,
- g and h be functions on an open interval I
- containing c such that for all x in I ,
-
- f(x)\leq g(x) \leq h(x)
- .
-
- limit Squeeze Theorem
- Squeeze Theorem
-
- If
-
- \lim_{x\to c} f(x) = L = \lim_{x\to c} h(x)
- ,
- then
-
- \lim_{x\to c} g(x) = L
- .
-
-
-
-
Explaining the Squeeze Theorem
diff --git a/ptx/sec_limit_def.ptx b/ptx/sec_limit_def.ptx
index 3e71ee97b..540780d17 100644
--- a/ptx/sec_limit_def.ptx
+++ b/ptx/sec_limit_def.ptx
@@ -806,8 +806,8 @@
We can then set \delta to be the minimum of
\abs{\ln(1-\varepsilon)} and \ln(1+\varepsilon) ; ,
- \delta = \min\{\abs{\ln(1-\varepsilon)}, \ln(1+\varepsilon)\} = \ln(1+\varepsilon)
- .
+ \delta = \min\{\abs{\ln(1-\varepsilon)}, \ln(1+\varepsilon)\} = \ln(1+\varepsilon)\text{.} \quad \text{(See marginal note.)}
+
From 34854d2f3922abfe5ff6331d4ed63fd11ed28121 Mon Sep 17 00:00:00 2001
From: sean-fitzpatrick
Date: Mon, 12 May 2025 10:19:44 -0600
Subject: [PATCH 6/9] don't use smallcaps for intialization, etc
---
xsl/apex-latex-print-color.xsl | 7 +++++++
xsl/apex-latex-print-style.xsl | 7 +++++++
xsl/apex-latex-style.xsl | 7 +++++++
3 files changed, 21 insertions(+)
diff --git a/xsl/apex-latex-print-color.xsl b/xsl/apex-latex-print-color.xsl
index 7d9108fae..d15d27eb0 100644
--- a/xsl/apex-latex-print-color.xsl
+++ b/xsl/apex-latex-print-color.xsl
@@ -349,6 +349,13 @@ https://tex.stackexchange.com/questions/605955/can-i-avoid-indentation-of-margin
+
+
+
+
+ {#1}
+
+
diff --git a/xsl/apex-latex-print-style.xsl b/xsl/apex-latex-print-style.xsl
index e2b604d7f..60f9194c5 100644
--- a/xsl/apex-latex-print-style.xsl
+++ b/xsl/apex-latex-print-style.xsl
@@ -354,6 +354,13 @@ https://tex.stackexchange.com/questions/605955/can-i-avoid-indentation-of-margin
+
+
+
+
+ {#1}
+
+
diff --git a/xsl/apex-latex-style.xsl b/xsl/apex-latex-style.xsl
index 5935906cf..2da652ac3 100644
--- a/xsl/apex-latex-style.xsl
+++ b/xsl/apex-latex-style.xsl
@@ -347,6 +347,13 @@ https://tex.stackexchange.com/questions/605955/can-i-avoid-indentation-of-margin
+
+
+
+
+ {#1}
+
+
From fb25c33c4fa807e05a9059798f69dd07160c88a2 Mon Sep 17 00:00:00 2001
From: sean-fitzpatrick
Date: Mon, 12 May 2025 10:20:10 -0600
Subject: [PATCH 7/9] add xref
---
ptx/sec_limit_infty.ptx | 4 ++--
1 file changed, 2 insertions(+), 2 deletions(-)
diff --git a/ptx/sec_limit_infty.ptx b/ptx/sec_limit_infty.ptx
index b4182b3f7..e85f7722c 100644
--- a/ptx/sec_limit_infty.ptx
+++ b/ptx/sec_limit_infty.ptx
@@ -13,7 +13,7 @@
As a motivating example, consider f(x) = 1/x^2 ,
as shown in .
Note how, as x approaches 0, f(x) grows very,
- very large in fact, it grows without bound.
+ very large in fact, it grows without bound.
It seems appropriate, and descriptive, to state that
\lim_{x\to 0} \frac1{x^2}=\infty
@@ -284,7 +284,7 @@
- Evaluating \lim\limits_{x\to 0}\frac1x
+ Evaluating \lim\limits_{x\to 0}\frac1x in
From 638644439fabc8feed1bdb4f09b473181712b2e7 Mon Sep 17 00:00:00 2001
From: sean-fitzpatrick
Date: Mon, 12 May 2025 10:49:25 -0600
Subject: [PATCH 8/9] remaining edits unrelated to figure placement,etc
---
ptx/sec_differentials.ptx | 13 +++--
ptx/sec_limit_analytically.ptx | 10 ++--
ptx/sec_optimization.ptx | 13 +++--
ptx/sec_related_rates.ptx | 91 ++++++++++++++--------------------
4 files changed, 57 insertions(+), 70 deletions(-)
diff --git a/ptx/sec_differentials.ptx b/ptx/sec_differentials.ptx
index e0699d4e9..d60fdea8a 100644
--- a/ptx/sec_differentials.ptx
+++ b/ptx/sec_differentials.ptx
@@ -419,7 +419,7 @@
- For instance, consider the task of steering a self driving car.
+ For instance, consider the task of steering a self driving car.
If the vehicle drifts to the left, how much should the wheel be turned to the right to correct the path?
Too little and the problem may not be corrected at all;
too much and the car will overshoot its target and extreme corrective action will need to be taken.
@@ -676,7 +676,7 @@
Terms and Concepts
-
+
+
+
+ Given a differentiable function y=f(x) ,
+ we are generally free to choose a value for dx ,
+ which then determines the value of dy .
+
+
diff --git a/ptx/sec_limit_analytically.ptx b/ptx/sec_limit_analytically.ptx
index 0c54ce582..c84cc419e 100644
--- a/ptx/sec_limit_analytically.ptx
+++ b/ptx/sec_limit_analytically.ptx
@@ -477,17 +477,17 @@
- Squeeze Theorem
+ Squeeze Theorem
+
- Let f ,
- g and h be functions on an open interval I
+ Let f , g , and h be functions on an open interval I
containing c such that for all x in I ,
f(x)\leq g(x) \leq h(x)
.
- limit Squeeze Theorem
- Squeeze Theorem
+ limit Squeeze Theorem
+ Squeeze Theorem
If
diff --git a/ptx/sec_optimization.ptx b/ptx/sec_optimization.ptx
index be9e93832..eff966781 100644
--- a/ptx/sec_optimization.ptx
+++ b/ptx/sec_optimization.ptx
@@ -253,9 +253,9 @@
Create equations relevant to the context of the problem,
- using the information given. (One of these should describe the
+ using the information given. One of these should describe the
quantity to be optimized.
- We'll call this the fundamental equation. )
+ We'll call this the fundamental equation.
@@ -264,8 +264,7 @@
If the fundamental equation defines the quantity to be optimized as
a function of more than one variable,
reduce it to a single variable function using substitutions derived
- from the other equations
- (we'll call these constraint equations).
+ from the other equations, which we call the constraint equations.
@@ -284,7 +283,7 @@
- Identify the values of all relevant quantities of the problem.
+ Identify the values of all relevant quantities of the problem and write a full sentence conclusion.
@@ -933,7 +932,7 @@
A rancher has feet of fencing in which to construct
- adjacent, equally sized rectangular pens.
+ adjacent, equally sized rectangular pens, as shown below.
What dimensions should these pens have to maximize the enclosed area?
@@ -1032,7 +1031,7 @@
- The #10 can is a standard sized can used by the restaurant
+ The #10 can is a standard sized can used by the restaurant
industry that holds about
with a diameter of
6\,\frac{3}{16}\,\text{in}
diff --git a/ptx/sec_related_rates.ptx b/ptx/sec_related_rates.ptx
index 221cae607..7eb20cda5 100644
--- a/ptx/sec_related_rates.ptx
+++ b/ptx/sec_related_rates.ptx
@@ -38,12 +38,12 @@
-
+
- This section relies heavily on implicit differentiation,
+ Note: This section relies heavily on implicit differentiation,
so referring back to may help.
-
+
We demonstrate the concepts of related rates through examples.
@@ -124,58 +124,39 @@
-
-
- Read the problem carefully and identify the quantities that are
- changing with time.
- (There may be many quantities that change with time,
- try to identify which variables are important to your goal and only
- focus on these quantities.)
-
-
-
-
+
- Draw a diagram
- (if applicable)
- and assign mathematical variables to each quantity that is changing
- with time.
- (If you are given a particular value of a quantity that is also
- changing with time, do not include these values on your diagram.
- We will call these instantaneous values of the variable.)
+ Understand the problem.
+ Clearly identify the quantity whose rate of change you need to determine.
+ Make sketch, if helpful.
-
- Relate the important variables using a mathematical model.
- (Typical models are known formulas for area,
- perimeter, the Pythagorean Theorem or Trigonometric Ratios.)
- It may be necessary to use more than one technique
- (such as similar triangles)
- to reduce your model down to one that only involves the variables
- of interest.
+ Identify other quantities relevant to the context of the problem
+ and create an equation that relates them to the quantity identified in Step .
+ If values for certain quantities are already known, do not substitute these values into your equation yet.
+ These instantaneous values will be used in Step .
- Implicitly differentiate both sides of the equation found in
- Step with respect to t .
+ Implicitly differentiate both sides of the equation found in Step
+ with respect to t .
-
+
- Substitute in the known values of rates and known instantaneous
- values of the variables.
+ Substitute in the known values of rates and known instantaneous values of the variables.
- Solve for the unknown rate.
+ Solve for the unknown rate identified in Step .
@@ -510,6 +491,25 @@
We want to find \lz{B}{t} .
+
+ Practicality
+
+
+ is both interesting and impractical.
+ It highlights the difficulty in using radar in a nonlinear fashion,
+ and explains why in real life
+ the police officer would follow the other driver to determine their speed,
+ and not pull out pencil and paper.
+
+
+
+ The principles here are important, though.
+ Many automated vehicles make judgments about other moving objects based on
+ perceived distances,
+ radar-like measurements and the concepts of related rates.
+
+
+
We have values for everything except \lz{B}{t} .
Solving for this we have:
@@ -534,25 +534,6 @@
-
- Practicality
-
-
- is both interesting and impractical.
- It highlights the difficulty in using radar in a nonlinear fashion,
- and explains why in real life
- the police officer would follow the other driver to determine their speed,
- and not pull out pencil and paper.
-
-
-
- The principles here are important, though.
- Many automated vehicles make judgments about other moving objects based on
- perceived distances,
- radar-like measurements and the concepts of related rates.
-
-
-
Studying related rates
@@ -1124,7 +1105,7 @@
(note the lower elevation here).
- How fast must the gun be able to turn to accurately track the
+ How fast (in radians per second) must the gun be able to turn to accurately track the
aircraft when the plane is:
From f4978a0f4f1bd1aa7f9078bd30ae5eec4e97729a Mon Sep 17 00:00:00 2001
From: sean-fitzpatrick
Date: Mon, 12 May 2025 14:59:48 -0600
Subject: [PATCH 9/9] make consistent use of parentheses for sin etc
---
ptx/appendix_back_reference.ptx | 178 ++++++++++++++++----------------
ptx/sec_ABC.ptx | 2 +-
ptx/sec_FTC.ptx | 2 +-
ptx/sec_Separable.ptx | 14 +--
ptx/sec_graph_incr_decr.ptx | 2 +-
ptx/sec_greensthm.ptx | 32 +++---
ptx/sec_lagrange.ptx | 8 +-
ptx/sec_limit_intro.ptx | 4 +-
ptx/sec_line_int_intro.ptx | 34 +++---
ptx/sec_line_int_vf.ptx | 26 ++---
ptx/sec_parametric_surfaces.ptx | 60 +++++------
ptx/sec_polarcalc.ptx | 14 +--
ptx/sec_shell_method.ptx | 2 +-
ptx/sec_stokes_divergence.ptx | 34 +++---
ptx/sec_surface_integral.ptx | 24 ++---
ptx/sec_taylor_poly.ptx | 2 +-
ptx/sec_vector_fields.ptx | 22 ++--
17 files changed, 230 insertions(+), 230 deletions(-)
diff --git a/ptx/appendix_back_reference.ptx b/ptx/appendix_back_reference.ptx
index 016a48550..01d4d4f74 100644
--- a/ptx/appendix_back_reference.ptx
+++ b/ptx/appendix_back_reference.ptx
@@ -61,51 +61,51 @@
- \lzo{x}(\sin x)=\cos x
+ \lzo{x}(\sin(x))=\cos(x)
- \lzo{x}(\cos x)=-\sin x
+ \lzo{x}(\cos(x))=-\sin(x)
- \lzo{x}(\csc x)=-\csc x\cot x
+ \lzo{x}(\csc(x))=-\csc(x)\cot(x)
- \lzo{x}(\sec x)=\sec x\tan x
+ \lzo{x}(\sec(x))=\sec(x)\tan(x)
- \lzo{x}(\tan x)=\sec^2 x
+ \lzo{x}(\tan(x))=\sec^2(x)
- \lzo{x}(\cot x)=-\csc^2 x
+ \lzo{x}(\cot(x))=-\csc^2(x)
- \lzo{x}(\cosh x)=\sinh x
+ \lzo{x}(\cosh=(x))=\sinh(x)
- \lzo{x}(\sinh x)=\cosh x
+ \lzo{x}(\sinh(x))=\cosh(x)
- \lzo{x}(\sech x)=-\sech x\tanh x
+ \lzo{x}(\sech(x))=-\sech(x)\tanh(x)
- \lzo{x}(\tanh x)=\sech^2 x
+ \lzo{x}(\tanh(x))=\sech^2(x)
- \lzo{x}(\csch x)=-\csch x\coth x
+ \lzo{x}(\csch(x))=-\csch(x)\coth(x)
- \lzo{x}(\coth x)=-\csch^2 x
+ \lzo{x}(\coth(x))=-\csch^2(x)
@@ -114,51 +114,51 @@
Derivatives of Inverse Functions
- \lzo{x}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}
+ \lzo{x}(\sin^{-1}(x))=\frac{1}{\sqrt{1-x^2}}
- \lzo{x}(\cos^{-1}x)=\frac{-1}{\sqrt{1-x^2}}
+ \lzo{x}(\cos^{-1}(x))=\frac{-1}{\sqrt{1-x^2}}
- \lzo{x}(\csc^{-1}x)=\frac{-1}{\abs{x}\sqrt{x^2-1}}
+ \lzo{x}(\csc^{-1}(x))=\frac{-1}{\abs{x}\sqrt{x^2-1}}
- \lzo{x}(\sec^{-1}x)=\frac{1}{\abs{x}\sqrt{x^2-1}}
+ \lzo{x}(\sec^{-1}(x))=\frac{1}{\abs{x}\sqrt{x^2-1}}
- \lzo{x}(\tan^{-1}x)=\frac{1}{1+x^2}
+ \lzo{x}(\tan^{-1}(x))=\frac{1}{1+x^2}
- \lzo{x}(\cot^{-1}x)=\frac{-1}{1+x^2}
+ \lzo{x}(\cot^{-1}(x))=\frac{-1}{1+x^2}
- \lzo{x}(\cosh^{-1}x)=\frac1{\sqrt{x^2-1}}
+ \lzo{x}(\cosh^{-1}(x))=\frac1{\sqrt{x^2-1}}
- \lzo{x}(\sinh^{-1}x)=\frac1{\sqrt{x^2+1}}
+ \lzo{x}(\sinh^{-1}(x))=\frac1{\sqrt{x^2+1}}
- \lzo{x}(\sech^{-1}x)=\frac{-1}{x\sqrt{1-x^2}}
+ \lzo{x}(\sech^{-1}(x))=\frac{-1}{x\sqrt{1-x^2}}
- \lzo{x}(\csch^{-1}x)=\frac{-1}{\abs{x}\sqrt{1+x^2}}
+ \lzo{x}(\csch^{-1}(x))=\frac{-1}{\abs{x}\sqrt{1+x^2}}
- \lzo{x}(\tanh^{-1}x)=\frac1{1-x^2}
+ \lzo{x}(\tanh^{-1}(x))=\frac1{1-x^2}
- \lzo{x}(\coth^{-1}x)=\frac1{1-x^2}
+ \lzo{x}(\coth^{-1}(x))=\frac1{1-x^2}
@@ -196,11 +196,11 @@
- \int \ln x\,dx=x\ln x -x +C
+ \int \ln(x)\,dx=x\ln(x) -x +C
- \int a^x\,dx=\frac{1}{\ln a}\cdot a^x+C
+ \int a^x\,dx=\frac{1}{\ln(a)}\cdot a^x+C
@@ -217,51 +217,51 @@
Integrals Involving Trigonometric Functions
- \int \cos x\,dx=\sin x+C
+ \int \cos(x)\,dx=\sin(x)+C
- \int \sin x\,dx=-\cos x+C
+ \int \sin(x)\,dx=-\cos(x)+C
- \int \tan x\,dx=-\ln \abs{\cos x}+C
+ \int \tan(x)\,dx=-\ln \abs{\cos(x)}+C
- \int \sec x\,dx=\ln \abs{\sec x+\tan x}+C
+ \int \sec(x)\,dx=\ln \abs{\sec(x)+\tan(x)}+C
- \int \csc x\,dx=-\ln \abs{\csc x+\cot x}+C
+ \int \csc(x)\,dx=-\ln \abs{\csc(x)+\cot(x)}+C
- \int \cot x\,dx=\ln \abs{\sin x}+C
+ \int \cot(x)\,dx=\ln \abs{\sin(x)}+C
- \int \sec^2 x\,dx=\tan x+C
+ \int \sec^2(x)\,dx=\tan(x)+C
- \int \csc^2x\,dx=-\cot x+C
+ \int \csc^2(x)\,dx=-\cot(x)+C
- \int \sec x\tan x\,dx=\sec x+C
+ \int \sec(x)\tan(x)\,dx=\sec(x)+C
- \int \csc x\cot x\,dx=-\csc x+C
+ \int \csc(x)\cot(x)\,dx=-\csc(x)+C
- \int \cos^2x\,dx=\frac12x+\frac14\sin\big(2x\big)+C
+ \int \cos^2(x)\,dx=\frac12x+\frac14\sin\big(2x\big)+C
- \int \sin^2x\,dx=\frac12x-\frac14\sin\big(2x\big)+C
+ \int \sin^2(x)\,dx=\frac12x-\frac14\sin\big(2x\big)+C
@@ -282,19 +282,19 @@
Integrals Involving Hyperbolic Functions
- \int \cosh x\,dx=\sinh x+C
+ \int \cosh(x)\,dx=\sinh(x)+C
- \int \sinh x\,dx=\cosh x+C
+ \int \sinh(x)\,dx=\cosh(x)+C
- \int \tanh x\,dx=\ln(\cosh x)+C
+ \int \tanh(x)\,dx=\ln(\cosh(x))+C
- \int \coth x\,dx=\ln \abs{\sinh x}+C
+ \int \coth(x)\,dx=\ln \abs{\sinh(x)}+C
@@ -397,22 +397,22 @@
- \sin \theta = y |
- \cos \theta = x |
+ \sin(\theta) = y |
+ \cos(\theta) = x |
|
- \ds\csc \theta = \frac1y |
- \ds\sec \theta = \frac1x |
+ \ds\csc(\theta) = \frac1y |
+ \ds\sec(\theta) = \frac1x |
|
- \ds\tan \theta = \frac yx |
- \ds\cot \theta = \frac xy |
+ \ds\tan(\theta) = \frac yx |
+ \ds\cot(\theta) = \frac xy |
@@ -439,22 +439,22 @@
- \ds\sin \theta = \frac{\text{O} }{\text{H} } |
- \ds\csc \theta = \frac{\text{H} }{\text{O} } |
+ \ds\sin(\theta) = \frac{\text{O} }{\text{H} } |
+ \ds\csc(\theta) = \frac{\text{H} }{\text{O} } |
|
- \ds\cos \theta = \frac{\text{A} }{\text{H} } |
- \ds\sec \theta = \frac{\text{H} }{\text{A} } |
+ \ds\cos(\theta) = \frac{\text{A} }{\text{H} } |
+ \ds\sec(\theta) = \frac{\text{H} }{\text{A} } |
|
- \ds\tan \theta = \frac{\text{O} }{\text{A} } |
- \ds\cot \theta = \frac{\text{A} }{\text{O} } |
+ \ds\tan(\theta) = \frac{\text{O} }{\text{A} } |
+ \ds\cot(\theta) = \frac{\text{A} }{\text{O} } |
@@ -467,28 +467,28 @@
Pythagorean Identities
- \sin ^2x+\cos ^2x= 1
- \tan^2x+ 1 = \sec^2 x
- 1 + \cot^2x=\csc^2 x
+ \sin^2(x)+\cos^2(x)= 1
+ \tan^2(x)+ 1 = \sec^2(x)
+ 1 + \cot^2(x)=\csc^2(x)
Double Angle Formulas
- \sin 2x = 2\sin x\cos x
+ \sin(2x) = 2\sin(x)\cos(x)
- \cos 2x \amp = \cos^2x - \sin^2 x \amp \amp
- \amp = 2\cos^2x-1 \amp \amp
- \amp = 1-2\sin^2x \amp \amp
+ \cos(2x) \amp = \cos^2(x) - \sin^2(x) \amp \amp
+ \amp = 2\cos^2(x)-1 \amp \amp
+ \amp = 1-2\sin^2(x) \amp \amp
- \tan 2x = \frac{2\tan x}{1-\tan^2 x}
+ \tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}
@@ -497,33 +497,33 @@
Cofunction Identities
- \sin\left(\frac{\pi}{2}-x\right) = \cos x
+ \sin\left(\frac{\pi}{2}-x\right) = \cos(x)
- \cos\left(\frac{\pi}{2}-x\right) = \sin x
+ \cos\left(\frac{\pi}{2}-x\right) = \sin(x)
- \tan\left(\frac{\pi}{2}-x\right) = \cot x
- \csc\left(\frac{\pi}{2}-x\right) = \sec x
+ \tan\left(\frac{\pi}{2}-x\right) = \cot(x)
+ \csc\left(\frac{\pi}{2}-x\right) = \sec(x)
- \sec\left(\frac{\pi}{2}-x\right) = \csc x
+ \sec\left(\frac{\pi}{2}-x\right) = \csc(x)
- \cot\left(\frac{\pi}{2}-x\right) = \tan x
+ \cot\left(\frac{\pi}{2}-x\right) = \tan(x)
Even/Odd Identities
- \sin(-x) = -\sin x
+ \sin(-x) = -\sin(x)
- \cos (-x) = \cos x
+ \cos (-x) = \cos(x)
- \tan (-x) = -\tan x
+ \tan (-x) = -\tan(x)
- \csc(-x) = -\csc x
+ \csc(-x) = -\csc(x)
- \sec (-x) = \sec x
+ \sec (-x) = \sec(x)
- \cot (-x) = -\cot x
+ \cot (-x) = -\cot(x)
@@ -532,24 +532,24 @@
Power-Reducing Formulas
- \sin^2 x = \frac{1-\cos 2x}{2}
+ \sin^2(x) = \frac{1-\cos(2x)}{2}
- \cos^2 x = \frac{1+\cos 2x}{2}
+ \cos^2(x) = \frac{1+\cos(2x)}{2}
- \tan^2x = \frac{1-\cos 2x}{1+\cos 2x}
+ \tan^2(x) = \frac{1-\cos(2x)}{1+\cos(2x)}
Sum to Product Formulas
- \sin x+\sin y = 2\sin \left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)
+ \sin(x)+\sin(y) = 2\sin\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)
- \sin x-\sin y = 2\sin \left(\frac{x-y}2\right)\cos\left(\frac{x+y}2\right)
+ \sin(x)-\sin(y) = 2\sin\left(\frac{x-y}2\right)\cos\left(\frac{x+y}2\right)
- \cos x+\cos y = 2\cos \left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)
+ \cos(x)+\cos(y) = 2\cos\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)
- \cos x-\cos y = -2\sin \left(\frac{x+y}2\right)\sin\left(\frac{x-y}2\right)
+ \cos(x)-\cos(y) = -2\sin\left(\frac{x+y}2\right)\sin\left(\frac{x-y}2\right)
@@ -557,22 +557,22 @@
Product to Sum Formulas
- \sin x\sin y = \frac12 \big(\cos(x-y) - \cos (x+y)\big)
+ \sin(x)\sin(y) = \frac12 \big(\cos(x-y) - \cos (x+y)\big)
- \cos x\cos y = \frac12\big(\cos (x-y) +\cos (x+y)\big)
+ \cos(x)\cos(y) = \frac12\big(\cos (x-y) +\cos (x+y)\big)
- \sin x\cos y = \frac12 \big(\sin(x+y) + \sin (x-y)\big)
+ \sin(x)\cos(y) = \frac12 \big(\sin(x+y) + \sin (x-y)\big)
Angle Sum/Difference Formulas
- \sin (x\pm y) = \sin x\cos y \pm \cos x\sin y
+ \sin (x\pm y) = \sin(x)\cos(y) \pm \cos(x)\sin(y)
- \cos (x\pm y) = \cos x\cos y \mp \sin x\sin y
+ \cos (x\pm y) = \cos(x)\cos(y) \mp \sin(x)\sin(y)
- \tan (x\pm y) = \frac{\tan x\pm \tan y}{1\mp \tan x\tan y}
+ \tan (x\pm y) = \frac{\tan(x)\pm \tan(y)}{1\mp \tan(x)\tan(y)}
@@ -588,7 +588,7 @@
diff --git a/ptx/sec_ABC.ptx b/ptx/sec_ABC.ptx
index 91461afb4..15461061b 100644
--- a/ptx/sec_ABC.ptx
+++ b/ptx/sec_ABC.ptx
@@ -1558,7 +1558,7 @@
The functions f(x) = \cos (x) and
- g(x) = \sin x intersect infinitely many times,
+ g(x) = \sin(x) intersect infinitely many times,
forming an infinite number of repeated, enclosed regions.
Find the areas of these regions.
diff --git a/ptx/sec_FTC.ptx b/ptx/sec_FTC.ptx
index 71b574d1e..b5b6b7011 100644
--- a/ptx/sec_FTC.ptx
+++ b/ptx/sec_FTC.ptx
@@ -2548,7 +2548,7 @@
- Explain why \ds\int_{a}^{a+2\pi} \sin t\, dt = 0 for all values of a .
+ Explain why \ds\int_{a}^{a+2\pi} \sin(t)\, dt = 0 for all values of a .
diff --git a/ptx/sec_Separable.ptx b/ptx/sec_Separable.ptx
index 2ab03775e..4d90a399b 100644
--- a/ptx/sec_Separable.ptx
+++ b/ptx/sec_Separable.ptx
@@ -424,7 +424,7 @@
- \displaystyle (y + 3)\yp + (\ln(x)) \yp - x\sin y = (y+3)\ln(x)
+ \displaystyle (y + 3)\yp + (\ln(x)) \yp - x\sin(y) = (y+3)\ln(x)
@@ -436,13 +436,13 @@
- \displaystyle \yp -x^2\cos y + y = \cos y - x^2 y
+ \displaystyle \yp -x^2\cos(y) + y = \cos(y) - x^2 y
Separable.
- \displaystyle \frac{1}{\cos y - y}\,dy = (x^2+1)\,dx
+ \displaystyle \frac{1}{\cos(y) - y}\,dy = (x^2+1)\,dx
@@ -562,13 +562,13 @@
- \displaystyle \yp = \frac{\sin(x)}{\cos y} ,
+ \displaystyle \yp = \frac{\sin(x)}{\cos(y)} ,
with y(0) = \displaystyle \frac{\pi}{2}
- \sin y + \cos(x) = 2
+ \sin(y) + \cos(x) = 2
@@ -636,12 +636,12 @@
- \displaystyle \yp = (\cos^2x)(\cos^2 2y) , with y(0) = 0
+ \displaystyle \yp = (\cos^2(x))(\cos^2 (2y)) , with y(0) = 0
- 2\tan 2y = 2x + \sin 2x
+ 2\tan(2y) = 2x + \sin(2x)
diff --git a/ptx/sec_graph_incr_decr.ptx b/ptx/sec_graph_incr_decr.ptx
index 67d5e1099..9893cda4a 100644
--- a/ptx/sec_graph_incr_decr.ptx
+++ b/ptx/sec_graph_incr_decr.ptx
@@ -1301,7 +1301,7 @@
- A function f has derivative \fp(x) = (\sin x +2)e^{x^2+1} ,
+ A function f has derivative \fp(x) = (\sin(x) +2)e^{x^2+1} ,
where \fp(x) \gt 1 for all x .
Is f increasing, decreasing,
or can we not tell from the given information? Why or why not?
diff --git a/ptx/sec_greensthm.ptx b/ptx/sec_greensthm.ptx
index 6696b8ca3..ab601aaeb 100644
--- a/ptx/sec_greensthm.ptx
+++ b/ptx/sec_greensthm.ptx
@@ -472,13 +472,13 @@
\vec r_1(t) = \la 1,0\ra + t\la -1,1\ra = \la 1-t, t\ra ,
for 0\leq t\leq 1 .
We parametrize C_2 with the familiar
- \vec r_2(t) = \la \cos t,\sin t\ra on 0\leq t\leq \pi/2 .
+ \vec r_2(t) = \la \cos(t),\sin(t)\ra on 0\leq t\leq \pi/2 .
For reference later, we give each function and its derivative below:
\vec r_1(t) = \la 1-t, t\ra, \vrp_1(t) = \la -1,1\ra
.
- \vec r_2(t) = \la \cos t, \sin t\ra, \vrp_2(t) = \la -\sin t ,\cos t\ra
+ \vec r_2(t) = \la \cos(t), \sin(t)\ra, \vrp_2(t) = \la -\sin(t) ,\cos(t)\ra
.
@@ -496,12 +496,12 @@
Over C_2 ,
- we have M = y = \sin t and N = -x+1 = 1-\cos t .
+ we have M = y = \sin(t) and N = -x+1 = 1-\cos(t) .
Thus the flux across C_2 is:
\int_{C_1} \vec F\cdot \vec n\, ds \amp = \int_{C_1} \Big(M\,g'(t) - N\,\fp(t)\Big)\, dt
- \amp = \int_0^{\pi/2}\Big((\sin t)(\cos t) - (1-\cos t)(-\sin t)\Big)\, dt
- \amp = \int_0^{\pi/2} \sin t\, dt
+ \amp = \int_0^{\pi/2}\Big((\sin(t))(\cos(t)) - (1-\cos(t))(-\sin(t))\Big)\, dt
+ \amp = \int_0^{\pi/2} \sin(t)\, dt
\amp =1
.
@@ -525,12 +525,12 @@
Over C_2 ,
- we have M = -x = -\cos t and N = 2y-x = 2\sin t-\cos t .
+ we have M = -x = -\cos(t) and N = 2y-x = 2\sin(t)-\cos(t) .
Thus the flux across C_2 is:
\int_{C_1} \vec F\cdot \vec n\, ds \amp = \int_{C_1} \Big(M\,g'(t) - N\,\fp(t)\Big)\, dt
- \amp = \int_0^{\pi/2}\Big((-\cos t)(\cos t) - (2\sin t-\cos t)(-\sin t)\Big)\, dt
- \amp = \int_0^{\pi/2} \big(2\sin^2 t-\sin t\cos t-\cos^2t\big)\, dt
+ \amp = \int_0^{\pi/2}\Big((-\cos(t))(\cos(t)) - (2\sin(t)-\cos(t))(-\sin(t))\Big)\, dt
+ \amp = \int_0^{\pi/2} \big(2\sin^2 t-\sin(t)\cos(t)-\cos^2t\big)\, dt
\amp =\pi/4 - 1/2\approx 0.285
.
@@ -765,8 +765,8 @@
Using Green's Theorem
- Let \vec F = \la \sin x,\cos y\ra and let R be the region enclosed by the curve C parametrized by
- \vec r(t) = \la 2\cos t+ \frac1{10}\cos(10t),2\sin t+\frac1{10}\sin(10t)\ra on 0\leq t\leq 2\pi ,
+ Let \vec F = \la \sin(x),\cos(y)\ra and let R be the region enclosed by the curve C parametrized by
+ \vec r(t) = \la 2\cos(t)+ \frac1{10}\cos(10t),2\sin(t)+\frac1{10}\sin(10t)\ra on 0\leq t\leq 2\pi ,
as shown in .
Find the circulation around C .
@@ -849,7 +849,7 @@
Computing the circulation directly using the line integral looks difficult,
as the integrand will include terms like
- \sin\big(2\cos t + \frac1{10}\cos(10t)\big) .
+ \sin\big(2\cos(t) + \frac1{10}\cos(10t)\big) .
@@ -1101,11 +1101,11 @@
- \la 2\cos t,2\sin t\ra , 0\leq t\leq 2\pi .
+ \la 2\cos(t),2\sin(t)\ra , 0\leq t\leq 2\pi .
The flux across C is
\oint_C \vec F\cdot \vec n\, ds \amp = \oint_C\big(M\gp(t)-N\fp(t)\big)\, dt
- \amp = \int_0^{2\pi} \big((2\cos t-2\sin t)(2\cos t) - (2\cos t+2\sin t)(-2\sin t)\big)\, dt
+ \amp = \int_0^{2\pi} \big((2\cos(t)-2\sin(t))(2\cos(t)) - (2\cos(t)+2\sin(t))(-2\sin(t))\big)\, dt
\amp = \int_0^{2\pi} 4\, dt = 8\pi
.
@@ -1558,7 +1558,7 @@
C is the ellipse parametrized by
- \vec r(t) = \langle 4\cos t,3\sin t\rangle on 0\leq t\leq 2\pi .
+ \vec r(t) = \langle 4\cos(t),3\sin(t)\rangle on 0\leq t\leq 2\pi .
@@ -1575,7 +1575,7 @@
C is the curve parametrized by
- \vec r(t) = \langle \cos t,\sin (2t)\rangle on -\pi/2\leq t\leq \pi/2 .
+ \vec r(t) = \langle \cos(t),\sin (2t)\rangle on -\pi/2\leq t\leq \pi/2 .
@@ -1610,7 +1610,7 @@
C is the curve parametrized by
- \vec r(t) = \langle 2\cos t+\frac1{10}\cos(10t),2\sin t+\frac1{10}\sin (10t)\rangle on 0\leq t\leq 2\pi .
+ \vec r(t) = \langle 2\cos(t)+\frac1{10}\cos(10t),2\sin(t)+\frac1{10}\sin (10t)\rangle on 0\leq t\leq 2\pi .
diff --git a/ptx/sec_lagrange.ptx b/ptx/sec_lagrange.ptx
index b52a3670b..816bbafbd 100644
--- a/ptx/sec_lagrange.ptx
+++ b/ptx/sec_lagrange.ptx
@@ -45,10 +45,10 @@
Next, we look for extreme values on the boundary.
The boundary of our region is the circle x^2+y^2=4 ,
- which we can parametrize using x=2\cos t , y=2\sin t ,
+ which we can parametrize using x=2\cos(t) , y=2\sin(t) ,
for t\in [0,2\pi] . For (x,y) on the boundary, we have
- f(x,y) = x^2-8x-3y^2 = 4\cos^2t-16\cos t-12\sin^2t = h(t)
+ f(x,y) = x^2-8x-3y^2 = 4\cos^2(t)-16\cos(t)-12\sin^2(t) = h(t)
,
a function of one variable, with domain [0,2\pi] .
@@ -57,9 +57,9 @@
We learned how to find the extreme values of such a function back in our first course in calculus:
see . We have h(0)=h(2\pi)=-12 , and
- h'(t) = -8\cos t\sin t+16\sin t-24\sin t\cos t = 16\sin t (1-2\cos t)
+ h'(t) = -8\cos(t)\sin(t)+16\sin(t)-24\sin(t)\cos(t) = 16\sin(t) (1-2\cos(t))
.
- Thus, h'(t)=0 if \sin t = 0 (t=0,\pi,2\pi ) or \cos =\frac12 (t=\pi/3, 5\pi/3 ).
+ Thus, h'(t)=0 if \sin(t) = 0 (t=0,\pi,2\pi ) or \cos =\frac12 (t=\pi/3, 5\pi/3 ).
We have already checked that h(0)=h(2\pi)=-12 , so we check the remaining points:
h(\pi) \amp = 4(-1)^2-16(-1) = 20
diff --git a/ptx/sec_limit_intro.ptx b/ptx/sec_limit_intro.ptx
index b524b5333..217769dc3 100644
--- a/ptx/sec_limit_intro.ptx
+++ b/ptx/sec_limit_intro.ptx
@@ -1541,7 +1541,7 @@
When x is near 0 ,
- \dfrac{\sin x}{x} is near what value?
+ \dfrac{\sin(x)}{x} is near what value?
@@ -1550,7 +1550,7 @@
Try values of x close to 0 , such as 0.0001 .
- A calculator reveals that \dfrac{\sin 0.0001}{0.0001}\approx0.999999998\ldots .
+ A calculator reveals that \dfrac{\sin(0.0001)}{0.0001}\approx0.999999998\ldots .
This is near 1 .
diff --git a/ptx/sec_line_int_intro.ptx b/ptx/sec_line_int_intro.ptx
index 77881e055..816695261 100644
--- a/ptx/sec_line_int_intro.ptx
+++ b/ptx/sec_line_int_intro.ptx
@@ -765,12 +765,12 @@
The curve C is the unit circle,
which we will describe with the parametrization
- \vrt = \langle \cos t, \sin t\rangle for 0\leq t\leq 2\pi .
+ \vrt = \langle \cos(t), \sin(t)\rangle for 0\leq t\leq 2\pi .
We find \norm{\vrp(t)} = 1 , so ds = 1 dt .
- We find the values of f over C as f(x,y) = f(\cos t, \sin t) = \cos^2t-\sin^2t+3 .
+ We find the values of f over C as f(x,y) = f(\cos(t), \sin(t)) = \cos^2t-\sin^2t+3 .
Thus the area we seek is (note the use of the \oint f(s) ds notation):
\oint_C f(s)\, ds \amp = \int_0^{2\pi}\big(\cos^2t-\sin^2t+3\big)\, dt
@@ -828,7 +828,7 @@
Evaluating a line integral: area under a curve in space
- Find the area above the xy -plane and below the helix parametrized by \vrt = \langle \cos t,2\sin t,t/\pi\rangle ,
+ Find the area above the xy -plane and below the helix parametrized by \vrt = \langle \cos(t),2\sin(t),t/\pi\rangle ,
for 0\leq t\leq 2\pi ,
as shown in .
@@ -897,7 +897,7 @@
We use the given vector-valued function \vec r(t) to determine the curve C in the xy -plane by simply using the first two components of \vec r(t):
- \vec c(t) = \langle \cos t,2\sin t\rangle .
+ \vec c(t) = \langle \cos(t),2\sin(t)\rangle .
Thus ds = \norm{\vec c\,'(t)}\,dt = \sqrt{\sin^2t + 4\cos^2t}\,dt .
@@ -1157,7 +1157,7 @@
A thin wire follows the path
- \vrt = \langle 1+\cos t,1+\sin t, 1+ \sin(2t)\rangle ,
+ \vrt = \langle 1+\cos(t),1+\sin(t), 1+ \sin(2t)\rangle ,
0\leq t\leq 2\pi .
The density of the wire is determined by its position in space:
\delta(x,y,z) = y+z gm/cm.
@@ -1227,7 +1227,7 @@
We compute the density of the wire as
- \delta(x,y,z) = \delta\big(1+\cos t,1+\sin t, 1+\sin(2t)\big) = 2+\sin t+\sin(2t)
+ \delta(x,y,z) = \delta\big(1+\cos(t),1+\sin(t), 1+\sin(2t)\big) = 2+\sin(t)+\sin(2t)
.
@@ -1241,7 +1241,7 @@
Thus the mass is
- M = \oint_C \delta(s)\, ds = \int_0^{2\pi} \big(2+\sin t+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt \approx 21.08\text{ gm }
+ M = \oint_C \delta(s)\, ds = \int_0^{2\pi} \big(2+\sin(t)+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt \approx 21.08\text{ gm }
.
@@ -1249,13 +1249,13 @@
We compute the moments about the coordinate planes:
M_{yz} \amp = \oint_C x\delta(s)\, ds
- \amp = \int_0^{2\pi}(1+\cos t)\big(2+\sin t+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt
+ \amp = \int_0^{2\pi}(1+\cos(t))\big(2+\sin(t)+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt
\amp \approx 21.08.
M_{xz} \amp = \oint_C y\delta(s)\, ds
- \amp = \int_0^{2\pi}(1+\sin t)\big(2+\sin t+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt
+ \amp = \int_0^{2\pi}(1+\sin(t))\big(2+\sin(t)+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt
\amp \approx 26.35
M_{xy} \amp = \oint_C z\delta(s)\, ds
- \amp = \int_0^{2\pi}\big(1+\sin(2 t)\big)\big(2+\sin t+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt
+ \amp = \int_0^{2\pi}\big(1+\sin(2 t)\big)\big(2+\sin(t)+\sin(2t)\big)\sqrt{1+4\cos^2(2t)}\, dt
\amp \approx 25.40
@@ -1416,7 +1416,7 @@
- C is the curve given by \vec r(t) = \langle \cos t+t\sin t, \sin t-t\cos t\rangle on [0,2\pi] ;
+ C is the curve given by \vec r(t) = \langle \cos(t)+t\sin(t), \sin(t)-t\cos(t)\rangle on [0,2\pi] ;
the function is f(x,y)=5 .
@@ -1449,7 +1449,7 @@
C is the piecewise curve composed of the line segment joining the points (0,0) and (1,1) ,
- along with the quarter-circle parametrized by \langle \cos t,-\sin t+1\rangle on
+ along with the quarter-circle parametrized by \langle \cos(t),-\sin(t)+1\rangle on
[0,\pi/2] (which starts at the point (1,1) and ends at (0,0) ;
the function is f(x,y)=x^2+y^2 .
@@ -1492,7 +1492,7 @@
- C is the portion of the curve y=\sin x on [0,\pi] ;
+ C is the portion of the curve y=\sin(x) on [0,\pi] ;
the function is f(x,y)=x .
@@ -1506,7 +1506,7 @@
- C is the ellipse given by \vec r(t) = \langle 2\cos t,\sin t\rangle on [0,2\pi] ;
+ C is the ellipse given by \vec r(t) = \langle 2\cos(t),\sin(t)\rangle on [0,2\pi] ;
the function is f(x,y)=10-x^2-y^2 .
@@ -1556,7 +1556,7 @@
- C : \vec r(t) = \langle \cos t,\sin t, \sin(2t)+1\rangle for 0\leq t\leq 2\pi .
+ C : \vec r(t) = \langle \cos(t),\sin(t), \sin(2t)+1\rangle for 0\leq t\leq 2\pi .
@@ -1569,7 +1569,7 @@
- C : \vec r(t) = \langle 3\cos t,3\sin t, t^2\rangle for 0\leq t\leq 2\pi .
+ C : \vec r(t) = \langle 3\cos(t),3\sin(t), t^2\rangle for 0\leq t\leq 2\pi .
@@ -1606,7 +1606,7 @@
- C : \vec r(t) = \langle \cos t,\sin t, t\rangle for 0\leq t\leq 4\pi ;
+ C : \vec r(t) = \langle \cos(t),\sin(t), t\rangle for 0\leq t\leq 4\pi ;
\delta(x,y,z) = z .
diff --git a/ptx/sec_line_int_vf.ptx b/ptx/sec_line_int_vf.ptx
index c3b916783..87fa4b7bf 100644
--- a/ptx/sec_line_int_vf.ptx
+++ b/ptx/sec_line_int_vf.ptx
@@ -146,11 +146,11 @@
Notation note: the above Definition and Key Idea implicitly evaluate \vec F along the curve C ,
which is parametrized by \vrt .
For instance,
- if \vec F = \langle x+y, x-y\rangle and \vrt = \langle t^2,\cos t\rangle ,
+ if \vec F = \langle x+y, x-y\rangle and \vrt = \langle t^2,\cos(t)\rangle ,
then evaluating \vec F along C means substituting the x - and y -components of \vrt in for x and y ,
respectively, in \vec F .
Therefore, along C ,
- \vec F = \langle x+y,x-y\rangle = \la t^2+\cos t, t^2-\cos t\ra .
+ \vec F = \langle x+y,x-y\rangle = \la t^2+\cos(t), t^2-\cos(t)\ra .
Since we are substituting the output
of \vrt for the input of \vec F ,
we write this as \vec F\big(\vrt\big) .
@@ -400,20 +400,20 @@
For C_2 ,
it is probably simplest to parametrize the half circle using sine and cosine.
- Recall that \vec r(t) = \la \cos t, \sin t\ra is a parametrization of the unit circle on 0\leq t\leq 2\pi ;
+ Recall that \vec r(t) = \la \cos(t), \sin(t)\ra is a parametrization of the unit circle on 0\leq t\leq 2\pi ;
we add 1 to the second component to shift the circle up one unit,
then restrict the domain to
\pi\leq t\leq 2\pi to obtain only the lower half,
- giving \vec r_2(t) = \la \cos t, \sin t+1\ra ,
+ giving \vec r_2(t) = \la \cos(t), \sin(t)+1\ra ,
\pi\leq t\leq 2\pi ,
- and hence \vrp_2(t) = \la -\sin t, \cos t\ra and \vec F\big(\vec r_2(t)\big) = \la y,x\ra = \la \sin t+1,\cos t\ra .
+ and hence \vrp_2(t) = \la -\sin(t), \cos(t)\ra and \vec F\big(\vec r_2(t)\big) = \la y,x\ra = \la \sin(t)+1,\cos(t)\ra .
Computing the work along C_2 , we have:
- \int_{C_2} \vec F\cdot d\vec r_2 \amp = \int_{\pi}^{2\pi} \la \sin t+1,\cos t\ra\cdot\la -\sin t,\cos t\ra\, dt
- \amp = \int_{\pi}^{2\pi} \big(-\sin^2t-\sin t+\cos^2t\big)\, dt = 2 \text{ ft-lbs }
+ \int_{C_2} \vec F\cdot d\vec r_2 \amp = \int_{\pi}^{2\pi} \la \sin(t)+1,\cos(t)\ra\cdot\la -\sin(t),\cos(t)\ra\, dt
+ \amp = \int_{\pi}^{2\pi} \big(-\sin^2t-\sin(t)+\cos^2t\big)\, dt = 2 \text{ ft-lbs }
.
@@ -651,7 +651,7 @@
Let \vec F = \la -y, x, 1\ra ,
- and let C be the portion of the helix given by \vrt = \langle \cos t,\sin t, t/(2\pi)\rangle on [0,2\pi] ,
+ and let C be the portion of the helix given by \vrt = \langle \cos(t),\sin(t), t/(2\pi)\rangle on [0,2\pi] ,
as shown in .
Evaluate \int_C\vec F\cdot d\vec r .
@@ -768,10 +768,10 @@
- We have \vec F\big(\vec r(t)\big) = \la -\sin t, \cos t, 1\ra and \vrp(t) = \la -\sin t, \cos t, 1/(2\pi)\ra .
+ We have \vec F\big(\vec r(t)\big) = \la -\sin(t), \cos(t), 1\ra and \vrp(t) = \la -\sin(t), \cos(t), 1/(2\pi)\ra .
Thus
- \int_C \vec F\cdot d\vec r \amp = \int_0^{2\pi} \la -\sin t, \cos t, 1\ra\cdot \la -\sin t, \cos t, 1/(2\pi)\ra dt
+ \int_C \vec F\cdot d\vec r \amp = \int_0^{2\pi} \la -\sin(t), \cos(t), 1\ra\cdot \la -\sin(t), \cos(t), 1/(2\pi)\ra dt
\amp = \int_0^{2\pi} \Big(\sin^2t+\cos^2t + \frac1{2\pi}\Big)dt
\amp = 2\pi + 1 \approx 7.28
@@ -1480,7 +1480,7 @@
0 . (One parametrization for C is
- \vec r(t) = \langle \cos t,\sin t\rangle on 0\leq t\leq \pi .)
+ \vec r(t) = \langle \cos(t),\sin(t)\rangle on 0\leq t\leq \pi .)
@@ -1519,7 +1519,7 @@
\vec F = \langle y+z,x+z,x+y\rangle ;
- C is the helix \vec r(t) = \langle \cos t,\sin t,t/(2\pi)\rangle on 0\leq t\leq 2\pi .
+ C is the helix \vec r(t) = \langle \cos(t),\sin(t),t/(2\pi)\rangle on 0\leq t\leq 2\pi .
@@ -1762,7 +1762,7 @@
\vec F = \langle 2x, 2y, 2z\rangle ,
C is curve parametrized by
- \vec r(t) = \langle \cos t,\sin t, \sin (2t)\rangle on 0\leq t\leq 2\pi .
+ \vec r(t) = \langle \cos(t),\sin(t), \sin (2t)\rangle on 0\leq t\leq 2\pi .
diff --git a/ptx/sec_parametric_surfaces.ptx b/ptx/sec_parametric_surfaces.ptx
index 032c288dd..c211c5d7f 100644
--- a/ptx/sec_parametric_surfaces.ptx
+++ b/ptx/sec_parametric_surfaces.ptx
@@ -340,11 +340,11 @@
- We can parametrize the circular boundary of R with the vector-valued function \la 2\cos u,2\sin u\ra ,
+ We can parametrize the circular boundary of R with the vector-valued function \la 2\cos(u),2\sin(u)\ra ,
where 0\leq u\leq 2\pi .
We can obtain the interior of R by scaling this function by a variable amount,
, by multiplying by v :
- \la 2v\cos u,2v\sin u\ra ,
+ \la 2v\cos(u),2v\sin(u)\ra ,
where 0\leq v\leq 1 .
@@ -421,15 +421,15 @@
Thus far, we have determined the x and y components of our parametrization of the surface:
- x=2v\cos u and y=2v\sin u .
+ x=2v\cos(u) and y=2v\sin(u) .
We find the z component simply by using z = f(x,y) = x^2+2y^2 :
- z = (2v\cos u)^2+2(2v\sin u)^2 = 4v^2\cos^2u+8v^2\sin^2u
+ z = (2v\cos(u))^2+2(2v\sin(u))^2 = 4v^2\cos^2u+8v^2\sin^2u
.
- Thus \vec r(u,v) = \langle 2v\cos u,2v\sin u,4v^2\cos^2u+8v^2\sin^2u\rangle ,
+ Thus \vec r(u,v) = \langle 2v\cos(u),2v\sin(u),4v^2\cos^2u+8v^2\sin^2u\rangle ,
0\leq u\leq 2\pi , 0\leq v\leq 1 ,
which is graphed in .
The way that this graphic was generated highlights how the surface was parametrized.
@@ -861,7 +861,7 @@
We can parametrize this ellipse using sines and cosines;
our parametrization can begin with
- \vec r(u,v) = \la \cos u, \text{ ??? } , 2\sin u\ra, 0\leq u\leq 2\pi
+ \vec r(u,v) = \la \cos(u), \text{ ??? } , 2\sin(u)\ra, 0\leq u\leq 2\pi
,
where we still need to determine the y component.
@@ -873,7 +873,7 @@
we can use another variable, v , to describe y .
Our final answer is
- \vec r(u,v) = \la \cos u, v, 2\sin u\ra, 0\leq u\leq 2\pi, -1\leq v\leq 2
+ \vec r(u,v) = \la \cos(u), v, 2\sin(u)\ra, 0\leq u\leq 2\pi, -1\leq v\leq 2
.
@@ -953,11 +953,11 @@
We can parametrize the x component of our surface with
- x=2z\cos u and the y component with y=3z\sin u ,
+ x=2z\cos(u) and the y component with y=3z\sin(u) ,
where 0\leq u\leq 2\pi .
Putting all components together, we have
- \vec r(u,v) = \la 2v\cos u, 3v\sin u, v\ra, 0\leq u\leq 2\pi, -2\leq v\leq 3
+ \vec r(u,v) = \la 2v\cos(u), 3v\sin(u), v\ra, 0\leq u\leq 2\pi, -2\leq v\leq 3
.
@@ -1153,13 +1153,13 @@
Substituting u for \theta and v for \varphi ,
we have
- \vec r(u,v) = \langle 5\sin u\cos v, \sin u\sin v,2\cos u\rangle
+ \vec r(u,v) = \langle 5\sin(u)\cos(v), \sin(u)\sin(v),2\cos(u)\rangle
,
where we still need to determine the ranges of u and v .
- Note how the x and y components of \vec r have \cos v and \sin v terms,
+ Note how the x and y components of \vec r have \cos(v) and \sin(v) terms,
respectively.
This hints at the fact that ellipses are drawn parallel to the xy -plane as v varies,
which implies we should have v range from 0 to 2\pi .
@@ -1167,15 +1167,15 @@
One may be tempted to let 0\leq u\leq 2\pi as well,
- but note how the z component is 2\cos u .
- We only need \cos u to take on values between -1 and 1 once,
+ but note how the z component is 2\cos(u) .
+ We only need \cos(u) to take on values between -1 and 1 once,
therefore we can restrict u to 0\leq u\leq \pi .
The final parametrization is thus
- \vec r(u,v) = \langle 5\sin u\cos v, \sin u\sin v,2\cos u\rangle, 0\leq u\leq\pi, 0\leq v\leq 2\pi
+ \vec r(u,v) = \langle 5\sin(u)\cos(v), \sin(u)\sin(v),2\cos(u)\rangle, 0\leq u\leq\pi, 0\leq v\leq 2\pi
.
@@ -1581,15 +1581,15 @@
In ,
- we parametrized the surface as \vec r(u,v) = \la 2v\cos u, 2v\sin u, 4v^2\cos^2u+8v^2\sin^2u\ra ,
+ we parametrized the surface as \vec r(u,v) = \la 2v\cos(u), 2v\sin(u), 4v^2\cos^2u+8v^2\sin^2u\ra ,
for 0\leq u\leq 2\pi , 0\leq v\leq 1 .
To find the surface area using ,
we need \snorm{\vec r_u\times\vec r_v} .
We find:
- \vec r_u \amp = \la -2v\sin u, 2v\cos u, 8v^2\cos u\sin u\ra
- \vec r_v \amp = \la 2\cos u, 2\sin v, 8v\cos^2 u+16v\sin^2u\ra
- \vec r_u\times\vec r_v \amp = \la 16v^2\cos u, 32v^2\sin u, -4v\ra
+ \vec r_u \amp = \la -2v\sin(u), 2v\cos(u), 8v^2\cos(u)\sin(u)\ra
+ \vec r_v \amp = \la 2\cos(u), 2\sin(v), 8v\cos^2 u+16v\sin^2u\ra
+ \vec r_u\times\vec r_v \amp = \la 16v^2\cos(u), 32v^2\sin(u), -4v\ra
\snorm{\vec r_u\times\vec r_v} \amp = \sqrt{256v^4\cos^2u+1024v^4\sin^2u+16v^2}
.
@@ -1721,7 +1721,7 @@
- \vec r(u,v) = \langle 3v\cos u+1, 3v\sin u+2, 3(3v\cos u+1)^2(3v\sin u+2)\rangle ,
+ \vec r(u,v) = \langle 3v\cos(u)+1, 3v\sin(u)+2, 3(3v\cos(u)+1)^2(3v\sin(u)+2)\rangle ,
on 0\leq u\leq 2\pi , 0\leq v\leq 1 .
@@ -1792,7 +1792,7 @@
- \vec r(u,v) = \langle 4v\cos u, 3v\sin u, 16v\cos u+2(3v\sin u)^2\rangle ,
+ \vec r(u,v) = \langle 4v\cos(u), 3v\sin(u), 16v\cos(u)+2(3v\sin(u))^2\rangle ,
on 0\leq u\leq 2\pi , 0\leq v\leq 1 .
@@ -1806,7 +1806,7 @@
- \vec r(u,v) = \langle v\cos u, v\sin u, 4v\cos u + 2(v\sin u)^2\rangle on
+ \vec r(u,v) = \langle v\cos(u), v\sin(u), 4v\cos(u) + 2(v\sin(u))^2\rangle on
0\leq u\leq 2\pi , 2\leq v\leq 5 .
@@ -1864,7 +1864,7 @@
- \vec r(u,v) = \langle 3\sin u\cos v, 2\sin u\sin v, 4\cos u\rangle with
+ \vec r(u,v) = \langle 3\sin(u)\cos(v), 2\sin(u)\sin(v), 4\cos(u)\rangle with
0\leq u\leq \pi , 0\leq v\leq 2\pi .
@@ -1880,7 +1880,7 @@
Answers may vary;
- one solution is \vec r(u,v) = \langle v\cos u, v, 4v\sin u\rangle with
+ one solution is \vec r(u,v) = \langle v\cos(u), v, 4v\sin(u)\rangle with
0\leq u\leq 2\pi , -1\leq v\leq 5 .
@@ -2428,19 +2428,19 @@
For x^2+y^2/9=1 :
- \vec r(u,v) = \langle \cos u, 3\sin u, v\rangle with
+ \vec r(u,v) = \langle \cos(u), 3\sin(u), v\rangle with
0\leq u\leq 2\pi and 1\leq v\leq 3 .
For z=1 :
- \vec r(u,v) = \langle v\cos u, 3v\sin u, 1\rangle with
+ \vec r(u,v) = \langle v\cos(u), 3v\sin(u), 1\rangle with
0\leq u\leq 2\pi and 0\leq v\leq 1 .
For z=3 :
- \vec r(u,v) = \langle v\cos u, 3v\sin u, 3\rangle with
+ \vec r(u,v) = \langle v\cos(u), 3v\sin(u), 3\rangle with
0\leq u\leq 2\pi and 0\leq v\leq 1 .
@@ -2519,13 +2519,13 @@
For x^2+y^2=(z-1)^2 :
- \vec r(u,v) = \langle v\cos u, v\sin u, 1-v\rangle with
+ \vec r(u,v) = \langle v\cos(u), v\sin(u), 1-v\rangle with
0\leq u\leq 2\pi and 0\leq v\leq 1 .
For z=0 :
- \vec r(u,v) = \langle v\cos u, v\sin u, 0\rangle with
+ \vec r(u,v) = \langle v\cos(u), v\sin(u), 0\rangle with
0\leq u\leq 2\pi and 0\leq v\leq 1 .
@@ -2721,13 +2721,13 @@
For z=4-x^2-4y^2 :
- \vec r(u,v) = \langle 2v\cos u,v\sin u,4-(2v\cos u)^2-4(v\sin u)^2\rangle with
+ \vec r(u,v) = \langle 2v\cos(u),v\sin(u),4-(2v\cos(u))^2-4(v\sin(u))^2\rangle with
0\leq u\leq 2\pi and 0\leq v\leq 1 .
For z=0 :
- \vec r(u,v) = \langle 2v\cos u,v\sin u,0\rangle with
+ \vec r(u,v) = \langle 2v\cos(u),v\sin(u),0\rangle with
0\leq u\leq 2\pi and 0\leq v\leq 1 .
diff --git a/ptx/sec_polarcalc.ptx b/ptx/sec_polarcalc.ptx
index 8f8043718..d41c0af60 100644
--- a/ptx/sec_polarcalc.ptx
+++ b/ptx/sec_polarcalc.ptx
@@ -222,7 +222,7 @@
the 4 th and 1 st quadrants.
Again using reference angles, we have:
- \sin\theta = \frac{-1+\sqrt{33}}8 \Rightarrow \theta = 0.6349,\,2.5067 \text{ radians }
+ \sin(\theta) = \frac{-1+\sqrt{33}}8 \Rightarrow \theta = 0.6349,\,2.5067 \text{ radians }
and
@@ -1115,9 +1115,9 @@
giving us our bounds of integration.
Applying , we have
- L \amp = \int_0^{2\pi} \sqrt{(2\cos\theta)^2+(1+2\sin\theta)^2}\, d\theta
- \amp = \int_0^{2\pi} \sqrt{4\cos^2\theta+4\sin^2\theta +4\sin\theta+1}\, d\theta
- \amp = \int_0^{2\pi} \sqrt{4\sin\theta+5}\, d\theta
+ L \amp = \int_0^{2\pi} \sqrt{(2\cos(\theta))^2+(1+2\sin(\theta))^2}\, d\theta
+ \amp = \int_0^{2\pi} \sqrt{4\cos^2\theta+4\sin^2\theta +4\sin(\theta)+1}\, d\theta
+ \amp = \int_0^{2\pi} \sqrt{4\sin(\theta)+5}\, d\theta
\amp \approx 13.3649
.
@@ -2309,7 +2309,7 @@
- Use the arc length formula to compute the arc length of r=\cos \theta+\sin \theta .
+ Use the arc length formula to compute the arc length of r=\cos(\theta)+\sin(\theta) .
@@ -2322,9 +2322,9 @@
- Use the arc length formula to compute the arc length of the cardioid r=1+\cos\theta .
+ Use the arc length formula to compute the arc length of the cardioid r=1+\cos(\theta) .
(Hint: apply the formula, simplify,
- then use a Power-Reducing Formula to convert 1+\cos \theta into a square.)
+ then use a Power-Reducing Formula to convert 1+\cos(\theta) into a square.)
diff --git a/ptx/sec_shell_method.ptx b/ptx/sec_shell_method.ptx
index 80b612e13..571e31b66 100644
--- a/ptx/sec_shell_method.ptx
+++ b/ptx/sec_shell_method.ptx
@@ -1415,7 +1415,7 @@
Note that in order to use the Washer Method,
- we would need to solve y=\sin x for x ,
+ we would need to solve y=\sin(x) for x ,
requiring the use of the arcsine function.
We leave it to the reader to verify that the outside radius function is
R(y) = \pi-\arcsin y and the inside radius function is r(y)=\arcsin y .
diff --git a/ptx/sec_stokes_divergence.ptx b/ptx/sec_stokes_divergence.ptx
index 835e86c73..e88c94587 100644
--- a/ptx/sec_stokes_divergence.ptx
+++ b/ptx/sec_stokes_divergence.ptx
@@ -773,20 +773,20 @@
We begin by parametrizing C and then find the circulation.
- A unit circle centered at (1,1) can be parametrized with x=\cos t+1 ,
- y=\sin t+1 on 0\leq t\leq 2\pi ;
+ A unit circle centered at (1,1) can be parametrized with x=\cos(t)+1 ,
+ y=\sin(t)+1 on 0\leq t\leq 2\pi ;
to put this curve on the surface f ,
make the z component equal f(x,y) :
- z = 7-2(\cos t+1)-2(\sin t+1) = 3-2\cos t - 2\sin t .
+ z = 7-2(\cos(t)+1)-2(\sin(t)+1) = 3-2\cos(t) - 2\sin(t) .
All together,
- we parametrize C with \vec r(t) = \la \cos t+1, \sin t+1, 3-2\cos t-2\sin t\ra .
+ we parametrize C with \vec r(t) = \la \cos(t)+1, \sin(t)+1, 3-2\cos(t)-2\sin(t)\ra .
The circulation of \vec F around C is
\oint_C\vec F\cdot \, d\vec r \amp = \int_0^{2\pi}\vec F\big(\vec r(t)\big)\cdot \vrp(t)\, dt
- \amp = \int_0^{2\pi}\big(2\sin^3t-2\cos t\sin^2t+3\sin^2t-3\cos t\sin t\big)\, dt
+ \amp = \int_0^{2\pi}\big(2\sin^3t-2\cos(t)\sin^2t+3\sin^2t-3\cos(t)\sin(t)\big)\, dt
\amp = 3\pi
.
@@ -794,7 +794,7 @@
We now parametrize \surfaceS . (We reuse the letter r
for our surface as this is our custom.) Based on the parametrization of C above,
- we describe \surfaceS with \vec r(u,v) = \la v\cos u+1, v\sin u+1, 3-2v\cos u-2v\sin u\ra ,
+ we describe \surfaceS with \vec r(u,v) = \la v\cos(u)+1, v\sin(u)+1, 3-2v\cos(u)-2v\sin(u)\ra ,
where 0\leq u\leq 2\pi and 0\leq v\leq 1 .
@@ -810,7 +810,7 @@
The surface integral of Stokes' Theorem is thus
\iint_\surfaceS (\curl\vec F)\cdot \vec n\, dS \amp = \iint_\surfaceS (\curl\vec F)\cdot (\vec r_u\times \vec r_v)\, dA
- \amp = \int_0^1\int_0^{2\pi} \langle 2v\sin u+2,0,-1\rangle\cdot\langle 2v,2v,v\rangle\, du\, dv
+ \amp = \int_0^1\int_0^{2\pi} \langle 2v\sin(u)+2,0,-1\rangle\cdot\langle 2v,2v,v\rangle\, du\, dv
\amp = 3\pi
,
which matches our previous result.
@@ -970,9 +970,9 @@
We begin by demonstrating that C lies on the surface z=6-x^2-y^2 .
- We can parametrize the x and y components of C with x=\cos t+1 ,
- y=\sin t+1 as before.
- Lifting these components to the surface z=6-x^2-y^2 gives the z component as z = 6-x^2-y^2 = 6-(\cos t+1)^2-(\sin t+1)^2 = 3-2\cos t-2\sin t ,
+ We can parametrize the x and y components of C with x=\cos(t)+1 ,
+ y=\sin(t)+1 as before.
+ Lifting these components to the surface z=6-x^2-y^2 gives the z component as z = 6-x^2-y^2 = 6-(\cos(t)+1)^2-(\sin(t)+1)^2 = 3-2\cos(t)-2\sin(t) ,
which is the same z component as found in .
Thus the curve C lies on the surface z=6-x^2-y^2 ,
as illustrated in .
@@ -987,18 +987,18 @@
We parametrize \surfaceS with
- \vec r(u,v) = \langle v\cos u+1,v\sin u+1, 6-(v\cos u+1)^2-(v\sin u+1)^2\rangle
+ \vec r(u,v) = \langle v\cos(u)+1,v\sin(u)+1, 6-(v\cos(u)+1)^2-(v\sin(u)+1)^2\rangle
,
where 0\leq u\leq 2\pi and 0\leq v\leq 1 ,
and leave it to the reader to confirm that
- \vec r_u\times \vec r_v = \la 2v\big(v\cos u+1\big), 2v\big(v\sin u+1\big),v\ra
+ \vec r_u\times \vec r_v = \la 2v\big(v\cos(u)+1\big), 2v\big(v\sin(u)+1\big),v\ra
,
which also conforms to the right-hand rule with regard to the orientation of C .
With \curl \vec F = \langle 2y,0,-1\rangle as before, we have
\amp\iint_\surfaceS (\curl\vec F)\cdot \vec n\, dS
- \quad\quad\amp = \int_0^1\int_0^{2\pi} \la 2v\sin u+2,0,-1\ra\cdot \la 2v\big(v\cos u+1\big), 2v\big(v\sin u+1\big),v\ra\, du\, dv
+ \quad\quad\amp = \int_0^1\int_0^{2\pi} \la 2v\sin(u)+2,0,-1\ra\cdot \la 2v\big(v\cos(u)+1\big), 2v\big(v\sin(u)+1\big),v\ra\, du\, dv
\quad\quad\amp =3\pi
.
@@ -1551,7 +1551,7 @@
C is the curve parametrized by
- \vec r(t) = \langle \cos t, \sin t, 1\rangle and \surfaceS is the portion of
+ \vec r(t) = \langle \cos(t), \sin(t), 1\rangle and \surfaceS is the portion of
z=x^2+y^2 enclosed by C ;
\vec F = \langle z,-x,y\rangle .
@@ -1625,7 +1625,7 @@
C is the curve parametrized by
- \vec r(t) = \langle \cos t, \sin t, e^{-1}\rangle and \surfaceS is the portion of
+ \vec r(t) = \langle \cos(t), \sin(t), e^{-1}\rangle and \surfaceS is the portion of
z=e^{-x^2-y^2} enclosed by C ;
\vec F = \langle -y,x,1\rangle .
@@ -2301,7 +2301,7 @@
C is the curve whose x - and y -values are given by
- \vec r(t) = \langle 2\cos t,2\sin t\rangle and the z -values are determined by the function z=x^2+y^3-3y+1 ;
+ \vec r(t) = \langle 2\cos(t),2\sin(t)\rangle and the z -values are determined by the function z=x^2+y^3-3y+1 ;
\vec F = \langle -y,x,z\rangle .
@@ -2379,7 +2379,7 @@
C is the curve whose x - and y -values are given by
- \vec r(t) = \langle \cos t,3\sin t\rangle and the z -values are determined by the function z=5-2x-y ;
+ \vec r(t) = \langle \cos(t),3\sin(t)\rangle and the z -values are determined by the function z=5-2x-y ;
\vec F = \langle -\frac13y,3x,\frac23y-3x\rangle .
diff --git a/ptx/sec_surface_integral.ptx b/ptx/sec_surface_integral.ptx
index 7b82a705f..94175036f 100644
--- a/ptx/sec_surface_integral.ptx
+++ b/ptx/sec_surface_integral.ptx
@@ -564,11 +564,11 @@
The boundary of the unit disk in the xy -plane is the unit circle,
which can be described with
- \langle \cos u,\sin u,0\rangle , 0\leq u\leq 2\pi .
+ \langle \cos(u),\sin(u),0\rangle , 0\leq u\leq 2\pi .
To obtain the interior of the circle as well,
we can scale by v , giving
- \vec r_1(u,v) = \langle v\cos u,v\sin u, 0\rangle, 0\leq u\leq 2\pi 0\leq v\leq 1
+ \vec r_1(u,v) = \langle v\cos(u),v\sin(u), 0\rangle, 0\leq u\leq 2\pi 0\leq v\leq 1
.
@@ -579,7 +579,7 @@
we just need a different z component.
With z = 1-x^2-y^2 , we have
- \vec r_2(u,v) = \langle v\cos u,v\sin u, 1-v^2\cos^2u-v^2\sin^2u\rangle = \langle v\cos u,v\sin u, 1-v^2\rangle
+ \vec r_2(u,v) = \langle v\cos(u),v\sin(u), 1-v^2\cos^2u-v^2\sin^2u\rangle = \langle v\cos(u),v\sin(u), 1-v^2\rangle
,
where 0\leq u\leq 2\pi and 0\leq v\leq 1 .
@@ -590,8 +590,8 @@
For \vec n_1 :
- \vec r_{1u}= \langle -v\sin u, v\cos u,0\rangle ,
- \vec r_{1v} = \langle \cos u,\sin u,0\rangle , so
+ \vec r_{1u}= \langle -v\sin(u), v\cos(u),0\rangle ,
+ \vec r_{1v} = \langle \cos(u),\sin(u),0\rangle , so
\vec n_1 = \vec r_{1u}\times \vec r_{1v} = \langle 0,0,-v\rangle
.
@@ -606,17 +606,17 @@
Similarly, \vec n_2 :
- \vec r_{2u}= \langle -v\sin u, v\cos u,0\rangle ,
- \vec r_{2v} = \langle \cos u,\sin u,-2v\rangle , so
+ \vec r_{2u}= \langle -v\sin(u), v\cos(u),0\rangle ,
+ \vec r_{2v} = \langle \cos(u),\sin(u),-2v\rangle , so
- \vec n_2 = \vec r_{2u}\times \vec r_{2v} = \langle -2v^2\cos u,-2v^2\sin u,-v\rangle
+ \vec n_2 = \vec r_{2u}\times \vec r_{2v} = \langle -2v^2\cos(u),-2v^2\sin(u),-v\rangle
.
Again, this normal vector has a negative z -component so we use
- \vec n_2 = \vec r_{2v}\times \vec r_{2u} = \langle 2v^2\cos u,2v^2\sin u,v\rangle
+ \vec n_2 = \vec r_{2v}\times \vec r_{2u} = \langle 2v^2\cos(u),2v^2\sin(u),v\rangle
.
@@ -631,7 +631,7 @@
.
\text{ Flux across } \surfaceS_2 \amp = \iint_{\surfaceS_2} \vec F_1\cdot \vec n_2\, dS
- \amp = \iint_R\langle 0,0,1\rangle\cdot\langle 2v^2\cos u,2v^2\sin u,v\rangle\, dA
+ \amp = \iint_R\langle 0,0,1\rangle\cdot\langle 2v^2\cos(u),2v^2\sin(u),v\rangle\, dA
\amp = \int_0^1\int_0^{2\pi} (v)\, du\, dv
\amp = \pi
.
@@ -680,7 +680,7 @@
\text{ Flux across } \surfaceS_2 \amp = \iint_{\surfaceS_2} \vec F_2\cdot \vec n_2\, dS .
Over \surfaceS_2 , \vec F_2 = \vec F_2\big(\vec r_2(u,v)\big) = \langle 0,0,1-v^2\rangle . Therefore,
- \amp = \iint_R\langle 0,0,1-v^2\rangle\cdot\langle 2v^2\cos u,2v^2\sin u,v\rangle\, dA
+ \amp = \iint_R\langle 0,0,1-v^2\rangle\cdot\langle 2v^2\cos(u),2v^2\sin(u),v\rangle\, dA
\amp = \int_0^1\int_0^{2\pi} (v^3-v)\, du\, dv
\amp = \pi/2
.
@@ -1075,7 +1075,7 @@
9\pi/8 ; the flux over
\surfaceS_1 is 3\pi/4 (use
- \vec r(u,v) = \langle \sin u\cos v,\sin u\sin v,\cos u\rangle on \pi/3\leq u\leq \pi ,
+ \vec r(u,v) = \langle \sin(u)\cos(v),\sin(u)\sin(v),\cos(u)\rangle on \pi/3\leq u\leq \pi ,
0\leq v\leq 2\pi ) and the flux over
\surfaceS_2 is 3\pi/8 (use \vec r(u,v) = \langle v\sqrt{3}\cos (u)/2, v\sqrt{3}\sin(u)/2,1/2\rangle for
0\leq u\leq 2\pi , 0\leq v\leq 1 .
diff --git a/ptx/sec_taylor_poly.ptx b/ptx/sec_taylor_poly.ptx
index f384c4806..f8aa738b3 100644
--- a/ptx/sec_taylor_poly.ptx
+++ b/ptx/sec_taylor_poly.ptx
@@ -2367,7 +2367,7 @@
Find a formula for the n th term of the Maclaurin polynomial
- for f(x)=\sin x .
+ for f(x)=\sin(x) .
diff --git a/ptx/sec_vector_fields.ptx b/ptx/sec_vector_fields.ptx
index 491d853fb..613c9699c 100644
--- a/ptx/sec_vector_fields.ptx
+++ b/ptx/sec_vector_fields.ptx
@@ -559,11 +559,11 @@
Now apply the del operator \nabla to vector fields.
- Let \vec F = \langle x+\sin y,y^2+z,x^2\rangle .
+ Let \vec F = \langle x+\sin(y),y^2+z,x^2\rangle .
We can use vector operations and find the dot product of \nabla and \vec F :
- \nabla \cdot \vec F \amp = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y},\frac{\partial}{\partial z}\right\rangle\cdot \langle x+\sin y,y^2+z,x^2\rangle
- \amp = \frac{\partial}{\partial x}(x+\sin y)+ \frac{\partial}{\partial y}(y^2+z) + \frac{\partial}{\partial z}(x^2)
+ \nabla \cdot \vec F \amp = \left\langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y},\frac{\partial}{\partial z}\right\rangle\cdot \langle x+\sin(y),y^2+z,x^2\rangle
+ \amp = \frac{\partial}{\partial x}(x+\sin(y))+ \frac{\partial}{\partial y}(y^2+z) + \frac{\partial}{\partial z}(x^2)
\amp =1+2y
.
@@ -571,8 +571,8 @@
We can also compute their cross products:
- {\nabla\times \vec F }\amp = {\left\langle \frac{\partial}{\partial y}\big(x^2\big)-\frac{\partial}{\partial z}\big(y^2+z\big),\frac{\partial}{\partial z}\big(x+\sin y\big)-\frac{\partial}{\partial x}\big(x^2\big),\frac{\partial}{\partial x}\big(y^2+z\big)-\frac{\partial}{\partial y}\big(x+\sin y\big)\right\rangle }
- \amp =\langle -1,-2x,-\cos y\rangle
+ {\nabla\times \vec F }\amp = {\left\langle \frac{\partial}{\partial y}\big(x^2\big)-\frac{\partial}{\partial z}\big(y^2+z\big),\frac{\partial}{\partial z}\big(x+\sin(y)\big)-\frac{\partial}{\partial x}\big(x^2\big),\frac{\partial}{\partial x}\big(y^2+z\big)-\frac{\partial}{\partial y}\big(x+\sin(y)\big)\right\rangle }
+ \amp =\langle -1,-2x,-\cos(y)\rangle
.
@@ -766,7 +766,7 @@
- \vec F = \langle \cos y, \sin x\rangle (see )
+ \vec F = \langle \cos(y), \sin(x)\rangle (see )
@@ -1112,14 +1112,14 @@
Instead of trying to rationalize a guess,
we compute the divergence:
- \divv\vec F = \nabla \cdot \vec F = M_x + N_y = \frac{\partial}{\partial x}(\cos y) + \frac{\partial}{\partial y}(\sin x) = 0
+ \divv\vec F = \nabla \cdot \vec F = M_x + N_y = \frac{\partial}{\partial x}(\cos(y)) + \frac{\partial}{\partial y}(\sin(x)) = 0
.
Perhaps surprisingly, the divergence is 0.
With all the loops of different directions in the field,
one is apt to reason the curl is variable.
Indeed, it is:
- \curl \vec F = \nabla\times\vec F = N_x-M_y = \frac{\partial}{\partial x}(\sin x) - \frac{\partial}{\partial y}(\cos y) = \cos x + \sin y
+ \curl \vec F = \nabla\times\vec F = N_x-M_y = \frac{\partial}{\partial x}(\sin(x)) - \frac{\partial}{\partial y}(\cos(y)) = \cos(x) + \sin(y)
.
Depending on the values of x and y ,
the curl may be positive, negative, or zero.
@@ -1970,7 +1970,7 @@
- \vec F = \langle \cos (xy), \sin (xy)\rangle
+ \vec F = \langle \cos(xy), \sin(xy)\rangle
@@ -2072,12 +2072,12 @@
- \vec F = \nabla f , where f(x,y,z) = x^2y+\sin z .
+ \vec F = \nabla f , where f(x,y,z) = x^2y+\sin(z) .
- \divv \vec F = 2y-\sin z
+ \divv \vec F = 2y-\sin(z)