diff --git a/ptx/sec_deriv_chainrule.ptx b/ptx/sec_deriv_chainrule.ptx
index 0e6f985a0..a1b10aebc 100644
--- a/ptx/sec_deriv_chainrule.ptx
+++ b/ptx/sec_deriv_chainrule.ptx
@@ -166,7 +166,7 @@
- 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 |
diff --git a/ptx/sec_differentials.ptx b/ptx/sec_differentials.ptx
index 3296eaf01..d60fdea8a 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,
@@ -661,7 +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_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,
diff --git a/ptx/sec_graph_extreme_values.ptx b/ptx/sec_graph_extreme_values.ptx
index 00bdd690d..5db253ce5 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,
@@ -423,22 +439,6 @@
-
- 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..9893cda4a 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) .
@@ -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_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] ,
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_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..c84cc419e 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.)}
+
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
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_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_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_related_rates.ptx b/ptx/sec_related_rates.ptx
index 7546dc22a..48a3397bd 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 .
@@ -611,6 +592,25 @@
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:
@@ -635,25 +635,6 @@
-
- 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
@@ -1291,7 +1272,7 @@
(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:
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)
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}
+
+