diff --git a/ptx/docinfo.ptx b/ptx/docinfo.ptx
index 9473103f7..17bc991f6 100644
--- a/ptx/docinfo.ptx
+++ b/ptx/docinfo.ptx
@@ -1,6 +1,6 @@
-
+ APEXExercisePart
diff --git a/ptx/review-exercises-limits.ptx b/ptx/review-exercises-limits.ptx
index 8c14ddf4f..09dd87313 100644
--- a/ptx/review-exercises-limits.ptx
+++ b/ptx/review-exercises-limits.ptx
@@ -272,7 +272,7 @@
For a numerical approximation, make a table:
-
+ x
diff --git a/ptx/sec_FTC.ptx b/ptx/sec_FTC.ptx
index 189840c64..79d218056 100644
--- a/ptx/sec_FTC.ptx
+++ b/ptx/sec_FTC.ptx
@@ -129,7 +129,7 @@
x \geq 1 is given by A(x)=\frac12 (x)(2x)-\frac12 (1)(2)=x^2-1.
-
+
The area of the shaded region is F(x) = \int_1^x 2t\, dt
Adjust the initial condition in this interactive figure to answer the following.
@@ -1073,7 +1072,7 @@
Slope field for the logistic differential equation \yp = y(1-y) from
- Graph of slope field for the logistic differential equation y’=y(1-y) from the example.
+ Graph of slope field for the logistic differential equation y'=y(1-y) from the example.
@@ -1161,7 +1160,7 @@
with a few representative solution curves
- Graph of slope field for the logistic differential equation y’=y(1-y) with representative solution curves.
+ Graph of slope field for the logistic differential equation y'=y(1-y) with representative solution curves.
@@ -2114,7 +2113,7 @@
-
+
Match each slope field below with the appropriate differential equation.
diff --git a/ptx/sec_Modeling.ptx b/ptx/sec_Modeling.ptx
index 29391f743..654bed141 100644
--- a/ptx/sec_Modeling.ptx
+++ b/ptx/sec_Modeling.ptx
@@ -42,7 +42,7 @@
\draw [firstcolor] (-1.5,0) node [text width=60pt,align=center] (a) { \centering The rate of change of the population};
\draw [firstcolor,->] (a) -- (.3,.7);
- \draw [firstcolor] (2,.25) node [text width=60pt,align=center] (b) { \centering the population.};
+ \draw [firstcolor] (2,0) node [text width=60pt,align=center] (b) { \centering the\\ population.};
\draw [firstcolor,->] (b) -- (1.6,.8);
\draw [firstcolor] (.5,2) node [text width=32pt,align=center] (c) { \centering is};
diff --git a/ptx/sec_antider.ptx b/ptx/sec_antider.ptx
index ce04b90f1..8715ddf21 100644
--- a/ptx/sec_antider.ptx
+++ b/ptx/sec_antider.ptx
@@ -839,8 +839,8 @@
Fill in the blanks:
- Inverse operations do the
- things in the order.
+ Inverse operations do the
+ things in the order.
@@ -882,7 +882,7 @@
-->
- The derivative of a position function is a/an function.
+ The derivative of a position function is a/an function.
@@ -904,7 +904,7 @@
-->
- An antiderivative of an acceleration function is a/an
+ An antiderivative of an acceleration function is a/an
function.
- is a measure of the turning force applied to an object.
+ is a measure of the turning force applied to an object.
diff --git a/ptx/sec_curvature.ptx b/ptx/sec_curvature.ptx
index df08b0ba1..66da1cb70 100644
--- a/ptx/sec_curvature.ptx
+++ b/ptx/sec_curvature.ptx
@@ -208,7 +208,7 @@
We find it with \vec r(2) = \la 1/5, 18/5\ra.
-
+
Graphing \vec r in with parameters t and s
@@ -687,7 +687,7 @@
Being able to think of curvature in terms of the radius of a circle is very useful.
-
+
Illustrating the osculating circles for the curve seen in
@@ -757,7 +757,7 @@
.
-
+
Examining the curvature of y=x^2
@@ -834,26 +834,10 @@
-
- While this is not a particularly nice formula,
- it does explicitly tell us what the curvature is at a given t value.
- To maximize \kappa(t),
- we should solve \kappa'(t)=0 for t.
- This is doable, but very time consuming.
- Instead, consider the graph of
- \kappa(t) as given in .
- We see that \kappa is maximized at two t values;
- using a numerical solver, we find these values are t\approx\pm 0.189.
- In we graph \vrt and indicate the points where curvature is maximized.
-
-
-
-
Understanding the curvature of a curve in space
-
-
-
The curvature of \vec{r}(t)
+
+
The curvature of \vec{r}(t) in
-
+ A plot of the curvature as a function of the parameter t.
@@ -885,11 +869,11 @@
-
-
A plot of the curve \vec{r}(t)=\la t, t^2, 2t^3\ra
+
+
A plot of the curve \vec{r}(t)=\la t, t^2, 2t^3\ra in
-
+ A plot of the vector-valued function in this example, with points of maximum curvature marked.
@@ -950,9 +934,19 @@
-
-
+
+ While this is not a particularly nice formula,
+ it does explicitly tell us what the curvature is at a given t value.
+ To maximize \kappa(t),
+ we should solve \kappa'(t)=0 for t.
+ This is doable, but very time consuming.
+ Instead, consider the graph of
+ \kappa(t) as given in .
+ We see that \kappa is maximized at two t values;
+ using a numerical solver, we find these values are t\approx\pm 0.189.
+ In we graph \vrt and indicate the points where curvature is maximized.
+
diff --git a/ptx/sec_deriv_interpret.ptx b/ptx/sec_deriv_interpret.ptx
index 545eb19a0..52babc70e 100644
--- a/ptx/sec_deriv_interpret.ptx
+++ b/ptx/sec_deriv_interpret.ptx
@@ -851,7 +851,7 @@
What is the instantaneous rate of change of position called?
- The gradient is to level curves.
+ The gradient is to level curves.
@@ -1347,7 +1347,7 @@
It is generally more informative to view the directional derivative not as the result of a limit,
- but rather as the result of a product.
+ but rather as the result of a product.
Illustrating the relationship between the angle between vectors and the sign of their dot product
-
-
-
- Image shows the relation between the angle between two vectors and the sign of their dot product.
-
-
-
- Each image has the two vectors \vec u and \vec v along with \theta shown.
- The dot product along with three possible cases of <, > and equal to 0 are shown.
- In the first image, the dot product has a positive value, the angle between the two vectors is acute.
- In the second image, the dot product is equal to 0, the angle is 90 degrees.
- In the third image, the dot product has a negative value and the angle is obtuse.
-
+ Each image has the two vectors \vec u and \vec v along with \theta shown.
+ The dot product along with three possible cases of <, > and equal to 0 are shown.
+ In the first image, the dot product has a positive value, the angle between the two vectors is acute.
+ In the second image, the dot product is equal to 0, the angle is 90 degrees.
+ In the third image, the dot product has a negative value and the angle is obtuse.
+
Match the correct elements on the left to the corresponding elements on the right,
so that \iint_R f(x,y)\, dA is correctly converted to polar coordinates.
+
+ When evaluating \iint_R f(x,y)\, dA using polar coordinates,
+ f(x,y) is replaced with and dA is replaced with .
+
An integral can be interpreted as giving the signed area over an interval;
- a double integral can be interpreted as giving the signed over a region.
+ a double integral can be interpreted as giving the signed over a region.
Let \vec F be a vector field and let C be a curve.
- Flow is a measure of the amount of \vec F going C;
- flux is a measure of the amount of \vec F going C.
+ Flow is a measure of the amount of \vec F going C;
+ flux is a measure of the amount of \vec F going C.
@@ -1461,7 +1461,7 @@
Green's Theorem states, informally,
that the circulation around a closed curve that bounds a region R is equal to the sum of the
- of \vec{F} across R.
+ of \vec{F} across R.
@@ -1478,7 +1478,7 @@
The Divergence Theorem states, informally,
that the outward flux across a closed curve that bounds a region R
- is equal to the sum of the of \vec{F} across R.
+ is equal to the sum of the of \vec{F} across R.
@@ -1500,7 +1500,7 @@
Let \vec F be a vector field and let C_1 and C_2
be any nonintersecting paths except that each starts at point A and ends at point B.
- If the of \vec{F} is 0,
+ If the of \vec{F} is 0,
then \int_{C_1} \vec F\cdot \vec T\, ds = \int_{C_2} \vec F\cdot \vec T\, ds.
@@ -1518,7 +1518,7 @@
Let \vec F be a vector field and let C_1 and C_2
be any nonintersecting paths except that each starts at point A and ends at point B.
- If the of \vec{F} is 0,
+ If the of \vec{F} is 0,
then \int_{C_1} \vec F\cdot \vec n\, ds = \int_{C_2} \vec F\cdot \vec n\, ds.
diff --git a/ptx/sec_hyperbolic.ptx b/ptx/sec_hyperbolic.ptx
index e6c1ef962..cfd78f0ff 100644
--- a/ptx/sec_hyperbolic.ptx
+++ b/ptx/sec_hyperbolic.ptx
@@ -781,7 +781,7 @@
Domains and ranges of the hyperbolic and inverse hyperbolic functions
-
+ FunctionDomainRange
diff --git a/ptx/sec_improper_integration.ptx b/ptx/sec_improper_integration.ptx
index ce7b28413..f75acc9f3 100644
--- a/ptx/sec_improper_integration.ptx
+++ b/ptx/sec_improper_integration.ptx
@@ -1325,7 +1325,7 @@
If \lim\limits_{b\to \infty} \int_0^b f(x)\, dx exists,
- then the integral \ds \int_0^\infty f(x)\, dx is said to .
+ then the integral \ds \int_0^\infty f(x)\, dx is said to .
@@ -1351,7 +1351,7 @@
If \ds \int_1^\infty f(x)\, dx=10,
and 0\leq g(x)\leq f(x) for all x,
- then we know that \ds \int_1^\infty g(x)\, dx.
+ then we know that \ds \int_1^\infty g(x)\, dx.
diff --git a/ptx/sec_int_comp_tests.ptx b/ptx/sec_int_comp_tests.ptx
index 5dc0ab996..af755b346 100644
--- a/ptx/sec_int_comp_tests.ptx
+++ b/ptx/sec_int_comp_tests.ptx
@@ -513,6 +513,18 @@
Since the limit on the left side diverges to \infty,
we can say that \lim\limits_{n \to \infty}\sum_{i=N}^n b_i also diverges to \infty.
When evaluating an iterated integral,
- we integrate from to ,
- then from to .
+ we integrate from to ,
+ then from to .
@@ -958,7 +958,7 @@
One understanding of an iterated integral is that
- \ds \int_a^b\int_{g_1(x)}^{g_2(x)} \, dy\, dx gives the of a plane region.
+ \ds \int_a^b\int_{g_1(x)}^{g_2(x)} \, dy\, dx gives the of a plane region.
Fill in the blanks: The Quotient Rule is applied to
- \ds \frac{f(x)}{g(x)} when taking ;
- l'Hospital's Rule is applied when taking certain .
+ \ds \frac{f(x)}{g(x)} when taking ;
+ l'Hospital's Rule is applied when taking certain .
Iterations of the Bisection Method of Root Finding
-
+ Iteration #IntervalMidpoint Sign
@@ -3274,7 +3274,7 @@
If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign.
-
+ IterationIntervalMidpoint Sign
@@ -3364,7 +3364,7 @@
If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign.
-
+ IterationIntervalMidpoint Sign
@@ -3454,7 +3454,7 @@
If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign.
-
+ IterationIntervalMidpoint Sign
@@ -3544,7 +3544,7 @@
If f(m)\gt 0 at the midpoint m, put + for the midpoint sign. If f(m)\lt 0, put - for the midpoint sign.
-
+ IterationIntervalMidpoint Sign
diff --git a/ptx/sec_limit_infty.ptx b/ptx/sec_limit_infty.ptx
index 55dc94d27..6fb8b198b 100644
--- a/ptx/sec_limit_infty.ptx
+++ b/ptx/sec_limit_infty.ptx
@@ -759,7 +759,7 @@
Fill in the blank: The single variable Chain Rule states
- \ds\frac{d}{dx}\Big(f\big(g(x)\big)\Big) = \fp\big(g(x)\big)\cdot.
+ \ds\frac{d}{dx}\Big(f\big(g(x)\big)\Big) = \fp\big(g(x)\big)\cdot.
@@ -921,7 +921,7 @@
- The Multivariable Chain Rule allows us to compute implicit derivatives easily by just computing two derivatives.
+ The Multivariable Chain Rule allows us to compute implicit derivatives easily by just computing two derivatives.
A table of c values and the corresponding radius r of the spheres of constant value in
-
+ cr
@@ -980,7 +980,7 @@
-->
- The graph of a function of two variables is a .
+ The graph of a function of two variables is a .
@@ -999,7 +999,7 @@
-->
- Most people are familiar with the concept of level curves in the context of maps.
+ Most people are familiar with the concept of level curves in the context of maps.
@@ -1037,7 +1037,7 @@
-->
- The analogue of a level curve for functions of three variables is a level .
+ The analogue of a level curve for functions of three variables is a level .
Values used to approximate \int_{-\frac{\pi}4}^{\frac{\pi}2}\sin(x^3)\, dx in
-
+ x_i\sin(x_i^3)
@@ -1971,7 +1971,7 @@
Speed data collected at 30 second intervals for
-
+ TimeSpeed
@@ -2191,7 +2191,7 @@
Simpson's Rule is based on approximating portions of a function with what type of function?
-
+
diff --git a/ptx/sec_par_calc.ptx b/ptx/sec_par_calc.ptx
index 5f91f4226..98fa296f5 100644
--- a/ptx/sec_par_calc.ptx
+++ b/ptx/sec_par_calc.ptx
@@ -371,7 +371,7 @@
any line that passes through the center of a circle intersects the circle at right angles.
-
+
Illustrating how a circle's normal lines pass through its center
@@ -661,7 +661,7 @@
-
+
Graphing the parametric equations in to demonstrate concavity
@@ -789,38 +789,12 @@
.
-
- The points of inflection are found by setting \frac{d^2y}{dx^2}=0.
- This is not trivial,
- as equations that mix polynomials and trigonometric functions generally do not have nice solutions.
-
-
-
- In we see a plot of the second derivative.
- It shows that it has zeros at approximately t=0.5,\,3.5,\,6.5,\,9.5,\,12.5 and 16.
- These approximations are not very good,
- made only by looking at the graph.
- Newton's Method provides more accurate approximations.
- Accurate to 2 decimal places, we have:
-
- t=0.65,\,3.29,\,6.36,\,9.48,\,12.61\,\text{ and } \,15.74
- .
-
-
-
- The corresponding points have been plotted on the graph of the parametric equations in .
- Note how most occur near the x-axis,
- but not exactly on the axis.
-
-
-
-
In (a), a graph of \frac{d^2y}{dx^2}, showing where it is approximately 0. In (b), graph of the parametric equations in along with the points of inflection
-
-
-
+
+
Graphing \frac{d^2y}{dx^2} in , showing where it is approximately 0
+
+
-
- Graph of the second derivative is a sinusoid with increasing amplitude.
+ Graph of the second derivative is a sinusoid with increasing amplitude.
The image shows the graph y = 2\cos(t)-4t\sin(t), which is a graph of \frac{d^2y}{dx^2}.
@@ -848,15 +822,40 @@
\end{tikzpicture}
-
+
-
-
+
+ The points of inflection are found by setting \frac{d^2y}{dx^2}=0.
+ This is not trivial,
+ as equations that mix polynomials and trigonometric functions generally do not have nice solutions.
+
+
+
+ In we see a plot of the second derivative.
+ It shows that it has zeros at approximately t=0.5,\,3.5,\,6.5,\,9.5,\,12.5 and 16.
+ These approximations are not very good,
+ made only by looking at the graph.
+ Newton's Method provides more accurate approximations.
+ Accurate to 2 decimal places, we have:
+
+ t=0.65,\,3.29,\,6.36,\,9.48,\,12.61\,\text{ and } \,15.74
+ .
+
+
+
+ The corresponding points have been plotted on the graph of the parametric equations in .
+ Note how most occur near the x-axis,
+ but not exactly on the axis.
+
+
+
+
A graph of the parametric equations in along with the points of inflection
+
+
-
- Graph of the parametric curve in this example, with points of inflection marked.
+ Graph of the parametric curve in this example, with points of inflection marked.
The graph shows a curve that appears to be sinusoidal, but with a frequency that increases with x.
@@ -892,9 +891,6 @@
-
-
-
@@ -1281,7 +1277,7 @@
Find the surface area if this shape is rotated about the x-axis,
as shown in .
-
+
The limaçon in with its tangent line at \theta=\pi/4 and points of vertical and horizontal tangency
@@ -940,7 +940,7 @@
Find the area bounded between the polar curves r=1 and r=2\cos(2\theta),
as shown in .
-
+
The region bounded by the functions in
A zoomed in view of a region bounded by a circle, a rose curve, and the x axis.
@@ -995,7 +995,7 @@
.
-
+
Breaking the region bounded by the functions in into its component parts
A zoomed in view of a polar region, showing it divided into two parts.
@@ -1157,7 +1157,7 @@
.
-
+
The limaçon in whose arc length is measured
diff --git a/ptx/sec_power_series.ptx b/ptx/sec_power_series.ptx
index 69b5f6f5a..b284f7502 100644
--- a/ptx/sec_power_series.ptx
+++ b/ptx/sec_power_series.ptx
@@ -246,43 +246,41 @@
Determining the Radius and Interval of Convergence
-
-
- Given the power series \ds \infser[0] a_n(x-c)^n,
- apply the ratio test to the series \ds \infser[0]\abs{a_n (x-c)^n}.
- The result will be L\abs{x-c}, where \ds L=\lim_{n\to\infty}\frac{\abs{a_{n+1}}}{\abs{a_n}}.
-
-
-
-
-
-
- If L=0, then the power series converges for every x
- by the ratio test, since L\abs{x-c}=0\lt 1.
-
-
+
+ Given the power series \ds \infser[0] a_n(x-c)^n,
+ apply the ratio test to the series \ds \infser[0]\abs{a_n (x-c)^n}.
+ The result will be L\abs{x-c}, where \ds L=\lim_{n\to\infty}\frac{\abs{a_{n+1}}}{\abs{a_n}}.
+
+
+
+
+
+
+ If L=0, then the power series converges for every x
+ by the ratio test, since L\abs{x-c}=0\lt 1.
+
+
-
-
- If L=\infty, then power series converges only when x=c.
-
-
+
+
+ If L=\infty, then power series converges only when x=c.
+
+
-
-
- If 0\lt L\lt \infty, then R=1/L is the radius of convergence:
- by the ratio test, the series converges when \abs{x-c}\lt R.
-
+
+
+ If 0\lt L\lt \infty, then R=1/L is the radius of convergence:
+ by the ratio test, the series converges when \abs{x-c}\lt R.
+
-
- To determine the interval of convergence, plug the endpoints (x=c-R and x=c+R)
- into the power series, and test the resulting series for convergence.
- If the series converges, we include the endpoint. If it diverges, we exclude the endpoint.
-
-
-
-
-
+
+ To determine the interval of convergence, plug the endpoints (x=c-R and x=c+R)
+ into the power series, and test the resulting series for convergence.
+ If the series converges, we include the endpoint. If it diverges, we exclude the endpoint.
+
+
+
+
@@ -924,7 +922,7 @@
-->
- We adopt the convention that x^0=,
+ We adopt the convention that x^0=,
regardless of the value of x.
@@ -995,7 +993,7 @@
\ds\infser[0] a_nx^n is 5,
then the radius of convergence of \ds\infser[0] (-1)^na_nx^n is
- .
+ .
- A fundamental calculus technique is to use
+ A fundamental calculus technique is to use
to refine approximations to get an exact answer.
A scatter plot showing a representative sample of points from the third sequence in this example.
@@ -1514,7 +1514,7 @@
-
+
a_{n+1}-a_n \amp = \frac{n+2}{n+1} - \frac{n+1}{n}
@@ -1523,84 +1523,13 @@
\amp \lt 0 \text{ for all \(n\). }
Since a_{n+1}-a_n\lt 0 for all n,
- we conclude that the sequence is decreasing.
-
-
-
-
-
-
- a_{n+1}-a_n \amp = \frac{(n+1)^2+1}{n+2} - \frac{n^2+1}{n+1}
- \amp = \frac{\big((n+1)^2+1\big)(n+1)- (n^2+1)(n+2)}{(n+1)(n+2)}
- \amp = \frac{n^2+3n}{(n+1)(n+2)}
- \amp \gt 0 \text{ for all \(n\). }
-
- Since a_{n+1}-a_n\gt 0 for all n,
- we conclude the sequence is increasing.
-
-
-
-
-
- We can clearly see in ,
- where the sequence is plotted, that it is not monotonic.
- However, it does seem that after the first 4 terms it is decreasing.
- To understand why, perform the same analysis as done before:
-
-
- a_{n+1}-a_n \amp = \frac{(n+1)^2-9}{(n+1)^2-10(n+1)+26} - \frac{n^2-9}{n^2-10n+26}
- \amp = \frac{n^2+2n-8}{n^2-8n+17}-\frac{n^2-9}{n^2-10n+26}
- \amp = \frac{(n^2+2n-8)(n^2-10n+26)-(n^2-9)(n^2-8n+17)}{(n^2-8n+17)(n^2-10n+26)}
- \amp = \frac{-10n^2+60n-55}{(n^2-8n+17)(n^2-10n+26)}
- .
+ we conclude that the sequence is decreasing, illustrated in .
-
- We want to know when this is greater than, or less than, 0.
- The denominator is always positive,
- therefore we are only concerned with the numerator.
- For small values of n,
- the numerator is positive.
- As n grows large,
- the numerator is dominated by -10n^2,
- meaning the entire fraction will be negative;
- , for large enough n, a_{n+1}-a_n \lt 0.
- Using the quadratic formula we can determine that the numerator is negative for n\geq 5.
- In short, the sequence is simply not monotonic,
- though it is useful to note that for n\geq 5,
- the sequence is monotonically decreasing.
-
-
-
-
-
-
- Again, the plot in
- shows that the sequence is not monotonic,
- but it suggests that it is monotonically decreasing after the first term.
- We perform the usual analysis to confirm this.
-
- a_{n+1}-a_n \amp = \frac{(n+1)^2}{(n+1)!} - \frac{n^2}{n!}
- \amp = \frac{(n+1)^2-n^2(n+1)}{(n+1)!}
- \amp = \frac{-n^3+2n+1}{(n+1)!}
-
- When n=1, the above expression is \gt 0;
- for n\geq 2, the above expression is \lt 0.
- Thus this sequence is not monotonic,
- but it is monotonically decreasing after the first term.
-
-
-
-
-
-
-
Plots of sequences in
-
-
-
-
+
+
Plot of the sequence in of
-
+ Plot of the first sequence in this example. It is decreasing and bounded below.
@@ -1631,11 +1560,24 @@
+
+
+
+
+
+ a_{n+1}-a_n \amp = \frac{(n+1)^2+1}{n+2} - \frac{n^2+1}{n+1}
+ \amp = \frac{\big((n+1)^2+1\big)(n+1)- (n^2+1)(n+2)}{(n+1)(n+2)}
+ \amp = \frac{n^2+3n}{(n+1)(n+2)}
+ \amp \gt 0 \text{ for all \(n\). }
+
+ Since a_{n+1}-a_n\gt 0 for all n,
+ we conclude the sequence is increasing, illustrated in .
+
-
-
+
+
Plot of the sequence in of
-
+ Scatter plot for the second sequence in this example. It is increasing but not bounded.
@@ -1667,13 +1609,43 @@
-
+
+
+
+
+ We can clearly see in ,
+ where the sequence is plotted, that it is not monotonic.
+ However, it does seem that after the first 4 terms it is decreasing.
+ To understand why, perform the same analysis as done before:
+
+
+ a_{n+1}-a_n \amp = \frac{(n+1)^2-9}{(n+1)^2-10(n+1)+26} - \frac{n^2-9}{n^2-10n+26}
+ \amp = \frac{n^2+2n-8}{n^2-8n+17}-\frac{n^2-9}{n^2-10n+26}
+ \amp = \frac{(n^2+2n-8)(n^2-10n+26)-(n^2-9)(n^2-8n+17)}{(n^2-8n+17)(n^2-10n+26)}
+ \amp = \frac{-10n^2+60n-55}{(n^2-8n+17)(n^2-10n+26)}
+ .
+
-
-
-
+
+ We want to know when this is greater than, or less than, 0.
+ The denominator is always positive,
+ therefore we are only concerned with the numerator.
+ For small values of n,
+ the numerator is positive.
+ As n grows large,
+ the numerator is dominated by -10n^2,
+ meaning the entire fraction will be negative;
+ , for large enough n, a_{n+1}-a_n \lt 0.
+ Using the quadratic formula we can determine that the numerator is negative for n\geq 5.
+ In short, the sequence is simply not monotonic,
+ though it is useful to note that for n\geq 5,
+ the sequence is monotonically decreasing.
+
+
+
+
Plot of the sequence in of
-
+ Scatter plot for the third sequence in this example. It is not monotonic.
@@ -1710,12 +1682,32 @@
-
-
+
+
+
+
+
+ Again, the plot in of
+ shows that the sequence is not monotonic,
+ but it suggests that it is monotonically decreasing after the first term.
+ We perform the usual analysis to confirm this.
+
+ a_{n+1}-a_n \amp = \frac{(n+1)^2}{(n+1)!} - \frac{n^2}{n!}
+ \amp = \frac{(n+1)^2-n^2(n+1)}{(n+1)!}
+ \amp = \frac{-n^3+2n+1}{(n+1)!}
+
+ When n=1, the above expression is \gt 0;
+ for n\geq 2, the above expression is \lt 0.
+ Thus this sequence is not monotonic,
+ but it is monotonically decreasing after the first term.
+
+
+
+
Plot of the sequence in of
-
+ Scatter plot for the last sequence in this example. It is not monotonic.
- The domain of a sequence is the numbers.
+ The domain of a sequence is the numbers.
diff --git a/ptx/sec_series.ptx b/ptx/sec_series.ptx
index 68de229a4..7fcb7c391 100644
--- a/ptx/sec_series.ptx
+++ b/ptx/sec_series.ptx
@@ -983,7 +983,7 @@
Partial sums of the series are plotted in .
-
+
Scatter plots relating to the series of
@@ -1087,7 +1087,7 @@
-
+
We can decompose the fraction 2/(n^2+2n) as
@@ -1121,47 +1121,11 @@
so \infser \frac1{n^2+2n} = \frac32.
This is illustrated in .
-
-
-
-
- We begin by writing the first few partial sums of the series:
-
- S_1 \amp = \ln\left(2\right)
- S_2 \amp = \ln\left(2\right)+\ln\left(\frac32\right)
- S_3 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)
- S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right)
-
- At first, this does not seem helpful,
- but recall the logarithmic identity:
- \ln(x) +\ln(y) = \ln(xy).
- Applying this to S_4 gives:
-
- S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right)
- \amp = \ln\left(\frac21\cdot\frac32\cdot\frac43\cdot\frac54\right) = \ln\left(5\right)
- .
- We can conclude that \{S_n\} = \big\{\ln(n+1)\big\}.
- This sequence does not converge,
- as \lim\limits_{n\to\infty}S_n=\infty.
- Therefore \ds\infser \ln\left(\frac{n+1}{n}\right)=\infty;
- the series diverges.
- Note in how the sequence of partial sums grows slowly;
- after 100 terms, it is not yet over 5.
- Graphically we may be fooled into thinking the series converges,
- but our analysis above shows that it does not.
-
-
-
-
-
-
-
Scatter plots relating to the series in
-
-
-
-
-
+
+
Scatter plots of the sequence and corresponding partial sums in of
+
+ Scatter plots of the sequence, and corresponding partial sums, for the first part of this example.
@@ -1204,13 +1168,42 @@
-
-
+
+
+
-
-
-
-
+
+
+ We begin by writing the first few partial sums of the series:
+
+ S_1 \amp = \ln\left(2\right)
+ S_2 \amp = \ln\left(2\right)+\ln\left(\frac32\right)
+ S_3 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)
+ S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right)
+
+ At first, this does not seem helpful,
+ but recall the logarithmic identity:
+ \ln(x) +\ln(y) = \ln(xy).
+ Applying this to S_4 gives:
+
+ S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right)
+ \amp = \ln\left(\frac21\cdot\frac32\cdot\frac43\cdot\frac54\right) = \ln\left(5\right)
+ .
+ We can conclude that \{S_n\} = \big\{\ln(n+1)\big\}.
+ This sequence does not converge,
+ as \lim\limits_{n\to\infty}S_n=\infty.
+ Therefore \ds\infser \ln\left(\frac{n+1}{n}\right)=\infty;
+ the series diverges.
+ Note in how the sequence of partial sums grows slowly;
+ after 100 terms, it is not yet over 5.
+ Graphically we may be fooled into thinking the series converges,
+ but our analysis above shows that it does not.
+
+
+
+
Scatter plots of the sequence and corresponding partial sums in of
+
+ Scatter plots of the sequence, and corresponding partial sums, for the second part of this example.
@@ -1255,8 +1248,10 @@
-
-
+
+
+
+
Video solution
@@ -1388,7 +1383,7 @@
-
+
We start by using algebra to break the series apart:
@@ -1398,31 +1393,10 @@
.
This is illustrated in .
-
-
-
-
- This looks very similar to the series that involves e in .
- Note, however,
- that the series given in this example starts with n=1 and not n=0.
- The first term of the series in the Key Idea is 1/0! = 1,
- so we will subtract this from our result below:
-
- \infser \frac{1000}{n!} \amp = 1000\cdot\infser \frac{1}{n!}
- \amp = 1000\cdot (e-1) \approx 1718.28
- .
- This is illustrated in .
- The graph shows how this particular series converges very rapidly.
-
-
-
-
Scatter plots relating to the series in
-
-
-
-
-
-
+
+
Scatter plots of the sequence and corresponding partial sums in of
+
+ Scatter plots of the sequence, and corresponding partial sums, for the first part of this example.
@@ -1468,13 +1442,28 @@
-
+
+
-
-
-
-
+
+
+ This looks very similar to the series that involves e in .
+ Note, however,
+ that the series given in this example starts with n=1 and not n=0.
+ The first term of the series in the Key Idea is 1/0! = 1,
+ so we will subtract this from our result below:
+
+ \infser \frac{1000}{n!} \amp = 1000\cdot\infser \frac{1}{n!}
+ \amp = 1000\cdot (e-1) \approx 1718.28
+ .
+ This is illustrated in .
+ The graph shows how this particular series converges very rapidly.
+
+
+
Scatter plots of the sequence and corresponding partial sums in of
+
+ Scatter plots of the sequence, and corresponding partial sums, for the first part of this example.
@@ -1514,13 +1503,11 @@
-
-
-
-
+
+
-
+
The denominators in each term are perfect squares;
we are adding \ds \sum_{n=4}^\infty \frac{1}{n^2}
diff --git a/ptx/sec_shell_method.ptx b/ptx/sec_shell_method.ptx
index 2185229f5..d7df5a394 100644
--- a/ptx/sec_shell_method.ptx
+++ b/ptx/sec_shell_method.ptx
@@ -1447,7 +1447,7 @@
In the plane,
- flux is a measurement of how much of the vector field passes across a ;
- in space, flux is a measurement of how much of the vector field passes across a .
+ flux is a measurement of how much of the vector field passes across a ;
+ in space, flux is a measurement of how much of the vector field passes across a .
@@ -830,7 +830,7 @@
When \surfaceS is a closed surface,
- we choose the normal vector so that it points to the of the surface.
+ we choose the normal vector so that it points to the of the surface.
@@ -847,7 +847,7 @@
If \surfaceS is a plane,
and \vec F is always parallel to \surfaceS,
- then the flux of \vec F across \surfaceS will be .
+ then the flux of \vec F across \surfaceS will be .
diff --git a/ptx/sec_tan_norm.ptx b/ptx/sec_tan_norm.ptx
index 098580294..1ae2e6201 100644
--- a/ptx/sec_tan_norm.ptx
+++ b/ptx/sec_tan_norm.ptx
@@ -68,7 +68,7 @@
since they are only length 1.)
-
+
Plotting unit tangent vectors in
@@ -382,7 +382,7 @@
These are sketched in .
-
+
Plotting unit tangent and normal vectors in
@@ -447,17 +447,7 @@
-
+
The previous example was once again
@@ -517,7 +507,7 @@
we compute the unit tangent and normal vectors for t=-1,0 and 1 and sketch them in .
-
+
Plotting unit tangent and normal vectors in
@@ -607,6 +597,18 @@
+
+
+ A brief consideration of this theorem may make one wonder: what if the graph of \vec r does
+ not have a concave side? What if \vec r is a line?
+
+
+ This exposes a shortcoming in our definition of \unitnormal(t), where
+ we require that \unittangent(t) be smooth, i.e., that \unittangentprime(t) \neq \vec 0,
+ a requirement that lines do not fulfill. One may still want to compute a normal vector for a given line, though.
+ For straight lines in the x,y plane, it is most common to orient the normal vector 90^\circ
+ counterclockwise from the tangent vector. For lines in three dimensions, there is no preferred choice of normal vector.
+
@@ -802,7 +804,7 @@
gives a graph of the path for reference.
-
+
Graphing \vec r(t) in
@@ -875,7 +877,7 @@
which we plot in .
-
+
Plotting the position of a thrown ball, with 1s increments shown
@@ -966,7 +968,7 @@
A table of values of a_T and a_N in
-
+ ta_\text{T}a_\text{N}
@@ -1044,10 +1046,7 @@
If \unittangent(t) is a unit tangent vector,
what is \norm{\unittangent(t)}?
-
-
-
-
+
@@ -1073,10 +1072,7 @@
If \unitnormal(t) is a unit normal vector,
what is \unitnormal(t)\cdot \vrp(t)?
-
-
-
-
+
@@ -1119,7 +1115,7 @@
-->
- a_\text{T} measures how much the acceleration is affecting the of an object.
+ a_\text{T} measures how much the acceleration is affecting the of an object.
@@ -750,7 +750,7 @@
The displacement of \vec r(t) on [-1,1] is thus \vec d = \la 0,1\ra - \la 0,-1\ra = \la 0,2\ra.
-
+
Graphing the displacement of a position function in
@@ -924,7 +924,7 @@
When sketching vector-valued functions,
- technically one isn't graphing points, but rather .
+ technically one isn't graphing points, but rather .
@@ -943,7 +943,7 @@
-->
- It can be useful to think of as a vector that points from a starting position to an ending position.
+ It can be useful to think of as a vector that points from a starting position to an ending position.
@@ -960,7 +960,7 @@
In the context of vector-valued functions,
- average rate of change is divided by time.
+ average rate of change is divided by time.
diff --git a/ptx/sec_vvf_calc.ptx b/ptx/sec_vvf_calc.ptx
index e63c127f2..ac5d300b4 100644
--- a/ptx/sec_vvf_calc.ptx
+++ b/ptx/sec_vvf_calc.ptx
@@ -19,7 +19,7 @@
The theorem following the definition shows that in practice,
taking limits of vector-valued functions is no more difficult than taking limits of real-valued functions.
-