fixed various typos in the exercises

integralsOfPolarFunctions/exercises/area4.tex

@@ -2,7 +2,7 @@
 
 \input{../../preamble.tex}
 
-\author{Jason Miller}
+\author{Jason Miller and Jim Talamo}
 \license{Creative Commons 3.0 By-bC}
 
 
@@ -39,7 +39,7 @@
   \end{tikzpicture}  
 \end{image} 
 
-We want to set up an integral that expresses the area of the shaded region $S$ that lies inside the curve $r=1+\cos(3\theta)$ but outside of the circle $r=1$ in the 1st quadrant. 
+We want to set up an integral that expresses the area of the shaded region $S$ that lies inside the curve $r=2\sin(3\theta)$ but outside of the circle $r=1$ in the 1st quadrant. 
 
 
 
@@ -48,7 +48,7 @@
 The area of the region $S$ is 
 
 \[
-\int_{\answer{\frac{\pi}{18}}}^{\answer{ \frac{5\pi}{18} } } \answer{ \frac{1}{2}((2\cos(3\theta))^2-1)   } \d \theta=\answer{ \frac{2\pi-3\sqrt{3}}{18}}
+\int_{\answer{\pi/18}}^{\answer{ 5\pi/18 } } \answer{ \frac{1}{2}((4\sin(3\theta))^2-1)   } \d \theta=\answer{ \frac{2\pi-3\sqrt{3}}{18}}
 \]
 
 
@@ -157,7 +157,7 @@
 
 Understanding how $r=1$ is traced out is the simplest. It is traced out in the standard fashion as $\theta$ varies. That is, for each value of $\theta$, the associated point on the curve is $(1, \theta)$ in polar coordinates. 
 
-Let's graph $r=2\cos(3\theta)$ on $r$ and $\theta$ axes. 
+Let's graph $r=2\sin(3\theta)$ on $r$ and $\theta$ axes. 
 
 \begin{image}  
   \begin{tikzpicture}  
@@ -183,13 +183,11 @@
 
 As $\theta$ goes from $0$ to $\frac{\pi}{3}$ we see that $r$ increase from $0$ to $2$ and then decreases from $2$ back down to $0$. This corresponds to the petal in the 1st quadrant. 
 
-But we don't want the entire petal.So we need to find for which $\theta$ values the $r$ values of the two curves coincide. 
+But we don't want the entire petal. So we need to find for which $\theta$ values the $r$ values of the two curves coincide. 
 
 We set $r=2\sin(3\theta)$ and $r=1$ equal. 
 
-Solving $2\sin(3\theta)=1$ for $\theta$ we get $\theta=\answer{\frac{\pi}{18}}$ and $\theta=\answer{\frac{5\pi}{18}}$ (list the $\theta$ values from smallest to largest and remember to use $\theta$ values from $[0, \pi)$). 
-
-
+Set $2\sin(3\theta)=1$.  Note that $\sin(\cdot) = 0$ when $(\cdot) = \frac{\pi}{2},\frac{5\pi}{2}, \ldots$  (meaning that we have to set the expression \emph{inside} the sine function equal to those values.  Doing this gives us that $3 \theta = \frac{\pi}{2},\frac{5\pi}{2}, \ldots$, so $\theta=\answer{\frac{\pi}{18}}$ and $\theta=\answer{\frac{5\pi}{18}}$ (list the $\theta$ values from smaller to larger. 
 
 
 In order to evaluate the integral, recall the trig identity $\sin^2(\theta)=\frac{1-\cos(2\theta)}{2}$. 

integralsOfPolarFunctions/exercises/area6.tex

@@ -2,7 +2,7 @@
 
 \input{../../preamble.tex}
 
-\author{Jason Miller}
+\author{Jason Miller and Jim Talamo}
 \license{Creative Commons 3.0 By-bC}
 
 
@@ -28,10 +28,10 @@
         every axis y label/.style={at=(current axis.above origin),anchor=south},  
         every axis x label/.style={at=(current axis.right of origin),anchor=west},axis on top
       ]  
-       \addplot[data cs=polar, very thick, mark=none, fill=fill3,  domain=0:60, samples=180, smooth] (x, {1+cos(x)}) -- (axis cs:0,0)[penColor4!50];
-         \addplot[data cs=polar, very thick, mark=none, fill=fill3,  domain=-90:-60, samples=180, smooth] (x, {3*cos(x)}) -- (axis cs:0,0)[penColor4!50];
-         \addplot[data cs=polar, very thick, mark=none, fill=fill3,  domain=-60:0, samples=180, smooth] (x, {1+cos(x)}) -- (axis cs:0,0)[penColor4!50];
-       \addplot [data cs=polar, very thick, mark=none,fill=penColor4!50,domain=60:90,samples=180,smooth] (x, {3*cos(x)});
+       \addplot[data cs=polar, very thick, mark=none, fill=fill3,  domain=0:60, samples=180, smooth] (x, {1+cos(x)}) -- (axis cs:0,0)[green];
+         \addplot[data cs=polar, very thick, mark=none, fill=fill3,  domain=-90:-60, samples=180, smooth] (x, {3*cos(x)}) -- (axis cs:0,0)[green];
+         \addplot[data cs=polar, very thick, mark=none, fill=fill3,  domain=-60:0, samples=180, smooth] (x, {1+cos(x)}) -- (axis cs:0,0)[green];
+       \addplot [data cs=polar, very thick, mark=none,fill=green,domain=60:90,samples=180,smooth] (x, {3*cos(x)});
       \addplot[data cs=polar,penColor2,domain=0:360,samples=360,smooth, thick] (x,{1+cos(x)}) ;
       \addplot[data cs=polar, penColor, domain=0:360, samples=360, smooth, thick] (x, {3*cos(x)});      
             \end{axis}  
@@ -61,7 +61,7 @@
 
 
 \[
-\int_{\answer{0}}^{\answer{ \frac{\pi}{3} } } \answer{ \frac{1}{2}(1+\cos(\theta))^2   } \d \theta +  \int_{\answer{\frac{\pi}{3}}}^{\answer{\frac{\pi}{2}}} \answer{ \frac{1}{2}  (3\cos(\theta))^2 } \d \theta
+\int_{0}^{\answer{ \frac{\pi}{3} } } \answer{ \frac{1}{2}(1+\cos(\theta))^2   } \d \theta +  \int_{\answer{\frac{\pi}{3}}}^{\answer{\frac{\pi}{2}}} \answer{ \frac{1}{2}  (3\cos(\theta))^2 } \d \theta
 \]
 
 The area of the entire shaded region is obtained by multiplying these
@@ -170,11 +170,37 @@
 \end{image}
 
 
-Suppose we take a ray $\theta=\theta_{1}$ betwen $0$ and $\\frac{\pi}{3}$. We see where such an arbitary ray hits the boundary of our region. So we see that $r_{outer}=1+\cos(\theta)$ ( the blue curve) and $r_{inner}=0$. 
-
+Suppose we take a ray $\theta=\theta_{1}$ between $0$ and $\frac{\pi}{3}$. We see where such an arbitrary ray hits the boundary of our region. So we see that $r_{outer}=1+\cos(\theta)$ ( the blue curve) and $r_{inner}=0$. 
 
 However once we move past $\theta=\frac{\pi}{3}$, the outer radius changes. That is, $r_{outer}=3\cos(\theta)$ (the red curve) while $r_{inner}=0$.
 
+We can summarize in the following picture.
+
+\begin{image}  
+  \begin{tikzpicture}  
+    \begin{axis}[  
+        xmin=-1.2,  
+        xmax=3.5,  
+        ymin=-2,  
+        ymax=2,  
+        axis lines=center,  
+        xlabel=$x$,  
+        ylabel=$y$,  
+        every axis y label/.style={at=(current axis.above origin),anchor=south},  
+        every axis x label/.style={at=(current axis.right of origin),anchor=west},axis on top
+      ]  
+       \addplot[data cs=polar, very thick, mark=none, fill=fill3,  domain=0:60, samples=180, smooth] (x, {1+cos(x)}) -- (axis cs:0,0)[penColor2!50];
+       \addplot [data cs=polar, very thick, mark=none,fill=penColor!50,domain=60:90,samples=180,smooth] (x, {3*cos(x)});
+      \addplot[data cs=polar,penColor2,domain=0:360,samples=360,smooth, thick] (x,{1+cos(x)}) ;
+      \addplot[data cs=polar, penColor, domain=0:360, samples=360, smooth, thick] (x, {3*cos(x)});      
+            \end{axis}  
+  \end{tikzpicture}  
+\end{image} 
+
+The red area is given by $\int_{0}^{\answer{\pi/3}} \frac{1}{2}\left(\answer{1+\cos(\theta)}\right)^2 \d \theta$ and the blue area is given by 
+$\int_{\answer{\pi/3}}^{\answer{\pi/2}} \frac{1}{2}\left(\answer{3\cos(\theta)}\right)^2 \d \theta$
+
+
 
 
 
@@ -224,7 +250,7 @@
 
 
 \[
-\int_{\answer{\frac{\pi}{3}}}^{\answer{ \frac{\pi}{2} } } \answer{ \frac{1}{2}((1+\cos(\theta))^2 -(3\cos(\theta))^2  } \d \theta +  \int_{\answer{\frac{\pi}{2}}}^{\answer{\pi}} \answer{ \frac{1}{2} ( (3\cos(\theta))^2 } \d \theta
+\int_{\answer{\frac{\pi}{3}}}^{\answer{ \frac{\pi}{2} } } \answer{ \frac{1}{2}(1+\cos(\theta))^2 -(3\cos(\theta))^2  } \d \theta +  \int_{\answer{\frac{\pi}{2}}}^{\answer{\pi}} \answer{ \frac{1}{2} ( (3\cos(\theta))^2 } \d \theta
 \]
 
 

introductionToPolarCoordinates/exercises/polarToCartesian1.tex

@@ -8,9 +8,9 @@
 \begin{document}
 \begin{exercise}
 
-To express the polar curve $r= 4\sin(\theta)$ using Cartesian coordinates:
+To express the polar curve $r= 4\sin(\theta)$ using Cartesian coordinates, do the following.
 
-First, not that the curve is a:
+First, note that the curve is a:
 
 \begin{multipleChoice}
 \choice{line}
@@ -19,15 +19,17 @@
 \choice{none of these}
 \end{multipleChoice}
 
+(go to desmos.com and graph the curve if you aren't sure).
+
 We can thus write the curve in the form:
 
 \begin{align*}
 (x-h)^2+(y-k)^2&=r^2\\
-\left(x- \answer{0}\right)^2+\left(y-\answer{2 }\right)^2 = \answer{4 }
+\left(x- \answer{0}\right)^2+\left(y-\answer{2 }\right)^2 = \answer{4 }.
 \end{align*}
 
 \begin{hint}
-Start by using:
+Start by using
 
 \begin{multipleChoice}
 \choice[correct]{$\cos(\theta) =\frac{x}{r}, \sin(\theta) = \frac{y}{r}$}
@@ -42,10 +44,10 @@
 \end{align*}
 
 \begin{question}
-We thus can multiply both sides by $r$ to obtain:
+We thus can multiply both sides by $r$ to obtain
 
 \[
-r^2 =4y
+r^2 =4y.
 \]
 Now, powers of $r$ are easy to convert by using the fact that $r^2 = x^2+y^2$.  Thus:
 

review/refreshFactorials/exponents.tex

@@ -47,7 +47,7 @@
 a_{k+1} = 5\left(\frac{1}{2}\right)^{2(k+1)} = 5\left(\frac{1}{2}\right)^{2k+2}.
 \]
 
-Thus, we can write out the desired ration and simplify.
+Thus, we can write out the desired ratio and simplify.
 
 \[
 \frac{a_{k+1}}{a_k} = \frac{5\left(\frac{1}{2}\right)^{2k+2}}{5\left(\frac{1}{2}\right)^{2k}}= \frac{\cancel{5\left(\frac{1}{2}\right)^{2k}}\cdot \left(\frac{1}{2}\right)^2}{\cancel{5\left(\frac{1}{2}\right)^{2k}}} = \frac{1}{4}

review/refreshFactorials/puttingItTogether.tex

@@ -15,7 +15,7 @@
 
 Now, let's try to simplify some expressions that involve polynomials, exponentials, and factorials.
 
-Suppose that $a_k = \frac{k^2 \cdot 3^k}{k!}$. Simplify $\frac{a_{n+1}}{a_n}$, then find $\lim_{n \to \infty} \frac{a_{n+1}}{a_n}$.
+Suppose that $a_k = \frac{5^{n^2}}{n!}$ (note that the $n^2$ is in the exponent).  Simplify $\frac{a_{n+1}}{a_n}$, then find $\lim_{n \to \infty} \frac{a_{n+1}}{a_n}$.
 
 \begin{align*}
 \frac{a_{n+1}}{a_n} = \answer{ \frac{5^{2n+1}}{n+1}} \\