Merge branch 'master' of github.com:mooculus/calculus

calculusAndVectorValuedFunctions/exercises/exerciseList.tex

@@ -6,6 +6,8 @@
 \begin{document}
 
 \part{Calulus and vector-valued functions}
+\practice{domainAndCont1.tex}
+\practice{tangentLinesTwoWays}%new
 \practice{velocityVersusAcceleration.tex}
 \practice{derivAndTan2.tex}
 \practice{interpretGraph1} %new 
@@ -29,8 +31,7 @@ \part{Calulus and vector-valued functions}
 \practice{derivAndTan1.tex} %new
 \practice{positionVersusVelocity.tex}
 \practice{vectorValuedLimitTrig.tex}
-\practice{domainAndCont1.tex}
-\practice{conceptualQuestionsDynamics}
+\practice{conceptualQuestionsDynamics}%new
 
 
 

calculusAndVectorValuedFunctions/exercises/exerciseListE.tex

@@ -6,20 +6,18 @@
 \begin{document}
 
 \part{Calculus and vector-valued functions}
+\practice{domainAndCont1.tex}
 \practice{vectorValuedLimitRational.ditto.tex}
-
-
+\practice{tangentLinesTwoWays}
 \practice{velocityVersusAcceleration.tex}
 \practice{derivAndTan2.tex}
-\practice{domainAndCont1.tex}
 \practice{deriv1.tex}
 \practice{sumRule.tex}
 \practice{linearCombination.tex}
 \practice{productRule.tex}
 \practice{deriv2.tex}
 \practice{antiderivative2.tex}
 \practice{interpretGraph1}
-
 \practice{velocityVersusAccelerationCircle.tex}
 \practice{vectorValuedLimitPolynomial.ditto.tex}
 \practice{domainAndCont2.tex}
@@ -31,6 +29,6 @@ \part{Calculus and vector-valued functions}
 \practice{interpretGraph2}
 \practice{conceptualQuestionsDynamics}
 
-
+%Need curve that does not intersect a surface but tangent line that does
 
 \end{document}

calculusAndVectorValuedFunctions/exercises/tangentLinesTwoWays.tex

@@ -0,0 +1,62 @@
+\documentclass{ximera}
+
+\input{../../preamble.tex}
+
+\outcome{Compute the derivative of a vector-valued function.}
+
+\author{Jim Talamo}
+
+\begin{document}
+\begin{exercise}
+
+Consider the curve $\mathcal{C} = \left\{(x,y) \in \R^2 ~ \bigg| ~ y=2x^2-7\right\}$ .
+
+Find a Cartesian representation of the tangent line at $x=2$.  Express you final answer in the form $y=mx+b$.
+
+\[
+y= \answer{8x-15}
+\]
+
+\begin{exercise}
+Now, require that $x(t) =t$.  
+
+A parametric equation that traces out $\mathcal{C}$ is given by
+
+\[
+\vec{r}(t) = \vector{\answer{t},\answer{2t^2-7}}.
+\]
+
+A parameterization of the tangent line to the curve where $x=2$ is
+
+\[
+\vec{l}(t)=\vector{\answer{t+2},\answer{8t+1}}.
+\]
+
+\begin{hint}
+The curve passes through $x=2$ when $t=\answer{2}$.  
+
+\begin{itemize}
+\item A vector $\vec{v}$ parallel to the tangent line at $x=2$ is found by evaluating \wordChoice{\choice{$\vec{r}(t)$}\choice[correct]{$\vec{r}'(t)$}} when $t=\answer{2}$.
+\item A point $P_0$ on the tangent line is found by evaluating \wordChoice{\choice[correct]{$\vec{r}(t)$}\choice{$\vec{r}'(t)$}} when $t=\answer{2}$.
+\end{itemize}
+
+A parametric description of the tangent line can be found from 
+
+\[\vec{l}(t) = \vec{v}t+\vec{P}_0\]
+
+\end{hint}
+
+\begin{feedback}[correct]
+Do the answers to Part I and Part II describe the same line?  Intuitively, they should since the tangent line at the point on the curve where $x=2$ should not depend on how we choose to describe the curve.
+
+To check this, we can determine if the set of parametric equations $x(t)=t+2$ and $y(t)=8t+1$ satisfy the equation $y=8x-15$ from the first part.
+
+\[
+8[x(t)]-15 = 8 [t+2] -15 = 8t+16-15 = 8t+1=y(t).
+\] 
+
+Thus, the two equations describe the same line.
+\end{feedback}
+\end{exercise}
+\end{exercise}
+\end{document}

commonCoordinates/digInSphericalCoordinates.tex

@@ -18,7 +18,7 @@
 
 \begin{definition}
   An ordered triple consisting of a radius, an angle, and a height
-  $(\rho,\varphi,\theta)$ can be graphed as
+  $(\rho,\theta, \varphi)$ can be graphed as
   \begin{align*}
     x &= \rho\cdot \cos(\theta)\sin(\varphi)\\
     y &= \rho\cdot \sin(\theta)\sin(\varphi)\\

crossProducts/exercises/jCrossProductAndOrthogonalDecomposition1.tex

@@ -14,28 +14,28 @@
 Find a vector $\vec{w}$ of magnitude 3 that is orthogonal to both $\vec{u}$ and $\vec{v}$.
 
 \[
-\vec{w} = \vector{ \answer{\frac{2}{\sqrt{5}}},\answer{\frac{-4}{\sqrt{5}}}, \answer{\frac{5}{\sqrt{5}}}   }
+\vec{w} = \vector{ \answer{\frac{-2}{\sqrt{5}}},\answer{\frac{4}{\sqrt{5}}}, \answer{\frac{5}{\sqrt{5}}}   }
 \]
 
 
 \begin{hint}
-To find a vector orthognal to both $\vec{u}$ and $\vec{v}$, we can:
+To find a vector orthogonal to both $\vec{u}$ and $\vec{v}$, we can:
 
 \begin{multipleChoice}
 \choice{Compute $\vec{u} \dotp \vec{v}$.}
 \choice[correct]{Compute $\vec{u} \cross \vec{v}$.}
 \end{multipleChoice}
 
-We find:
+We find that
 
 \[
-\vec{u} \cross \vec{v} = \vector{\answer{2},\answer{-4},\answer{5}}
+\vec{u} \cross \vec{v} = \vector{\answer{-2},\answer{4},\answer{5}}.
 \]
 
-To find a vector of magnitude 5 in this direction:
+To find a vector of magnitude 3 in this direction:
 
 \begin{multipleChoice}
-\choice{Multiply the above vector by 5.}
+\choice{Multiply the above vector by 3.}
 \choice[correct]{Compute the magnitude and scale the vector accordingly.}
 \end{multipleChoice}
 
@@ -44,18 +44,18 @@
 \[
 |\vec{u} \cross \vec{v} | = \answer{\sqrt{45}}
 \]
-We can make the vector $\vec{w}$ by finding a unit vector $\hat{w}$ in the appropriate direction:
+We can make the vector $\vec{w}$ by finding a unit vector $\uvec{w}$ in the appropriate direction.
 
 \[
-\hat{w} = \frac{\vec{u} \cross \vec{v}}{|\vec{u} \cross{v}|} = \vector{ \answer{\frac{2}{\sqrt{45}}},\answer{\frac{-4}{\sqrt{45}}}, \answer{\frac{5}{\sqrt{45}}}   }
+\uvec{w} = \frac{\vec{u} \cross \vec{v}}{|\vec{u} \cross{v}|} = \vector{ \answer{\frac{-2}{\sqrt{45}}},\answer{\frac{4}{\sqrt{45}}}, \answer{\frac{5}{\sqrt{45}}}   }
 \]
 
-Recall that if $a>0$ is a scalar, the length of $a\vec{u}$ is $a$ times the length of $\vec{u}$, so a vector $\vec{w}$ of magnitude 3 in the direction of $\hat{w}$ is $\answer{3}\hat{w}$. 
+Recall that if $a>0$ is a scalar, the length of $a\vec{u}$ is $a$ times the length of $\vec{u}$, so a vector $\vec{w}$ of magnitude 3 in the direction of $\uvec{w}$ is $\answer{3}\uvec{w}$. 
 \end{hint}
 
 %%%%%%%%%
 \begin{exercise}
-Find a vector $\vec{P}$ parallel to $\vec{u}$ and a vector $\vec{N}$ orthogonal to $\vec{v}$ so $\vec{u} = \vec{P} +\vec{N}$.
+Find a vector $\vec{P}$ parallel to $\vec{v}$ and a vector $\vec{N}$ orthogonal to $\vec{v}$ so $\vec{u} = \vec{P} +\vec{N}$.
 
 \begin{align*}
 \vec{P} = \vector{\answer{4},\answer{2}, \answer{0}} \\
@@ -71,6 +71,12 @@
  \end{multipleChoice}
  \end{hint}
 
-  \end{exercise}
+ \begin{feedback}[correct] We can check that the vectors $\vec{P}$ and $\vec{N}$ have the desired directions as follows.
+ \begin{itemize}
+ \item  Note that $\vec{P}$ is parallel to $\vec{v}$ because it is a scalar multiple of $\vec{v}$. 
+ \item We can check that $\vec{N}$ is orthogonal to $\vec{v}$ by noting that $\vec{v} \dotp \vec{N} =  \vector{2,1,0} \dotp \vector{1,-2,-2} = 2-2+0=0.$
+ \end{itemize}
+ \end{feedback}
+\end{exercise}
 \end{exercise}
 \end{document}

crossProducts/exercises/jVectorOrScalar1.tex

@@ -10,14 +10,14 @@
 
 \begin{document}
 \begin{exercise}
-Suppose that $\vec u $ and $\vec v $ are nonzero three dimensional vectors and $\cross$ denotes the cross product and $\dotp$ denotes the dot product.
+Suppose that $\vec u $ and $\vec v $ are nonzero three dimensional vectors.  Let $\cross$ denote the cross product, $\dotp$ denote the dot product, and $\cdot$ denote scalar multiplication.
 
 The quantity $(\vec u \dotp \vec v)+ |\vec v|$ is:
 
 \begin{multipleChoice}
-\choice {A Scalar}
-\choice{A Vector}
-\choice[correct] {Undefined}
+\choice {a scalar.}
+\choice{a vector.}
+\choice[correct] {undefined.}
 \end{multipleChoice}
 
 
@@ -26,59 +26,59 @@
 \end{hint}
 
 \begin{hint}
-Dot products of vectors is a scalar, and the Cross product of vectors is a vector.
+Dot products of vectors is a scalar, and the cross product of vectors is a vector.
 \end{hint}
 
 
 
-The quantity $(proj_{\vec v} \vec u)\cross \vec v$ is:
+The quantity $(\proj_{\vec v} \vec u)\cross \vec v$ is:
 
 \begin{multipleChoice}
-\choice{A Scalar}
-\choice[correct] {A Vector}
-\choice{Undefined}
+\choice{a scalar.}
+\choice[correct] {a vector.}
+\choice{undefined.}
 \end{multipleChoice}
 
 
 \begin{hint}
-Multilplying a vector by a scalar is defined. However, adding a vector and a scalar is undefined.
+Multiplying a vector by a scalar is defined. However, adding a vector and a scalar is undefined.
 \end{hint}
 
 
 The quantity $\vec u \cdot \vec v$ is:
 
 \begin{multipleChoice}
-\choice[correct]  {A Scalar}
-\choice{A Vector}
-\choice{Undefined}
+\choice[correct]  {a scalar.}
+\choice{a vector.}
+\choice{undefined.}
 \end{multipleChoice}
 
 \begin{hint}
 The dot and cross products are defined only for \emph{vectors}.
 \end{hint}
 \begin{hint}
-Dot products of vectors is a scalar, and the Cross product of vectors is a vector.
+Dot products of vectors is a scalar, and the cross product of vectors is a vector.
 \end{hint}
 
 
-The quantity $(\vec u + \vec v) \vec v$ is:
+The quantity $(\vec u + \vec v) \cdot  \vec v$ is:
 \begin{multipleChoice}
-\choice{A Scalar}
-\choice{A Vector}
-\choice[correct] {Undefined}
+\choice{a scalar.}
+\choice{a vector.}
+\choice[correct] {undefined.}
 \end{multipleChoice}
 
 \begin{hint}
-Multilplying a vector by a scalar is defined. However, adding a vector and a scalar is undefined.
+Multiplying a vector by a scalar is defined. However, adding a vector and a scalar is undefined.
 \end{hint}
 
 
-The quantity $proj_{\vec v} \vec u$ is:
+The quantity $\proj_{\vec v} \vec u$ is:
 
 \begin{multipleChoice}
-\choice{A Scalar}
-\choice[correct]{A Vector}
-\choice {Undefined}
+\choice{a scalar.}
+\choice[correct]{a vector.}
+\choice {undefined.}
 \end{multipleChoice}
 
 

crossProducts/exercises/jVectorOrScalar2.tex

@@ -10,47 +10,47 @@
 
 \begin{document}
 \begin{exercise}
-Suppose that $\vec u $ and $\vec v $ are nonzero three dimensional vectors and $\cross$ denotes the cross product and $\dotp$ denotes the dot product.
+Suppose that $\vec u $ and $\vec v $ are nonzero three dimensional vectors.  Let $\cross$ denote the cross product, $\dotp$ denote the dot product, and $\cdot$ denote scalar multiplication.
 
-The quantity $(\vec u \cdot \vec v) \times \vec u$ is:
+The quantity $(\vec u \cdot \vec v) \times \vec u$ is
 
 \begin{multipleChoice}
-\choice{A Scalar}
-\choice{A Vector}
-\choice[correct] {Undefined}
+\choice{a scalar.}
+\choice{a vector.}
+\choice[correct] {undefined.}
 \end{multipleChoice}
 
 
 
 The quantity $(\vec u \times \vec v)\times \vec v$ is:
 
 \begin{multipleChoice}
-\choice{A Scalar}
-\choice[correct]{A Vector}
-\choice {Undefined}
+\choice{a scalar.}
+\choice[correct]{a vector.}
+\choice {undefined.}
 \end{multipleChoice}
 
 The quantity $\frac {\vec u} {\vec v}$ is:
 
 \begin{multipleChoice}
-\choice{A Scalar}
-\choice{A Vector}
-\choice[correct] {Undefined}
+\choice{a Scalar.}
+\choice{a Vector.}
+\choice[correct] {Undefined.}
 \end{multipleChoice}
 
-The quantity $(scal_{\vec u} \vec v$ is:
+The quantity $(\scal_{\vec u} \vec v) \cdot \vec{v}$ is:
 \begin{multipleChoice}
-\choice{A Scalar}
-\choice [correct] {A Vector}
-\choice{Undefined}
+\choice{a scalar.}
+\choice [correct] {a vector.}
+\choice{undefined.}
 \end{multipleChoice}
 
-The quantity $(proj_{\vec v} \vec u)\cdot \vec v$ is:
+The quantity $(\proj_{\vec v} \vec u)\cdot \vec v$ is:
 
 \begin{multipleChoice}
-\choice [correct]{A Scalar}
-\choice{A Vector}
-\choice {Undefined}
+\choice [correct]{a scalar.}
+\choice{a vector.}
+\choice {undefined.}
 \end{multipleChoice}
 
 

definiteIntegral/digInTheDefiniteIntegral.tex

@@ -476,7 +476,7 @@
 \item $\int_a^c f(x)\d x + \int_c^b f(x)\d x = \int_a^b f(x)\d x$
 \item $\int_a^bf(x)\d x = -\int_b^a f(x)\d x$
 \item $\int_a^bk\cdot f(x)\d x = k\cdot\int_a^bf(x)\d x$
-\item $\int_a^b f(x)\pm g(x)\d x = \int_a^bf(x)\d x \pm \int_a^bg(x)\d x$
+\item $\int_a^b \left(f(x)\pm g(x)\right)\d x = \int_a^bf(x)\d x \pm \int_a^bg(x)\d x$
 \end{enumerate}
 \begin{explanation}
   We will address each property in turn:

integralsOfPolarFunctions/exercises/area2.tex

@@ -12,7 +12,7 @@
 \begin{exercise}
 
 
-Consider the polar curve $r=\cos(3\theta)$. The graph is shown below: 
+Consider the polar curve $r=\cos(3\theta)$. The graph is shown below.
 
 
 
@@ -42,11 +42,11 @@
 The area of the region $S$ is 
 
 \[
-\int_{\answer{\frac{\pi}{6}}}^{\answer{ \frac{\pi}{3} } } \answer{ \frac{1}{2}(\cos(3\theta))^2   } \d \theta=\answer{ \frac{\pi}{24}}
+\int_{\answer{\frac{\pi}{6}}}^{\answer{ \frac{\pi}{2} } } \answer{ \frac{1}{2}(\cos(3\theta))^2   } \d \theta=\answer{ \frac{\pi}{12}}
 \]
 
 
-The entire area enclosed by the curve (all three petals) is $\answer{  \frac{\pi}{8}}$. 
+The entire area enclosed by the curve (all three petals) is $\answer{  \frac{\pi}{4}}$. 
 
 
 \begin{hint}
@@ -132,7 +132,7 @@
 That means as $\theta$ varies from $\frac{\pi}{6}$ to $\frac{\pi}{3}$, we can think of $r$ being positive instead and $\theta$ varying over the interval $\pi + \frac{\pi}{6}$ to $\pi + \frac{\pi}{3}$. 
 
 
-In order to evaluate the integral, recall the trig identity $sin^2(\theta)=\frac{1-\cos(2\theta)}{2}$. 
+In order to evaluate the integral, recall the trig identity $\sin^2(\theta)=\frac{1-\cos(2\theta)}{2}$.