fixed some typos

continuityOfFunctionsOfSeveralVariables/exercises/limitDifferentLinearE.tex

@@ -23,7 +23,7 @@
   Is this enough to determine that $\Lim{x,y}{0,0} = 0$? \wordChoice{\choice{Yes}\choice[correct]{No}}
 
   \begin{exercise}
-     As $\point{x,y} \to \point{0,0}$ along $y=2x$, $F(x,y)$ approaches $\answer{-\frac{4}{27}}$.
+     As $\point{x,y} \to \point{0,0}$ along $y=2x$, $F(x,y)$ approaches $\answer{-\frac{4}{81}}$.
 
 \begin{multipleChoice}
 \choice{Yes; $\Lim{x,y}{0,0} F(x,y)$ exists.}

motionAndPathsInSpace/digInParameterizingByArcLength.tex

@@ -76,7 +76,7 @@
 Let's state this as a definition.
 
 \begin{definition}
-  A vector-valued function $\vec{g}(s)$ is \dfn{parameterized by arc
+  A curve traced out by a vector-valued function $\vec{g}(s)$ is \dfn{parameterized by arc
     length} if
   \[
   s = \int_0^s |\vec{g}'(t)|\d t.
@@ -165,7 +165,7 @@
 theorem should be quite sensible.
 
 \begin{theorem}
-  A vector-valued function $\vec{f}:\R\to \R^2$ is parameterized by
+  A curve traced out by a continuously differentiable vector-valued function $\vec{f}:\R\to \R^2$ is parameterized by
   arc length if and only if $|\vec{f}'| = 1$.
 \end{theorem}
 

motionAndPathsInSpace/exercises/arcLengthParameterizationGuided.tex

@@ -15,10 +15,10 @@
 \vec{r}(t) = \vector{2t^{3/2},4t+1}, 0 \leq t \leq 4.
 \] 
 
-In order to determine if the curve is parameterized by arclength, we could check either if
+In order to determine if this continuously differentiable curve is parameterized by arclength, we could check either if
 
 \begin{itemize}
-\item $\int_0^s \left|\vec{r}'(t)\right| \d s = s$
+\item $\int_0^t \left|\vec{r}'(\tau)\right| \d \tau = t$
 \item $\left|\vec{r}'(t)\right|=1$ for all $t$.
 \end{itemize}
 

motionAndPathsInSpace/exercises/arclengthFromGraph1.tex

@@ -37,8 +37,8 @@
 \begin{selectAll}
 \choice[correct]{$\left|\vec{p}'(t)\right|$ should be $1$ for all $t$.}
 \choice{$\left|\vec{p}'(t)\right|$ must be $1$ for some $t>0$, but not necessarily for all $t$.}
-\choice[correct]{$\int_0^t \left|\vec{p}'(t)\right| \d t = t$ for all $t$.}
-\choice{$\int_0^t \left|\vec{p}'(t)\right| \d t = t$ for some $t>0$, but not necessarily for all $t$.}
+\choice[correct]{$\int_0^t \left|\vec{p}'(\tau)\right| \d \tau = t$ for all $t$.}
+\choice{$\int_0^t \left|\vec{p}'(\tau)\right| \d \tau = t$ for some $t>0$, but not necessarily for all $t$.}
 \choice{$|\vec{p}(t)| = t$ for all $t$.}
 \choice{$|\vec{p}(t)| = t$ for some $t>0$, but not necessarily for all $t$.}
 \end{selectAll}

partialDerivativesAndTheGradientVector/exercises/computations3.tex

@@ -45,7 +45,7 @@
 
 \begin{itemize}
 \item $f_x(x,y) = \answer{\frac{24(2x+8y^4)^2 \sec(x) - 4\sec(x)\tan(x) (2x+8y^4)^3}{16 \sec^2(x)}  }$
-\item $f_y(x,y) = \answer{\frac{6y^3 \big(2x+8y^4\big)^2}{\sec(x)}}$
+\item $f_y(x,y) = \answer{\frac{24y^3 \big(2x+8y^4\big)^2}{\sec(x)}}$
 \end{itemize}
 
 

partialDerivativesAndTheGradientVector/exercises/exerciseListE.tex

@@ -13,7 +13,6 @@ \part{Partial derivatives and the gradient vector}
 \practice{computations2.tex}%new
 \practice{computations3.tex}%new
 \practice{contour1.tex}
-\practice{table1.tex}
 \practice{verifyClairaut1.tex}%new
 \practice{verifyClairaut2.tex}%new
 \practice{gradient1.tex}%new

partialDerivativesAndTheGradientVector/exercises/verifyClairaut2.tex

@@ -24,16 +24,16 @@
 
 To verify Clairaut's Theorem, we have to establish the equality of the mixed partial derivatives.
 
-Noting that $\Pp{^2F}{x \partial y}$ \wordChoice{\choice{$F_{xy}$} \choice[correct]{$F_{yx}$}}, we find that 
+Noting that $\Pp{^2F}{x \partial y}=$ \wordChoice{\choice{$F_{xy}$} \choice[correct]{$F_{yx}$}}, we find that 
 
 \[
-\Pp{^2F}{x \partial y} = \pp{x} \left[\answer{   e^{y/x} -\frac{ye^{y/x}}{x}   } \right] = \answer{-\frac{ye^{y/x}}{x^2}}.
+\Pp{^2F}{x \partial y} = \pp{x}  \left[\answer{e^{y/x}} \right]   = \answer{-\frac{ye^{y/x}}{x^2}}.
 \]
 
-Noting that $\Pp{^2F}{y \partial x}$ \wordChoice{\choice[correct]{$F_{xy}$} \choice{$F_{yx}$}} , we find that 
+Noting that $\Pp{^2F}{y \partial x}=$ \wordChoice{\choice[correct]{$F_{xy}$} \choice{$F_{yx}$}} , we find that 
 
 \[
-\Pp{^2F}{y \partial x}  = \pp{y} \left[\answer{e^{y/x}} \right] = \answer{-\frac{ye^{y/x}}{x^2}}.
+\Pp{^2F}{y \partial x}  = \pp{y} \left[\answer{   e^{y/x} -\frac{ye^{y/x}}{x}   } \right]= \answer{-\frac{ye^{y/x}}{x^2}}.
 \]
 
 Do the results agree?