%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                                                           %
%  Fundamental Theorem of Calculus Derivative Form          %
%                                                           %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\problemgrouptitle{Fundamental Theorem of Calculus (Derivative Form)}

\begin{problemblock}[2]{Compute the derivatives of the following functions using the Fundamental Theorem of Calculus.}

\item $F(x)=\displaystyle\int_0^x\sqrt{t^3+2t+1}\d{t}$
\answer{$\displaystyle\frac{dF}{dx}=\sqrt{x^3+2x+1}$}

\item $h(x)=\displaystyle\int_0^{x^4}\sqrt{t^3+2t+1}\d{t}$
\answer{$\displaystyle h'(x)=4x^3\sqrt{x^{12}+2x^4+1}$}

\item $\displaystyle g(x)=\int_{x}^{1}\left[\cos\left(t^3\right)\right]dt$
\answer{$g'(x)=-\left[\cos\left(x^3\right)\right]$}

\item $\displaystyle H(x)=\int_{2x}^{3x}\sqrt[3]{t^3+1}\,dt$
\answer{Noting that $H(x)=\int_{2x}^0\sqrt[3]{t^3+1}\,dt+\int_0^{3x}\sqrt[3]{t^3+1}\,dt=-\int_0^{2x}\sqrt[3]{t^3+1}\,dt+\int_0^{3x}\sqrt[3]{t^3+1}\,dt$, one gets
$H'(x)=-2\sqrt[3]{8x^3+1}+3\sqrt[3]{27x^3+1}$}

\end{problemblock}

\item The \emph{error function}, $\mathrm{erf}(x)$, is defined by $\displaystyle\mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{-\infty}^xe^{-t^2}\,dt$ where $e^u$ is the natural exponential function. If $f(x)=\mathrm{erf}\left(x^3\right)$ find $f'(x)$\,.
\answer{$\displaystyle f'(x)=\frac{2}{\sqrt{\pi}}e^{-(x^3)^2}\left(3x^2\right)=\frac{6x^2}{\sqrt{\pi}}e^{-x^6}$ (Use the Chain Rule and the Fundamental Theorem of Calculus.)}

% Petry: My following problem is obviously quite unrealistic but I thought
% it would be interesting to have something that uses the FTC.
%\lab{\clearpage}
\textbook{\clearpage} 
\item The left-side of a symmetrical glacial river valley has an approximate parabolic shape described by the curve $y=-\frac{1}{4}x^2+1$ (in km) as shown.
\begin{center}
\resizebox{14.0cm}{!}{\includegraphics{graphics/valleybridge}}
\end{center}
 Using calculus techniques the arc length $s$ from point $A$ at the top of the valley to the point $P(x,y)$ can be shown to equal
\[s=\int_0^x\sqrt{1+\frac{1}{4}t^2}\,dt\ .\]
Engineers wish to build a road connecting points $A$ and $B$ with a bridge spanning the valley at the point $P$ to the corresponding point $Q$ on the opposite side of the valley.  If the cost to build the bridge is 25\% more per kilometre than the cost of building the road (i.e. it is $5/4$ times as much per km), at what point $P(x,y)$ should they start the bridge to minimize the total cost?  (Hint: Due to the symmetry of the situation just minimize the cost to build from point $A$ to the middle of the bridge.)
\answer{$(x,y)=\left(\frac{3}{2}\ \mbox{km},\frac{7}{16}\ \mbox{km}\right)$}

%%% Local Variables:
%%% mode: latex
%%% TeX-master: "../master"
%%% End:
