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\begin{document}
\[ \Large\textbf{Math 1310: Engineering Calculus I} \]
\[ \Large\textbf{Lab \#7, Week \#9, Fall 2018} \]
{} \\
\noindent\fbox{\parbox{\textwidth}{\Large{\textbf{Instructions:} Please show all your work and explain your reasoning as appropriate. You are allowed to use any results from lecture as long as they are stated correctly. You are encouraged to work in groups, but your final written solutions must be in your own words.}}}
\hfill\\
{\Large
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\textbf{Name} & : \underline{\hspace{12cm}} \\
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\textbf{uID} & : \underline{\hspace{12cm}} \\
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\textbf{Group} & : \underline{\hspace{12cm}} \\
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\textbf{Due Date} & : \underline{\textbf{11/01/18}} \\
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\newpage
\phantom{x}\vfill
\begin{center}
\gradetable[v][questions]
\end{center}
\vfill\newpage
\begin{questions}
\titledquestion{Related Rates}\textbf{\thequestiontitle}
It is a windy, fall day and a student decides to take a break from studying to go fly a kite. %However, you realize that there is no break from Calculus as it's applications are all around us.
Suppose that when the kite is $80$ ft above the student's hand level, the wind is blowing it on a horizontal course at a speed of $10$\ ft/sec (to the right).
\begin{parts}
\part Draw a sketch of the kite in relation to the student flying it. Assume the string remains straight from hand to kite. Be sure to label the height, $h$, above hand level and the horizontal distance, $x$, from the kite flyer. \vfill
\begin{solution}
\[x^2+h^2=c^2 \hspace{3cm} \cot\theta=\frac{x}{h}\]
\end{solution}
\part[5] How fast is the student letting out the string when 100\ ft of string is out? \vfill\vfill
\begin{solution}
We are given $\frac{dx}{dt}$=10\ ft/sec and want to find $\frac{dc}{dt}$ when $c=100$\ ft.\\
Using the Pythagorean Theorem, we have
\[x^2+(80)^2=c^2\]
Differentiating both sides implicitly with respect to time, we get
\[2x\frac{dx}{dt}=2c\frac{dc}{dt}\]
Solving for the unknown rate of interest $dc/dt$, we find
\[\frac{dc}{dt}=\frac{x}{c}\frac{dx}{dt}\]
Substituting in the known quantities, we find
\[\frac{dc}{dt}=\bigg(\frac{60}{100}\bigg)\bigg(\frac{10\ ft}{\text{sec}}\bigg) = 6\ \text{ft/sec}\]
\end{solution}
\part[5] At what rate is the angle between the string and the horizontal decreasing when $100$ ft of string has been let out? \vfill\vfill
\begin{solution}
We are given $\frac{dx}{dt}$=10\ ft/sec and want to find $\frac{d\theta}{dt}$ when $c=100$\ ft.\\
Using trigonometry, we have
\[\cot \theta = \frac{x}{80}\]
Differentiating both sides implicitly with respect to time, we have
\[-80\csc^2 \theta\frac{d\theta}{dt}=\frac{dx}{dt}\]
Solving for the unknown rate of interest $d\theta/dt$, we have
\[\frac{d\theta}{dt}=-\frac{dx}{dt}\frac{\sin^2\theta}{80}\]
Substituting in the known quantities, we find
\begin{align*}
\frac{d\theta}{dt}=\frac{-10}{80}\bigg(\frac{80}{100}\bigg)^2=\frac{-2}{25} \ \text{rad/s}
\end{align*}
\end{solution}
\end{parts}
\newpage
\titledquestion{Related Rates}\textbf{\thequestiontitle}
\begin{parts}
\part[5] Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure $P$ and volume $V$ satisfy the equation \[PV=C,\] where $C$ is a constant. Suppose that at a certain instant in time the volume is $600$\ cm$^3$, the pressure is $120$ kPa, and the pressure is increasing at a rate of 20\ kPa/min. At what rate is the volume decreasing at this instant? \vfill
\begin{solution}
We are given $\frac{dP}{dt}=20\ \text{kPa/min}$. We want to find $\frac{dV}{dt}$ when $V=600\ \text{cm}^3$ and $P=120\ \text{kPa}.$\\
Using Boyle's Law and differentiating both sides implicitly with respect to time, we have
\[\frac{dP}{dt}V+P\frac{dV}{dt}=0\]
Solving for the unknown rate of interest, we find,
\[\frac{dV}{dt}=-\frac{\frac{dP}{dt}V}{P}\]
Substituting in the known quantities, we find
\[\frac{dV}{dt}=-\frac{20(600)}{120}=-100\ \text{cm}^3/\text{min}\]
\end{solution}
\part[5] Ohm's Law relates the voltage $V$ (in volts) across a resistor to the electrical current $I$ (in amperes, $A$) passing through the resistor and the resistance $R$ (in ohms, $\Omega$) as follows:
\[V=IR.\]
In a circuit with variable resistance, the quantities $V,$ $I,$ and $R$ might all depend on time. Suppose that the voltage across the resistor is held constant at 15 volts while the resistance is steadily increased at a rate of $0.3\, \Omega/s.$ At what rate is the current decreasing when the resistance is $5\ \Omega$?\vfill
\begin{solution}
We are given $\frac{dV}{dt}=0$ volts/sec and $\frac{dR}{dt}=0.3\, \Omega/\text{s}$. We want to find $\frac{dI}{dt}$ when $R=5\Omega.$\\
Using Ohm's Law, we find that
\[I=\frac{V}{R}=\frac{15}{5}=3A\]
Differentiating both sides of Ohm's Law implicitly with respect to time, we have
\[\frac{dV}{dt}=\frac{dI}{dt}R+I\frac{dR}{dt}\]
Solving for the unknown rate of interest, $dI/dt$, we find
\[\frac{dI}{dt}=\frac{\frac{dV}{dt}-I\frac{dR}{dt}}{R}\]
Substituting in the known quantities, we find
\[\frac{dI}{dt}=\frac{0-3(0.3)}{5}=-\frac{9}{50}=-0.18\ \text{A/s}\]
Since $dI/dt<0$, the current is decreasing at that time.
\end{solution}
%At what rate is the current changing at that time? \vfill
%What is the current through the resistor when the resistance reaches $5\Omega$? \vfill
%At what rate is the current changing at that time? Is the current increasing or decreasing? \vfill\vfill
\end{parts}
\newpage
\titledquestion{Mean and Extreme Values}\textbf{\thequestiontitle}
\begin{parts}
\part [5] Suppose a person driving on a freeway passed through toll booth A at 7:00pm and toll booth B at 8:00pm. The two toll booths are exactly 80 miles apart and the speed limit on the freeway is 75 mph. State and apply a theorem you have learned to show that the person was speeding at some point during the trip. \vfill
%State and apply a theorem you have learned to explain why a driver who traveled 112 miles in 2 hours must have exceeded the speed limit of 55 mi/hr at some point in time during the journey.\vfill
\part[7] Suppose an object with weight $W$ is dragged along a horizontal plane by a force acting along a rope attached to the object. If the rope makes an angle $\theta$ with the plane, then the magnitude of the force is
\[F=\frac{\mu W}{\mu \sin\theta+\cos\theta}\] where $\mu$ is a positive constant called the coefficient of friction and $0\leq\theta\leq\frac{\pi}{2}$. State and apply a theorem you have learned to show that the absolute minimum force occurs when $\tan \theta =\mu$. \vfill\vfill
\end{parts}
\newpage
\titledquestion{Graphing with Calculus}\textbf{\thequestiontitle}
Given the function %$f(x)=3x^{\frac{2}{3}}-2x$
%$f(x)=x^{5/3}-5x^{2/3} = x^{2/3}(x-5)$
$f(x) = 2x^{2/3}(3-x^{1/3})$.
%$t\sqrt{4-t^2}$
\begin{parts}
\part[2] Find the domain of $f$ as well as the $x$- and $y$-intercepts, if any. \vfill
\part[3] Find the critical numbers, if any, of $f$. \vfill
\part[3] Find the intervals over which $f$ is increasing or decreasing. \vfill
\part[2] Find the values of $x$, if any, at which $f$ has a {\textit{local}} maximum or minimum value. \vfill
\part[2] Find the {\textit{absolute}} maximum or minimum values, if any, of $f$. \vfill
\part[3] Find the intervals on which $f$ is concave up or concave down and any inflection points. \vfill
\part[3] Sketch a graph of $f$. Be sure to label any $x$- and $y$-intercepts, maxima and minima, inflection points and asymptotes. \vfill
\end{parts}
\end{questions}
\end{document}