Q. 87 - Calculus

Question

$$\begin{gathered} \mathrm{The} \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{circle} \mathrm{can} \mathrm{be} \mathrm{written} \mathrm{in} \mathrm{terms} \mathrm{of} \mathrm{its} \mathrm{radius} \mathrm{as} \mathrm{A} = {\mathrm{\pi r}}^2, \mathrm{where} \mathrm{both} \mathrm{A} \mathrm{and} \mathrm{r} \mathrm{are} \mathrm{functions} \mathrm{of} \mathrm{time}. \\ \mathrm{Suppose} \mathrm{a} \mathrm{circular} \mathrm{area} \mathrm{from} \mathrm{a} \mathrm{spotlight} \mathrm{on} \mathrm{a} \mathrm{stage} \mathrm{floor} \mathrm{is} \mathrm{slowly} \mathrm{expanding}. \\ \left(\mathrm{a}\right) \mathrm{Find} \frac{d\mathrm{A}}{d\mathrm{r}} \mathrm{and} \mathrm{explain} \mathrm{its} \mathrm{meaning} \mathrm{in} \mathrm{practical} \mathrm{terms}. \\ \left(\mathrm{b}\right) \mathrm{Does} \mathrm{the} \mathrm{rate} \frac{d\mathrm{A}}{d\mathrm{r}} \mathrm{depend} \mathrm{on} \mathrm{how} \mathrm{fast} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{is} \mathrm{increasing}? \\ \mathrm{Does} \mathrm{it} \mathrm{depend} \mathrm{on} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{circle}? \mathrm{Why} \mathrm{or} \mathrm{why} \mathrm{not}? \\ \left(\mathrm{c}\right) \mathrm{Find} \frac{d\mathrm{A}}{d\mathrm{t}} \mathrm{and} \mathrm{explain} \mathrm{its} \mathrm{meaning} \mathrm{in} \mathrm{practical} \mathrm{terms}. \\ \left(\mathrm{d}\right) \mathrm{Does} \mathrm{the} \mathrm{rate} \frac{d\mathrm{A}}{d\mathrm{t}} \mathrm{depend} \mathrm{on} \mathrm{how} \mathrm{fast} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{is} \mathrm{increasing}? \mathrm{Does} \mathrm{it} \mathrm{depend} \mathrm{on} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{circle}? \\ \left(\mathrm{e}\right) \mathrm{If} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{of} \mathrm{light} \mathrm{is} \mathrm{increasing} \mathrm{at} \mathrm{a} \mathrm{constant} \mathrm{rate} \mathrm{of} 2 \mathrm{inches} \mathrm{per} \mathrm{second}, \\ \mathrm{how} \mathrm{fast} \mathrm{is} \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{of} \mathrm{light} \mathrm{increasing} \mathrm{at} \mathrm{the} \mathrm{moment} \mathrm{that} \mathrm{the} \mathrm{spotlight} \mathrm{has} \mathrm{a} \mathrm{radius} \mathrm{of} 24 \mathrm{inches}? \end{gathered}$$

Step-by-Step Solution

Answer

$$\begin{gathered} \left(\mathrm{a}\right). \mathrm{The} \mathrm{value} \mathrm{of} \frac{d\mathrm{A}}{d\mathrm{r}} = 2\mathrm{\pi r} \\ \left(\mathrm{b}\right). \mathrm{The} \mathrm{rate} \frac{d\mathrm{A}}{d\mathrm{r}} \mathrm{does} \mathrm{not} \mathrm{depends} \mathrm{on} \mathrm{how} \mathrm{fast} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{circle} \mathrm{is} \mathrm{increasing} \mathrm{because} \mathrm{rate} \mathrm{of} \mathrm{change} \mathrm{of} \\ \mathrm{derivative} \mathrm{of} \mathrm{the} \mathrm{area} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{area} \mathrm{and} \mathrm{the} \mathrm{rate} \frac{d\mathrm{A}}{d\mathrm{r}} \mathrm{depends} \mathrm{on} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{circle} \mathrm{because} \mathrm{the} \mathrm{value} \\ \mathrm{will} \mathrm{increase} \mathrm{or} \mathrm{decrease} \mathrm{as} \mathrm{the} \mathrm{radius} \mathrm{changes}. \\ \left(\mathrm{c}\right). \mathrm{The} \mathrm{value} \mathrm{of} \frac{d\mathrm{A}}{d\mathrm{t}} = 2\mathrm{\pi r}\frac{d\mathrm{r}}{d\mathrm{t}} \\ \left(\mathrm{d}\right). \mathrm{The} \mathrm{rate} \frac{d\mathrm{A}}{d\mathrm{t}} \mathrm{depends} \mathrm{on} \mathrm{how} \mathrm{fast} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{circle} \mathrm{is} \mathrm{increasing} \mathrm{because} \mathrm{rate} \mathrm{of} \mathrm{change} \mathrm{of} \\ \mathrm{derivative} \mathrm{of} \mathrm{the} \mathrm{area} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{area} \mathrm{and} \mathrm{the} \mathrm{rate} \frac{d\mathrm{A}}{d\mathrm{t}} \mathrm{depends} \mathrm{on} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{circle}. \\ \left(\mathrm{e}\right). \mathrm{The} \mathrm{rate} \mathrm{of} \mathrm{change} \mathrm{of} \mathrm{area} \mathrm{of} \mathrm{circle} \mathrm{of} \mathrm{light} = 96\mathrm{\pi }. \end{gathered}$$

Step 1. Given Information

$\mathrm{A} = {\mathrm{\pi r}}^2$

Step 2. Solution (a) : value of d A d r

 $$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{equation} : \mathrm{A} = {\mathrm{\pi r}}^2 \\ \mathrm{Differentiate} \mathrm{above} \mathrm{equation} \\ \frac{d\mathrm{A}}{d\mathrm{r}} = \frac{d }{d\mathrm{r}}\left({\mathrm{\pi r}}^2\right) = \mathrm{\pi } \frac{d}{\mathrm{dr}}\left({\mathrm{r}}^2\right) = 2\mathrm{\pi r} \mathrm{inch} \\ \mathrm{The} \mathrm{practical} \mathrm{meaning} \mathrm{is} \mathrm{the} \mathrm{rate} \mathrm{of} \mathrm{change} \mathrm{of} \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{circle} \mathrm{with} \mathrm{radius}. \end{gathered}$$

Step 3. Solution (b) : How   value   of   d A d r   related   to   increase   in   radius   of   circle .

$$\begin{gathered} \mathrm{The} \mathrm{rate} \frac{d\mathrm{A}}{d\mathrm{r}} \mathrm{does} \mathrm{not} \mathrm{depends} \mathrm{on} \mathrm{how} \mathrm{fast} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{circle} \mathrm{is} \mathrm{increasing} \mathrm{because} \mathrm{rate} \mathrm{of} \mathrm{change} \mathrm{of} \\ \mathrm{derivative} \mathrm{of} \mathrm{the} \mathrm{area} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{area} \mathrm{and} \mathrm{the} \mathrm{rate} \frac{d\mathrm{A}}{d\mathrm{r}} \mathrm{depends} \mathrm{on} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{circle} \mathrm{because} \mathrm{the} \mathrm{value} \\ \mathrm{will} \mathrm{increase} \mathrm{or} \mathrm{decrease} \mathrm{as} \mathrm{the} \mathrm{radius} \mathrm{changes}. \end{gathered}$$

Step 4. Solution (c) : Value of d A d t

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{equation} : \mathrm{A} = {\mathrm{\pi r}}^2 \\ \mathrm{Differentiate} \mathrm{above} \mathrm{equation} : \\ \frac{d\mathrm{A}}{d\mathrm{t}} = \frac{d }{d\mathrm{t}}\left({\mathrm{\pi r}}^2\right) = \mathrm{\pi }\frac{d }{d\mathrm{t}}\left({\mathrm{r}}^2\right) = 2\mathrm{\pi r}\frac{d\mathrm{r}}{d\mathrm{t}} \mathrm{inch} \mathrm{per} \mathrm{seconds} \end{gathered}$$

Step 5. solution (d) : How value of d A d t depends on increase in radius

$$\begin{gathered} \mathrm{The} \mathrm{rate} \frac{d\mathrm{A}}{d\mathrm{t}} \mathrm{depends} \mathrm{on} \mathrm{how} \mathrm{fast} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{circle} \mathrm{is} \mathrm{increasing} \mathrm{because} \mathrm{rate} \mathrm{of} \mathrm{change} \mathrm{of} \\ \mathrm{derivative} \mathrm{of} \mathrm{the} \mathrm{area} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{area} \mathrm{and} \mathrm{the} \mathrm{rate} \frac{d\mathrm{A}}{d\mathrm{t}} \mathrm{depends} \mathrm{on} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{circle}. \end{gathered}$$

Step 6. Solution (e) : The rate of change of area of circle of light.

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{equation} : \mathrm{A} = {\mathrm{\pi r}}^2 \\ \mathrm{Differentiate} \mathrm{above} \mathrm{equation} \\ \frac{d\mathrm{A}}{d\mathrm{t}} = \frac{d}{d\mathrm{t}}\left({\mathrm{\pi r}}^2\right) = \mathrm{\pi }\frac{d}{d\mathrm{t}}\left({\mathrm{r}}^2\right) = 2\mathrm{\pi r}\frac{d\mathrm{r}}{d\mathrm{t}} \mathrm{inch} \mathrm{per} \mathrm{seconds} \\ \mathrm{Substitute} \mathrm{the} \mathrm{value} \mathrm{in} \mathrm{above} \mathrm{equation} \\ {\overline{)\frac{d\mathrm{A}}{d\mathrm{t}}}}_{\mathrm{r}=24} = 2\mathrm{\pi }\left(24\right)\left(2\right) = 96\mathrm{\pi } \end{gathered}$$