Q. 86 - Calculus

Question

$$\begin{gathered} \mathrm{If} \mathrm{you} \mathrm{drop} \mathrm{a} \mathrm{pebble} \mathrm{into} \mathrm{a} \mathrm{large} \mathrm{lake}, \mathrm{you} \mathrm{will} \mathrm{cause} \mathrm{a} \mathrm{circle} \mathrm{of} \mathrm{ripples} \mathrm{to} \mathrm{expand} \mathrm{outward}. \\ \mathrm{The} \mathrm{area} \mathrm{A} = \mathrm{A}\left(\mathrm{t}\right) \mathrm{and} \mathrm{radius} \mathrm{r} = \mathrm{r}\left(\mathrm{t}\right) \mathrm{are} \mathrm{clearly} \mathrm{functions} \mathrm{of} \mathrm{t} (\mathrm{they} \mathrm{change} \mathrm{over} \mathrm{time}) \\ \mathrm{and} \mathrm{are} \mathrm{related} \mathrm{by} \mathrm{the} \mathrm{formula} \mathrm{A} = {\mathrm{\pi r}}^2. \\ \left(\mathrm{a}\right) \mathrm{If} \mathrm{r} \mathrm{is} \mathrm{measured} \mathrm{in} \mathrm{inches} \mathrm{and} \mathrm{t} \mathrm{is} \mathrm{measured} \mathrm{in} \mathrm{seconds}, \mathrm{what} \mathrm{are} \mathrm{the} \mathrm{units} \mathrm{of} \frac{d\mathrm{A}}{d\mathrm{t}} ? \\ \mathrm{What} \mathrm{are} \mathrm{the} \mathrm{units} \mathrm{of} \frac{d\mathrm{A}}{d\mathrm{r}} ? \\ \left(\mathrm{b}\right) \mathrm{Find} \frac{d\mathrm{A}}{d\mathrm{r}} \mathrm{and} \mathrm{explain} \mathrm{the} \mathrm{practical} \mathrm{meaning} \mathrm{of} {\overline{)\frac{d\mathrm{A}}{d\mathrm{r}}}}_{\mathrm{r}=2} . \\ \left(\mathrm{c}\right) \mathrm{Find} \frac{d\mathrm{A}}{d\mathrm{t}} \mathrm{and} \mathrm{explain} \mathrm{the} \mathrm{practical} \mathrm{meaning} \mathrm{of} {\overline{)\frac{d\mathrm{A}}{d\mathrm{t}}}}_{\mathrm{r}=2} . \end{gathered}$$

Step-by-Step Solution

Answer

$$\begin{gathered} \left(\mathrm{a}\right). \mathrm{Unit} \mathrm{of} \frac{d\mathrm{A}}{d\mathrm{t} }=\mathrm{inche} \mathrm{per} \mathrm{seconds}, \mathrm{unit} \mathrm{of} \frac{d\mathrm{A}}{d\mathrm{r}}=\mathrm{inch}. \\ \left(\mathrm{b}\right). \mathrm{The} \mathrm{value} \mathrm{of} \frac{d\mathrm{A}}{d\mathrm{r}} = 2\mathrm{\pi r}. \\ \left(\mathrm{c}\right). \mathrm{The} \mathrm{value} \mathrm{of} \frac{d\mathrm{A}}{d\mathrm{t}} = 2\mathrm{\pi r}\frac{d\mathrm{r}}{d\mathrm{t}} \end{gathered}$$

Step 1. Given Information

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

Step 2. Solution (a) : The   units   of   d A d t   and   d A d r

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{equation} : \mathrm{A} = {\mathrm{\pi r}}^2 \\ \mathrm{Differentiate} \mathrm{above} \mathrm{equation} : \\ \frac{d\mathrm{A}}{d\mathrm{t}} = \frac{d }{d\mathrm{t}}\left({\mathrm{\pi r}}^2\right) = \mathrm{\pi }\frac{d }{d\mathrm{t}}\left({\mathrm{r}}^2\right) = 2\mathrm{\pi r}\frac{d\mathrm{r}}{d\mathrm{t}} \mathrm{inch} \mathrm{per} \mathrm{seconds} .... \left(1\right) \\ \mathrm{Consider} \mathrm{the} \mathrm{equation} : \mathrm{A} = {\mathrm{\pi r}}^2 \\ \mathrm{Differentiate} \mathrm{above} \mathrm{equation} \\ \frac{d\mathrm{A}}{d\mathrm{r}} = \frac{d }{d\mathrm{r}}\left({\mathrm{\pi r}}^2\right) = \mathrm{\pi } \frac{d}{\mathrm{dr}}\left({\mathrm{r}}^2\right) = 2\mathrm{\pi r} \mathrm{inch} ...... \left(2\right) \end{gathered}$$

Step 3. Solution (b) : value   of   d A d r .

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

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

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