Question

A stone dropped into a pond of still water sends out concentric circular waves from the point of disturbance of water at the rate of 4 cm/sec. Find the rate of change of disturbed area at the instant when the radius of the wave ring is 15 cm.

Answer 1

Hint: The derivative of Y with respect to X, written $\dfrac{{dy}}{{dx}}$, is just a description of how fast Y changes when X changes. It so happens that if \[Y = {X^N}\], then \[\dfrac{{dy}}{{dx}} = N{X^{N - 1}}\]. So, for example, if $Y = 5{X^3}$, then $\dfrac{{dy}}{{dx}} = 15{X^2}$. The area of a circle is $\pi {r^2}$, and the circumference is $2\pi r$, which is the derivative.
Chain rule:
To differentiate y = f(g(x)), let u = g(x). Then y = f(u) and
\[\dfrac{{dy}}{{dx}}{\text{ }} = {\text{ }}\dfrac{{dy}}{{du}}{\text{ }} \times {\text{ }}\dfrac{{du}}{{dx}}\]

Complete step-by-step answer:
Let the radius of the ring be r
So, according to question
Radius, r = 15 cm
If the radius is increasing at a constant rate of, \[\dfrac{{dr}}{{dt}} = 4cm/\sec \]
Let area be \[A = \pi {r^2}\]…………………………(1)
Differentiating the equation (1) with respect to, \[\dfrac{{dA}}{{dt}} = 2\pi r \times \dfrac{{dr}}{{dt}}\]
So, Rate of change of disturbed area = \[\dfrac{{dA}}{{dt}} = 2\pi r \times \dfrac{{dr}}{{dt}}\]
\[
   \Rightarrow \dfrac{{dA}}{{dt}} = 2\pi \left( {15} \right) \times \left( 4 \right) \\
   \Rightarrow \dfrac{{dA}}{{dt}} = 376.99c{m^2}/\sec \\
 \]
The rate of change of disturbed area is \[376.99c{m^2}/\sec \]at the instant when the radius of the wave ring is 15 cm.

Note: \[\dfrac{{dy}}{{dx}}\] is positive if y increases as x increases and is negative if y decreases as x increases. The same “derivative thing” holds up for the circumference vs. the area of a circle. The change in area, \[dA\], is $dA = (2\pi r)dR$. So,$\dfrac{{dA}}{{dR}} = 2\pi r$. That is, the derivative of the area is just the circumference.