Question

A stone is dropped into a quiet lake and circular ripples are formed. Circular wavefronts move at the speed of radius increasing at the rate of 5 cm/sec. How fast is the area increasing when the radius is 10cm?

Answer 1

Hint: Area of circle, \[A=\pi {r^2}\] where radius of the circle is $r$.
Since we have to find the rate of the increasing area so, we have to find the derivate of $A$ with respect to $t$.
Here the radius of the circle and the rate of change of increasing radius are given.
So we can differentiate $A$ with respect to $t$ and put the given values to get the rate of the increasing area.

Complete step by step solution:
Let “$r$” be the radius of the circle & A be the area of the circle.
When a stone is dropped into the lake, wave motion in a circle at the speed of \[5cm/sec\].
Thus,
 \[\dfrac{{dr}}{{dt}} = 5cm/\sec \] … (1)
We need to find how fast the area increasing with respect to the time when the radius is $10cm$.
That is we need to find \[\dfrac{{dA}}{{dt}}\] where $r=10cm$.
We know that,
Area of circle =\[\pi {r^2}\]
That \[A = \pi {r^2}\]
Let us differentiate both sides of the area equation with respect to time we get,
\[\dfrac{{dA}}{{dt}} = \dfrac{{d(\pi {r^2})}}{{dt}}\]
Since \[\pi \]is a constant we can write the above equation as,
\[\dfrac{{dA}}{{dt}} = \pi \dfrac{{d({r^2})}}{{dt}}\]
Let us multiply and divide\[dr\] by on the right hand side,
\[\dfrac{{dA}}{{dt}} = \pi \dfrac{{d({r^2})}}{{dt}} \times \dfrac{{dr}}{{dr}}\]
\[\dfrac{{dA}}{{dt}} = \pi \dfrac{{d({r^2})}}{{dr}} \times \dfrac{{dr}}{{dt}}\]
On differentiating the equations in the above term we get,
\[\dfrac{{dA}}{{dt}} = \pi \times 2r \times \dfrac{{dr}}{{dt}}\]
\[\dfrac{{dA}}{{dt}} = 2\pi r\dfrac{{dr}}{{dt}}\]
Let us put the value of r and \[\dfrac{{dr}}{{dt}}\]we get,
\[\dfrac{{dA}}{{dt}} = \pi \times 2 \times 10 \times 5\]
\[\dfrac{{dA}}{{dt}} = 100\pi \]
Let us substitute the value of \[\pi \], we get,
\[\dfrac{{dA}}{{dt}} = 100 \times \dfrac{{22}}{7} = 314.28\]
That is \[\dfrac{{dA}}{{dt}}\left| {_{r = 10}} \right. = 314.28\]
Since the area is in \[c{m^2}\] and time is in sec we get,
\[\dfrac{{dA}}{{dt}}\left| {_{r = 10}} \right. = 314.28\] \[c{m^2}/sec\]

$\therefore$ The area increasing with respect to time at a rate of \[314.28c{m^2}/sec\] when the radius is \[10cm\].

Note:
In the question it is given that we should find how fast the area is increasing, it means that we should find the rate of increase which is nothing but the differentiation of area with respect to the time. While differentiating the area we multiply and divide $dr$ by on the right-hand side which is the most important step in finding the required result. Since it is said to find the rate at a certain centimeter we should finally substitute the given value in the found the area.