Question

If a circular iron sheet of radius 30 cm is heated such that its area increases at the uniform rate of $6\pi $$c{m^2}$/hr , then the rate (in mm/hr) at which the radius of the circular sheet increases is.
A. 2.0
B. 1.0
C. 1.1
D. 0.1

Answer 1

Hint: This problem deals with the concepts in applications of derivatives. Here the derivatives are applied in order to determine the rate of change of quantities. Given that a circular iron sheet is heated. Initial radius of the sheet is given i.e, before heating. In the duration of heating the circular sheet expands, the rate of the increase in the area of the circular sheet is given. We have to find the rate of increase of the radius. The area of the circle formula is used here.
Here the area of the circle of radius $r$ is given by:
$ \Rightarrow A = \pi {r^2}$

Complete step-by-step solution:
Given that initially there is a circular iron sheet with radius 30 cm.
Given the radius of the iron sheet, $r = 30$.
The circular iron sheet is heated and hence its area is increased at a uniform rate of $6\pi $$c{m^2}$/hr.
Rate of change of the area of the circular iron sheet is given by:
$ \Rightarrow \dfrac{{dA}}{{dt}} = 6\pi $$c{m^2}$/hr.
We have to find the rate at which the radius of the circular sheet increases.
The area of the circle is given by:
$ \Rightarrow A = \pi {r^2}$
Now differentiate the equation on both sides with respect to t, as given below:
$ \Rightarrow \dfrac{{dA}}{{dt}} = \pi \left( {2r} \right)\dfrac{{dr}}{{dt}}$
As $\pi $is a constant, here area and the radius are changing.
$ \Rightarrow \dfrac{{dA}}{{dt}} = 2\pi r\dfrac{{dr}}{{dt}}$
Here given that the change in the area is given by $\dfrac{{dA}}{{dt}} = 6\pi $$c{m^2}$/hr, and radius $r = 30$, substituting these expressions in the above equation as given below:
$ \Rightarrow 6\pi = 2\pi \left( {30} \right)\dfrac{{dr}}{{dt}}$
$ \Rightarrow 6\pi = 60\pi \dfrac{{dr}}{{dt}}$
$ \Rightarrow \dfrac{{dr}}{{dt}} = \dfrac{1}{{10}}$
$\therefore \dfrac{{dr}}{{dt}} = \dfrac{1}{{10}}cm$/hr.
But we have to find the rate of change of the radius in mm/hr.
We know that 1cm = 10 mm.
Hence substituting the cm to mm conversion in the rate of change of radius expression as given below;
$ \Rightarrow \dfrac{{dr}}{{dt}} = \dfrac{1}{{10}} \times 10$mm/hr.
$\therefore \dfrac{{dr}}{{dt}} = 1$ mm/hr.
The rate (in mm/hr) at which the radius of the circular sheet increases is 1.

Option B is the correct answer.

Note: Please note that we are asked to find the rate of change of radius in mm/hr units, hence we converted the cm to mm to get the correct answer, else if asked in cm/hr then there is no need to change or convert the unit of measurements.
If given the rate of change in perimeter, we should be able to do it, by differentiating the perimeter equation, the perimeter of a circle is given by $2\pi r$.