Question

The radius of a circle is increasing at a rate of $0.7cm/\sec $. What is the rate of increase of its circumference and also the rate of increase of its area where $r=10cm$?

Answer 1

Hint: To solve this question we need to know the concept of application on derivatives as rate change application. To solve, we will be differentiating the circumference of the circle and the area of the circle with respect to time individually and then substituting the value on the expression which we get.

Complete step by step answer:
The question asks us to find the rate of increase of the circumference and area of the circle having radius as $10cm$ and rate of increase in the radius of the circle is given as $0.7cm/\sec $. Firstly we would be solving for the rate of increase in the circumference of the circle having rate of increase of the circle as $0.7cm/\sec $. The rate of change in the radius in mathematical form is denoted as:
$\Rightarrow \dfrac{dr}{dt}=0.7cm/\sec $
Now the formula of circumference will be written which is:
$\Rightarrow C=2\pi r$
Now we will be differentiating the circumference with respect to time, which means the rate of circumference will be found. So on differentiating we get:
$\Rightarrow \dfrac{dC}{dt}=2\pi \dfrac{dr}{dt}$
On substituting the values on the above expression we get:
$\Rightarrow \dfrac{dC}{dt}=2\pi \left( 0.7 \right)$
Putting the value of $\pi $ we get:
$\Rightarrow \dfrac{dC}{dt}=2\times \dfrac{22}{7}\left( 0.7 \right)$
On further calculation we get:
$\Rightarrow \dfrac{dC}{dt}=2\times 22\times 0.1$
$\Rightarrow \dfrac{dC}{dt}=4.4cm/\sec $
So the rate of increase in the circumference is $4.4cm/\sec $.
Secondly we would be solving for the rate of increase in the area of the circle having rate of increase of the circle as $0.7cm/\sec $. The rate of change in the radius in mathematical form is denoted as:
$\Rightarrow \dfrac{dr}{dt}=0.7cm/\sec $
Now the formula of area will be written which is:
$\Rightarrow A=\pi {{r}^{2}}$
Now we will be differentiating the area with respect to time, which means the rate of area will be found. The formula used for the differentiation is, if a function ${{x}^{n}}$ is differentiated with respect then the result is$n{{x}^{n-1}}$. So on differentiating we get:
$\Rightarrow \dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$
On substituting the values on the above expression we get:
$\Rightarrow \dfrac{dA}{dt}=2\pi \left( 10 \right)\left( 0.7 \right)$
Putting the value of $\pi $ we get:
$\Rightarrow \dfrac{dA}{dt}=2\times \dfrac{22}{7}\times 10\left( 0.7 \right)$
On further calculation we get:
$\Rightarrow \dfrac{dA}{dt}=2\times 22\times 10\times 0.1$
$\Rightarrow \dfrac{dA}{dt}=44c{{m}^{2}}/\sec $
So the rate of increase in the circumference is $4.4c{{m}^{2}}/\sec $.
$\therefore $ The rate of increase of its circumference and also the rate of increase of its area are $4.4cm/\sec $ and $44c{{m}^{2}}/\sec $ respectively.

Note: To solve the problem we need to know the formula for the differentiation. Do not forget to write the units after the calculation. In case if the rate of change of a function is a negative integer this means the function is decreasing with increase in time. In the question given above the rate of change of the radius is a positive integer which infer that the radius is increasing with time.