If the circumference of the circle is changing at the rate of 5 cm/s then what will be rate of change of area of the circle if the radius is 6cm

The correct option is 30 cm²/s.

To find the rate of change of the area of the circle, we can use the relationships between the radius, circumference, and area of a circle with respect to time.

Let $r$ be the radius, $C$ be the circumference, and $A$ be the area of the circle at any time $t$.

We are given the following:
1. The rate of change of the circumference,
$$\frac{dC}{dt}=5\text{ cm/s}$$
2. The radius of the circle,
$$r=6\text{ cm}$$

First, let's write the formula for the circumference of a circle:
$$C=2\pi r$$

Differentiating both sides of the circumference formula with respect to time $t$, we get:
$$\frac{dC}{dt}=2\pi \frac{dr}{dt}$$

Substitute the given value of $\frac{dC}{dt}=5$ into the equation to find the rate of change of the radius:
$$5=2\pi \frac{dr}{dt}$$
Solving for $\frac{dr}{dt}$ gives:
$$\frac{dr}{dt}=\frac{5}{2\pi }$$

Next, let's write the formula for the area of a circle:
$$A=\pi r^2$$

Differentiating both sides of the area formula with respect to time $t$ using the chain rule, we obtain:
$$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$$

Now, substitute the values of $r=6$ and $\frac{dr}{dt}=\frac{5}{2\pi }$ into the derivative of the area:
$$\frac{dA}{dt}=2\pi (6)\left(\frac{5}{2\pi }\right)$$

Simplifying the expression by cancelling out $2\pi$ in the numerator and denominator:
$$\frac{dA}{dt}=6\times 5=30{\text{ cm}}^2\text{/s}$$

Thus, the rate of change of the area of the circle when the radius is 6 cm is 30 cm²/s.