The radius of a circle increases at a rate of 0.5\(cms^{-1}\). Find the rate of change in the area of the circle with radius 7cm. \([\pi = \frac{22}{7}]\)
Question 1 Report
The radius of a circle increases at a rate of 0.5\(cms^{-1}\). Find the rate of change in the area of the circle with radius 7cm. \([\pi = \frac{22}{7}]\)
Answer Details
We know that the area of a circle is given by the formula: $$A = \pi r^2$$
We need to find the rate of change of the area of the circle, i.e., \(\frac{dA}{dt}\).
We can use differentiation to find the rate of change of the area with respect to time: $$\frac{dA}{dt} = \frac{d}{dt} (\pi r^2)$$
Since the radius is increasing at a rate of 0.5\(cms^{-1}\), we have: $$\frac{dr}{dt} = 0.5$$
Using the chain rule of differentiation, we have: $$\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = 2 \pi r \cdot 0.5 = \pi r$$
Substituting the given value of the radius, we get: $$\frac{dA}{dt} = \pi \cdot 7 = 22\(cm^{2}s^{-1}\)$$
Therefore, the rate of change in the area of the circle with radius 7cm is 22\(cm^{2}s^{-1}\).
Hence, the correct option is: 22\(cm^{2}s^{-1}\).
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