A stone is dropped into a quiet lake and waves move in circles at a speed of 4 cm/sec. At the instant when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing?

Asked by Aaryan | 1 year ago |  38

1 Answer

Solution :-

Given a stone is dropped into a quiet lake and waves move in circles at a speed of 4 cm/sec.

To find the instant when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing

Let r be the radius of the circle and A be the area of the circle

When stone is dropped into the lake waves moves in circle at speed of 4cm/sec. i.e., radius of the circle increases at a rate of 4cm/sec

\( \dfrac{dr}{dt}\) = 4cm/sec ...(i)

As we know that the area of the circle is πr2

\(\dfrac{dA}{dt}=2\pi r\times 4\)

Therefore, when the radius of the circular wave is 10 cm,

the above equation becomes \( \dfrac{dA}{dt}\) = 2π x 10 x 4 

\( \dfrac{dA}{dt}\)= 80 πcm2/sec

Thus, the enclosed area is increasing at the rate of 80 πcm2/sec. 

Answered by Sakshi | 1 year ago

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