Question

The radius of a circle is increasing at a nonzero rate, and at a certain instant, the rate of increase in the area of the circle is numerically equal to the rate of increase in its circumference. At this instant, the radius of the circle is numerically equal to half the rate of increase in its circumference. At this instant, what is the rate of change of the radius?
A. 1/2pi
B. 2
C. pi
D. 1/pi

Answer 1

t = time

Area:

A = r² π

dA / dt = 2 r π dr / dt

Circumference:

C = 2 r π

dC / dt = 2 π dr / dt

The rate of increase in the area of the circle is numerically equal to the rate of increase in its circumference.

dA / dt = dC / dt

2 r π dr / dt = 2 π dr / dt

Divide both sides by ( 2 π dr / dt )

r = 1

At this instant, the radius of the circle is numerically equal to half the rate of increase in its circumference.

r = ( dC / dt ) / 2

1 = ( dC / dt ) / 2

Multiply both sides by 2

2 = dC / dt

dC / dt = 2

Put this value in equation:

dC / dt = 2 π dr / dt

2 = 2 π dr / dt

Divide both sides by 2 π

2 / 2 π = dr / dt

1 / π = dr / dt

dr / dt = 1 / π

Answer 2

it is just 1