Help me with this Calculus problem.?

Question

The radius of a circle is increasing at a constant rate of 0.2 meters per second. What is the rate of increase in the area of the circle at the instant when the circumference of the circle is 20pi meters?

Answer 1

The radius of a circle is increasing at a rate of 0.2 meters per second, so we know that

dr/dt = 0.2

Rate of increase in area:

dA/dt = ?

Next, we relate A and r; they are related by the area formula of a circle,

A = (pi)r^2

Differentiate implicitly, and we get

dA/dt = (pi)(2r)(dr/dt)
dA/dt = 2pi(r) (dr/dt)

Plug in dr/dt = 0.2,

dA/dt = 2pi(r) [0.2]
dA/dt = (0.4)pi(r)

As all related rates questions, we require our "when" statement, and here it is:
When the circumference of the circle is 20pi meters, it follows that since C = 2pi(r), 20pi = 2pi(r), or r = 10. Now, we plug in
r = 10.

dA/dt = (0.4)pi(10)
dA/dt = 4pi

This tells us that the area of the circle is increasing at a rate of 4pi meters squared.

Answer 2

Well, you're looking for 2 rates of change: radius and area. So you can just use the formula for area of a cirlce and take the derivative with respect to time.

A = pi * r^2
dA/dt = 2pi * r * dr/dt

You're solving for dA/dt at the instant when the circumference is 20pi - since the formula for the circumference of a circle is 2pi*r, you can simply substitute 20pi into the equation.

dA/dt = 2pi * r * dr/dt
dA/dt = 20pi m * 0.2 m/s
dA/dt = 4pi m^2/s

Answer 3

given dr/dt = 0.2

area A = pi r^2

dA/dt = 2 pi r dr/dt = circumference * dr/dt

with the given conditions

dA/dt = 20 pi * 0.2 = 4pi sq. metres / sec

Answer 4

dr/dt = 0.2 m/s
A = π r ²
dA/dr = 2 π r
C = 2πr
20π = 2 π r
r = 10
dA/dt = (dA/dr).(dr/dt)
dA/dt = 2πr x 0.2
dA/dt = 20π x 0.2
dA/dt = 4π m ² / s
dA/dt = 12.6 m ² / s (to 1 dec. place)