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The radius of a circle is changing at the rate of 1/π inches per second. At what rate, in square inches per second, is the circle’s area changing when the radius is 5 inches?

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Answer:

10 square inches per second.

Explanation:

The radius of the circle is given by the equation:

r(t) = (1/π in/s)*t

Where time in seconds.

Remember that the area of a circle of radius R is written as:

A = π*R^2

Then the area of our circle will be:

A(t) = π*( (1/π in/s)*t)^2 = π*(1/π in/s)^2*(t)^2

Now we want to find the rate of change (the first derivation of the area) when the radius is equal to 5 inches.

Then the first thing we need to do is find the value of t such that the radius is equal to 5 inches.

r(t) = 5 in = (1/ in/s)*t

5in*(π s/in) = t

5*π s = t

So the radius will be equal to 5 inches after 5*π seconds, let's remember that.

Now let's find the first derivate of A(t)

dA(t)/dt = A'(t) = 2*(π*(1/π in/s)^2*t = (2*π*t)*(1/π in/s)^2

Now we need to evaluate this in the time such that the radius is equal to 5 inches, we will get:

A'(5*π s) = (2*π*5*π s)*((1/π in/s)^2

= (10*π^2 s)*(1/π^2 in^2/s^2) = 10 in^2/s

The rate of change is 10 square inches per second.

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