area of a semicircle is increasing at a constant rate. At 5 secs the area is 4pi. At 7 secs the area is 5pi. In terms of r at what rate is the radius changing?

In two seconds, the area increased by π. So,

da/dt = π/2

a = π/2 r2

At t=7, a=5π
5π = π/2 r2
10 = r2
r = √10

a = π/2 r2
da = πr dr/dt
π/2 = π√10 dr/dt
dr/dt = 1/2√10

To find the rate at which the radius is changing, we need to use the relationship between the area and the radius of a semicircle. The formula for the area of a semicircle is A = (π * r^2) / 2, where A is the area and r is the radius.

Given that the area of the semicircle is increasing at a constant rate, we can express this relationship with the derivative. Let's call the rate at which the area is changing as A'. The derivative of the area with respect to time (t) is given as dA/dt = A'.

Now, we are given the following information:
At 5 secs (t = 5), the area is 4π: A = 4π
At 7 secs (t = 7), the area is 5π: A = 5π

We can now find the derivative of the area with respect to time by taking the difference between the two areas and dividing it by the difference in time:

(A2 - A1) / (t2 - t1) = A' [where A2 = 5π, A1 = 4π, t2 = 7, t1 = 5]

(5π - 4π) / (7 - 5) = A'

Simplifying this expression, we have:

π / 2 = A'

So, the rate at which the area is changing is π / 2.

To find the rate at which the radius is changing, we can differentiate the area formula with respect to r:

dA/dr = (2 * π * r) / 2 = πr

Since we have the rate at which the area is changing (A' = π / 2), we can substitute this value into the equation:

A' = dA/dr = πr

π / 2 = πr

Simplifying this equation, we find:

r = 1/2

Therefore, the radius is changing at a rate of 1/2 units per second.