Posted by **David** on Thursday, October 20, 2011 at 1:16pm.

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?

- calculus -
**Steve**, Thursday, October 20, 2011 at 3:30pm
In two seconds, the area increased by π. So,

da/dt = π/2

a = π/2 r^{2}

At t=7, a=5π

5π = π/2 r^{2}

10 = r^{2}

r = √10

a = π/2 r^{2}

da = πr dr/dt

π/2 = π√10 dr/dt

dr/dt = 1/2√10

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