Asked by David
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?
Answers
Answered by
Steve
In two seconds, the area increased by π. So,
da/dt = π/2
a = π/2 r<sup>2</sup>
At t=7, a=5π
5π = π/2 r<sup>2</sup>
10 = r<sup>2</sup>
r = √10
a = π/2 r<sup>2</sup>
da = πr dr/dt
π/2 = π√10 dr/dt
dr/dt = 1/2√10
da/dt = π/2
a = π/2 r<sup>2</sup>
At t=7, a=5π
5π = π/2 r<sup>2</sup>
10 = r<sup>2</sup>
r = √10
a = π/2 r<sup>2</sup>
da = πr dr/dt
π/2 = π√10 dr/dt
dr/dt = 1/2√10
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