229. The area A(t) of a circular shape is growing at a constant rate. If the area increases from 4π units to 9π units between times t = 2 and t = 3, find the net change in the radius during that time.

Question

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Answer 1

A=πr², dA/dt=k where k is a constant, because rate of change of A is constant.

A=kt+C where C is a constant of integration, to be found.

4π=2k+C, 9π=3k+C, 5π=k, so k=5π (by subtracting these two equations), and 4π=10π+C, C=-6π (by substituting k=5π). We now know how A and t are related:

A=5πt-6π. But A=πr² so we can relate r and t:

πr²=5πt-6π, r²=5t-6.

When t=2, r²=4, r=2, and when t=3, r²=9, r=3 therefore r changes from 2 to 3, an increase of 1 as t goes from 2 to 3.

Another way of looking at it (without using calculus) is to look at the growth rate of A in the time interval of 3-2=1:

9π/4π=9/4. The growth rate of r is the square root of this because A is proportional to r², making r proportional to the square root of A. Therefore r changes from 2 to 3 as A changes from 4π to 9π. But this only means an increase of 50%, rather than a net increase (an actual value). But r²=A/π, so when A=4π, r²=4, r=2; when A=9π, r²=9, r=3; which tells us that r actually changes from 2 to 3 when A changes from 4π to 9π, confirming a net increase of 1.