Let A be the area of a circle with radius r. If dr/dt=3, find dA/dt when r=1.
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We know that `A=pi r^2,` and in this case the radius and thus the area are functions of `t,` so we use the chain rule and differentiate both sides with respect to `t.` The result is
`(dA)/(dt)=2pi r (dr)/(dt).`
Substituting `r=1` and `(dr)/(dt)=3` in the right side, we get
so the area is changing at the rate of `6pi` at the instant ` ``r=1`.
A = pi r^2
--> dA/dr = 2 pi r
Form the eqn dA/dt = dA/dr x dr/dt
Sub dr/dt = 3 and dA/dr = 2 pi r, where r=1,
dA/dt = 2 pi x 3 = 6 pi length^2/time
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