A sphere of radius $2 \text { cm}$ starts expanding with its radius $r \text { cm}$ increasing at a constant rate of $3 \text { mm} / \text s$. Find the rate at which the surface area $A \text { cm}^2$ of the sphere is increasing after $10 \text { seconds}$, if the initial radius was $1 \text { cm}$.
ANS: $960 \pi \text { mm}^2 / \text s$
What I've tried:
I know that the formula for the surface area of a sphere is $4 \pi r^2$.
And I know that $\frac {dr} {dt} = 3 \text { mm} /\text {s}$.
I am asked to find $\frac {dA} {dt}$ after $10 \text { seconds}$, right?
I differentiated the formula for the surface area of the sphere and got $8 \pi r$, then replaced $r \text { cm}$ with $2$. I then used chain rule for finding $ \frac {dA} {dt} = \frac {dA} {dr} \cdot \frac {dr} {dt} = 16 \text { cm} \cdot 3 \text { mm} / \text s = 160 \text { mm} \cdot 3 \text { mm} / \text s = 480 \text { mm}^2 / \text{s}$, which is wrong !
Best Answer
$r(0) = 2\\ r(t) = r(0) + 0.3t$
3 millimiters is 0.3 centimeters.
$r(10) = 5\ cm$
$A = 4\pi r^2 cm^2\\ \frac{dA}{dt} = 4\pi(2r)\frac {dr}{dt} \frac {cm^2}{s}\\ \frac{dA}{dt} = 4\pi(10)(0.3) \frac {cm^2}{s} = 12\pi\frac {cm^2}{s}$
There is conflicting information in your post. At one point you say the initial raduis was 2 cm, and in another you say 1.
$r(0) = 1\\ r(10) = 4$
$\frac{dA}{dt} = 4\pi(8)(0.3) \frac {cm^2}{s} = 9.6\pi\frac {cm^2}{s} = 960 \frac {mm^2}{s}$