Rate Of Change Surface Area of Sphere with expanding radius O level problem.

Question

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 !

Answer 1

$r(0) = 2\\ r(t) = r(0) + 0.3t$

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.

$r(0) = 1\\ r(10) = 4$

$\frac{dA}{dt} = 4\pi(8)(0.3) \frac {cm^2}{s} = 9.6\pi\frac {cm^2}{s} = 960 \frac {mm^2}{s}$

Answer 2

The problem statement seems to be faulty. You might want to check the original source. (It is possible the mistake is already there in the copy you read--we've seen errors in test-review books here before!)

A sphere of radius $2 \text { cm}$ starts expanding ...

The usual interpretation of this is that the radius is not constant throughout the problem, but that the radius at the beginning of the events described in the problem (the "initial" radius) is $2.$

... if the initial volume was $1 \text { cm}$.

The issue here is that $2 \neq 1,$ that is, the statement about the "initial radius" at the end of the problem statement seems to contradict the statement about the initial radius at the beginning of the problem statement.

The steps of your solution are generally correct, except for the value of $r$ (for which I cannot find anything in the problem to suggest that $r = 2$ after $10$ seconds have passed). But if we suppose that the words "radius $2 \text{ cm}$" were not meant to be included in the problem statement, and instead take $1\text{ cm}$ as the initial radius, with $r$ increasing at the rate of $0.3\text{ cm/s}$ from $t=0$ to $t=10,$ and use the value of $r$ at $t=10$ according to that interpretation, then we get the intended answer.