I believe I've encountered an error in the problem set of Larson Calculus 11e. I have contacted the publisher about this, and they insist that both the textbook and answer key are correct. However, their reasoning seems flawed.
I want to bounce it off some truly knowledgeable people to make sure my own reasoning isn't flawed.
The problem reads, verbatim:
The volume of oil in a cylindrical container is increasing at a rate of 150 cubic inches per second. The height of the cylinder is approximately ten times the radius. At what rate is the height of the oil changing when the oil is 35 inches high?
On the surface, this reads like a classic related rates problem, and the solutions manual solves it like one. However, doesn't $\frac{r}{h}=\frac{1}{10}$ only hold for the cylindrical container itself? When we use this 1:10 relationship, we can relate the variables as $V=\pi\frac{h^3}{100}$, but again, doesn't this only hold for the container itself? As the height of oil in the container changes, the radius remains constant, thus we can't rely on this equation to differentiate and obtain $\frac{dh}{dt}$ for the volume of oil.
I think this problem would be solvable if the container were conical instead of cylindrical, because then the proportion relating the radius and height would apply to both the container and the changing volume oil at any height.
Have I got it twisted?
You won't be able to solve this problem without the radius of the cylinder.
To start: you're right that $h=10r$ only applies to the container. This makes sense when you think about it: the height of the oil is constantly changing while its radius remains constant.
Now, normally a solution would proceed like this:
We know that $V_{oil}=\pi r^2h$, and that $\frac{dV_{oil}}{dt} = 150 \frac{in^3}{s}$. We shall thus take the derivative of the volume with respect to time:
$$\frac{dV_{oil}}{dt}=\pi(2r\frac{dr}{dt}+r^2\frac{dh}{dt})$$
Since the radius is constant, $\frac{dr}{dt}=0$, making the equation thus
$$\frac{dV_{oil}}{dt}=\pi(r^2\frac{dh}{dt})$$
Substituting for $\frac{dV_{oil}}{dt}$:
$$150 \frac{in^3}{s}=\pi(r^2\frac{dh}{dt})$$
$$\frac{dh}{dt}=\frac{150 \frac{in^3}{s}}{\pi(r^2)}$$
This is all we can do. Any further substitution by your textbook of $r=\frac{h}{10}$ is wrong.