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I'm working on this related rates problem and haven't been able to get a numerical answer. I feel like I must be missing something stupid (It's not homework.)

A cup in the form of a right circular cone with radius $r$ and height $h$ is being filled with water at the rate of 5 cu in./sec. How fast is the level of the water rising when the volume of the water is equal to one half the volume of the cup?

So far I have:

Let $v_0$ represent the volume of the water, $h_0$ represent the height of the water and $r_0$ represent the radius of the surface of the water.

$$v_0=\frac{1}{3}\pi r_0^2h_0$$ Before taking the derivative, I use similar triangles to get rid of $r_0$. $$\frac{dv_0}{dt}=\pi\left(\frac{r}{h}\right)^2h_0^2\frac{dh_0}{dt}$$ Finally, using similar triangles again: $$\frac{dv_0}{dt}=\pi r_0^2\frac{dh_0}{dt}$$

Which just leaves solving for $r_0$ at the appropriate time. I can of course find $r_0$ in terms of $r, h, \text{and } h_0$. Is there a numeric answer to this question? I feel like I'm missing some information, and I don't think I can solve for $r_0$ without knowing $r \text{ and } h$. Every other question in this chapter has a numeric solution.

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  • $\begingroup$ Now just set $\frac{1}{3}\pi r_{0}^{2}h_{0}=\frac{1}{2}(\frac{1}{3}\pi r^{2}h)$ and solve for $r_0$, and then substitute into your last equation. $\endgroup$
    – user84413
    Oct 2, 2014 at 22:42
  • $\begingroup$ As stated in the question, I am aware of this. This does not yield a numeric answer, simply the rate of change in terms of the other variables. $\endgroup$
    – user17137
    Oct 3, 2014 at 1:05
  • $\begingroup$ Sorry, I didn't realize you were well aware of how to proceed, and you're right that the answer involves r or h. $\endgroup$
    – user84413
    Oct 3, 2014 at 14:06

1 Answer 1

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It seems we should assume that the base of the cone is level, so that the cup can be filled without spilling. When the cup is half full, the surface of the water then is a disk that divides the cone into two pieces of equal volume; this disk also is parallel to the base of the cone.

The quantity $\pi r_0^2$ is just the area of the disk formed by the surface of the water.

Suppose we multiply $r$ and $h$ by two. That is, we scale up the entire cup by a factor of two. The ratio $h_0/h$ will not change, and so neither will the ratio $r_0/r.$ Since we doubled $r,$ we must also double $r_0.$ The area of the disk is multiplied by a factor of four, and since the rate of flow of the water is unchanged, the rate at which the water is rising is one quarter of what it was in the smaller cup.

So unless there are some other numeric data in the problem statement, you are not missing any information that would enable you to get a simple numeric answer. You can, however, get an answer in terms of just $r,$ without regard to $h$ or $h_0.$

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  • $\begingroup$ Thanks. I thought that this was the case, but also suspected I might be missing something as the rest have numeric solutions. $\endgroup$
    – user17137
    Oct 3, 2014 at 1:06

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