# Why do these two approaches to a related rates problem produce a different result?

The radius $$r$$ and height $$h$$ of a circular cone change at a rate of 2 cm/s. How fast is the volume of the cone increasing when $$r = 10$$ and $$h = 20$$?

Correct Approach

$$V = \frac{1}{3}\pi r^2h$$

So, $$\frac{dV}{dt} = \frac{2}{3}\pi rh\frac{dr}{dt} + \frac{1}{3}\pi r^2\frac{dh}{dt}$$

Now, substituting our known values:

$$\frac{dV}{dt} = \frac{2}{3}\pi (10)(20)(2) + \frac{1}{3}\pi (10)^2(2)$$

$$\frac{dV}{dt} = \frac{1000}{3}\pi$$

Incorrect Approach

The book introduces conical tank problems by relating $$r$$ and $$h$$ using similar triangles. I will show a similar approach below that yields the incorrect answer:

$$\frac{r}{h} = \frac{1}{2} \Rightarrow r = \frac{1}{2}h$$

So, $$V = \frac{1}{3}\pi r^2h \Rightarrow V = \frac{1}{3} \pi \frac{1}{4}h^3$$

Then,

$$\frac{dV}{dt} = \frac{1}{4}\pi h^2 \frac{dh}{dt}$$

Substituting for $$h$$ and $$\frac{dh}{dt}$$:

$$\frac{dV}{dt} = \frac{1}{4}\pi (20)^2 (2) = 200\pi$$

Note that if we substitute for $$h$$ instead of $$r$$, we also get a different answer.

Why?

Where I am getting stuck is explaining why the second approach yields an incorrect answer. I know it has something to do with the fact with $$\frac{dr}{dt}=\frac{dh}{dt}$$, but I am stuck figuring out a way to describe exactly why it is not working.

Note: The approach I labeled as incorrect works for a problem like this: "Water pours into a conical tank of height 10 m and radius 4 m at a rate of 6$$m^3$$/min. At what rate is the water level rising when the level is 5 m high?" If we substitute $$0.4h$$ for $$r$$ in the volume equation, we can differentiate $$V = \frac{1}{3}\pi h(0.4h)^2$$ find that the $$\frac{dh}{dt}$$ is about 0.48m/min.

• In the case of water pouring in a conical tank, the slope (or simply the ratio between $h$ and $R$) remains a constant. This means you can express $V$ in terms of only one variable since $\frac{h}{R} = \text{constant}$. Does the slope remain constant here? For example, let $h_0 = 5$ and $R_0 = 3$. If both increase by $2$ units each second, you get $h_1 = 7$ and $R_1 = 5$. The ratio changed, so you can't use similar triangles to simplify $V$ in terms of only one variable. Commented Nov 11, 2019 at 23:31
• KM101 explains it well. In this case, the cone is not expanding in the "normal" way, which would make dh/dt twice dr/dt. so, you cannot operate on the variables together Commented Nov 11, 2019 at 23:51

Others are saying this in a way that is perhaps less than crystal clear. You have that at the instant of interest, $$r(t) = 10$$ and $$h(t) = 20$$. You are also told $$\frac{\mathrm{d}}{\mathrm{d}t} r(t) = \frac{\mathrm{d}}{\mathrm{d}t} h(t) = 2 \,\frac{\mathrm{cm}}{\mathrm{s}}$$. Suppose we label the time of the instant of interest $$t = 0$$. Then $$r(t) = (10 + 2t) \,\mathrm{cm}$$ and $$h(t) = (20 + 2t) \,\mathrm{cm}$$. This means the ratio you intended to write is $$\frac{r(t)}{h(t)} = \frac{10 + 2t}{20 + 2t} \text{,}$$ which is not constantly $$1/2$$. It's $$0$$ when $$t = -5 \,\mathrm{s}$$ and undefined when $$t = -10 \,\mathrm{s}$$. In the limit as $$t \rightarrow \infty$$, this ratio approaches $$1$$. From this, $$r'(t) = \frac{5h(t) + (t^2 + 15t + 50)h'(t)}{(t+10)^2} \text{,}$$ which is almost never $$(1/2)h'$$.
• Thanks! This helps a lot. I am still confused about the shape of the cone, and here is a question I left on another comment: Do we assume there's already water in the tank when it starts to be filled? If it always increases at 2cm/s for $h$ and $r$, then 5 seconds before $r=10$ and $h=20$, the radius would be 0 and the height 10? That part still doesn't make sense, but maybe I'm thinking about it in the wrong way again. Commented Nov 12, 2019 at 14:08