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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.

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    $\begingroup$ 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. $\endgroup$
    – KM101
    Commented Nov 11, 2019 at 23:31
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    $\begingroup$ 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 $\endgroup$ Commented Nov 11, 2019 at 23:51

1 Answer 1

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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 $(1/2)h'$.

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  • $\begingroup$ 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. $\endgroup$
    – rtpw123
    Commented Nov 12, 2019 at 14:08
  • $\begingroup$ @rtpw123 : There is no water in this cone. Somehow we have a cone whose radius and height are both changing with time. Maybe its a conical pile of beans or something.... It's likely that in a real world setting the modelled claim about the rates of change of the radius and height is only true for a second or two before and after the instant of interest. We don't actually know how far back in the past this history of this scenario actually goes. $\endgroup$ Commented Nov 12, 2019 at 14:28
  • $\begingroup$ Thanks! That helps. Sorry, I just assumed a scenario when trying to understand the problem. $\endgroup$
    – rtpw123
    Commented Nov 12, 2019 at 14:38

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