Calculus - Fluid Leaving Cone Vessel

Suppose that fluid flows out of the bottom of a cone-shaped vessel at the rate of 3 cu ft/min. If the height of the cone is 3 times the radius, how fast is the height of the fluid decreasing when the fluid is 6 inches deep in the middle?

I'm unsure whether my approach is correct and whether my answer is right.

My approach to solving this was the following:

Since we're trying to find how fast the height of the cone decreases, we need to find $$\frac{dh}{dt}$$. To do this, we can use the chain rule.

$$\frac{dh}{dt} = \frac{dh}{dr} \frac{dr}{dt}$$

Where $$r$$ is the radius and $$t$$ is the time. Since the height is 3 times the radius, $$\frac{dh}{dr} = 3$$. The derivative of the radius with respect to time can be solved by finding:

$$\frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt}$$

$$\frac{dV}{dt} = 3$$ and $$\frac{dV}{dr} = 3\pi r^2$$ $$\therefore \frac{dr}{dt} = \frac{1}{\pi r^2}$$

Plugging in, $$\frac{dh}{dt} = 3 \biggl(\frac{1}{\pi r^2}\biggl) = \frac{3}{\pi r^2}$$

So, if the height is 6 inches (0.5 feet) deep in the middle, we know the radius is $$\frac{1}{6}$$ feet or 2 inches. Therefore, the height decreases at a rate of $$\frac{3}{\pi * 2^2} = \frac{3}{4 \pi}$$ in / min.

Just to clarify, I would like to know whether this is the right answer and, if not, then whether my approach was correct or not. The book I got this problem out of is called Calculus by Morris Kline.

Thanks for the help.

• The height is three times the radius; you have the radius is three times the height... – DJohnM Dec 29 '18 at 23:55
• Thanks, I'll be sure to fix that. – S. Sharma Dec 29 '18 at 23:57
• Not only that, you're mixing cu ft with inches. dh/dt will be $\frac{108}{\pi}$ ft/min. – Phil H Dec 30 '18 at 2:21

Well, let's see . . .

If I recall correctly, the volume $$V$$ of a cone of height $$h$$ and radius $$r$$ is given by

$$V = \dfrac{\pi r^2 h}{3}; \tag 1$$

and here we have

$$h = 3r, \tag 2$$

that is,

$$r = \dfrac{h}{3}; \tag 3$$

then, in terms of $$h$$,

$$V = \dfrac{\pi h (h/3)^2}{3} = \dfrac{\pi h^3}{27}; \tag 4$$

now if the volume of the cone of fluid changes in such a way as to maintain the proportionality (2), (3) 'twixt $$r$$ and $$h$$, we have, by the chain rule,

$$\dfrac{dV}{dt} = \dfrac{\pi h^2}{9} \dfrac{dh}{dt}, \tag 5$$

and we merely need to plug in the numbers, taking care that the units are consistent: we have

$$V = 3 \; \text{cu. ft./min.}; \; h = 6 \; \text{in.}; \tag 6$$

now let's see,

$$1 \; \text{cu. ft.} = (12)^3 \; \text{cu. in.} = 1728 \; \text{cu. in.}; \tag 7$$

$$3 \; \text{cu. ft.} = 3 \times (12)^3 \; \text{cu. in.} = 3 \times 1728 \; \text{cu. in.} = 5184 \; \text{cu. in.}; \tag 8$$

from (5),

$$\dfrac{dh}{dt} = \dfrac{9}{\pi h^2} \dfrac{dV}{dt}, \tag 9$$

and plugging in the numbers

$$\dfrac{dh}{dt} = \dfrac{9}{\pi(6 \; \text{in.})^2} (5184 \; \text{cu. in./min.}) \cong \dfrac{1296}{\pi} \; \text{in./min.} \cong412.74 \; \text{in./min.}; \tag{10}$$

that's a pretty rapid decrease of depth; indeed, at that rate $$h = 0$$ in about

$$\Delta t \cong \dfrac{6}{412.74} \; \text{min.} \cong .015 \; \text{min.} \cong .90 \; \text{sec.} \tag{11}$$

The basic approach to such problems, using the chain rule and geometrical information to relate the rates of different variables, is correct and indeed often done.

• Your answer is correct. An easy check for this kind of problem is to simply divide the flow rate ($3$) by the surface area at the specified depth ($1/2$). This is $\frac{3}{\pi r^2} = \frac{3}{\frac{\pi}{36}} = \frac{108}{\pi} = 34.3775$ ft/min or $412.53$ in/min. This is indeed a rapid rate but consider that the volume of fluid at this depth is only $.01454$ cu ft and hence, emptying at a rate of $3$ cu ft/min, does so in $0.29$ sec. The rate at which h decreases, increases over time as the cone empties. – Phil H Dec 30 '18 at 16:32