How Fast Does Water Rise Up a Cone?

share calculus img.png

Rate-of-change of height of water in a rectangular container

Let's say that we have a rectangular box whose

 
Calculus thumbnail.png

width and depth are \(2×2\) square feet and whose height is \(h\). Let's say that water is being poured into the container at a constant rate of \(4ft^3/minute\); if we let \(V\) denote the volume of the water inside the container, and if \(4ft^3/minute\) is how much more water we keep adding to the container every minute, then the amount \(4ft^3/minute\) must be the rate-of-change of the volume of water inside of the container. Or, written mathematically,

$$\frac{dV}{dt}=4\frac{ft^3}{minute}.$$

 

 Figure 1: Water being poured into a rectangular box with dimensions \(4ft×4ft×h\) at a rate of \(\frac{dV}{dt}=4\frac{ft^3}{min}\). The container at time \(t=0\) starts out empty.

Figure 1: Water being poured into a rectangular box with dimensions \(4ft×4ft×h\) at a rate of \(\frac{dV}{dt}=4\frac{ft^3}{min}\). The container at time \(t=0\) starts out empty.

If we let the variable \(y\) denote the height of the water at time \(t\), then what is \(dy/dt\), the rate-of-change of the height of the water with time? Well, if the container starts out empty (\(V=0\)), after one minute we will have poured \(4\) cubic feet of water inside of the container. Since the width and depth of the container \(2×2ft^2\), then the height of the \(4\) cubic feet of water must be \(1\) foot. In other words, after one minute the height of the water changed by one foot. After another minute goes by, an additional \(4\) cubic feet of water will get added inside of the container so that after a total of two minutes there will be a total volume of \(8\) cubic feet of water inside of the container. Since the dimensions of the base of the container are \(2×2ft^2\), the dimensions of the entire volume of water must be \(2×2×2ft^3\) where the height of the water is \(2\) feet. Thus, after an additional minute went by, the height \(y\) of the water changed by another foot. This means that the change in height of the water with respect to time must be \(1ft/minute\). Or, written mathematically,

$$\frac{dy}{dt}=1\frac{ft}{minute}.$$

In this example, it was easy to determine \(dy/dt\) given the following two initial conditions: \(dV/dt\) and the dimensions of the container. No calculus was needed since the quantity \(dy/dt\) is a constant that does not change with time.


Rate-of-change of height of water in a conical container

"More free lessons at: http://www.khanacademy.org/video?v=Xe6YlrCgkIo"

Figure 2: Water is poured into a cone of height \(h\) and radius \(r\) at a rate of \(\frac{dV}{dt}=4\frac{ft^3}{min}. The height of the water is given by \(y\). This image was modified under a Create Commons license. Image credit: https://www.mathalino.com/blog/romel-verterra/calculator-technique-solving-volume-flow-rate-problems-calculus

But let's suppose, instead, that water was being poured into a cone of height \(h\) and radius \(r=h/3\) at a constant rate of \(dV/dt=4ft^3/minute\) and I asked you: find the rate-of-change of the height \(y\) of the volume of water with respect to time. This problem isn't as straightforward as the previous problem because the quantity \(dy/dt\) is clearly not a constant and is different at different heights. For example, if \(4\) cubic feet of water is being poured inside of the cone at the bottom of the cone, then the height of the water will change more rapidly than if it were poured at the top. To find \(dy/dt\), we must use calculus—and, in particular, differentiation using the chain rule.

The volume of a cone is given by the formula

$$V=\frac{1}{3}πr^2h,$$

where \(V\) is a function of \(r\) which we'll represent by writing \(V(r)\). Since \(dV/dt=4ft^3/minute\), we can write

$$\frac{d}{dt}V(r)=\frac{d}{dt}\biggl(\frac{1}{3}πr^2h\biggr).$$

Substituting \(r=h/3\), we have

$$\frac{d}{dt}V(r)=\frac{d}{dt}(\frac{1}{27}πh^3).\tag{1}$$
As you can see from Equation (1), if we had never learned about th echain rule, then the derivative in Equation (1) would be an insurmountable problem since we would have to evaluate \(dV/dt\) but \(V\) cannot be expressed as a function o ftime. However, according to the chain rule, we can write

$$\frac{dV}{dt}=\frac{dV}{dh}\frac{dh}{dt}\tag{2}$$

where \(dh/dt\) represents the rate-of-change of the height \(h\) of the water inside of the cone. From Equations (1) and (2), we can write

$$4\frac{ft^3}{minute}=\frac{d}{dh}\biggl(\frac{π}{27}h^3\biggr)\frac{dh}{dt},$$

or

$$4\frac{ft^3}{minute}=\frac{3π}{27}h^2\frac{dh}{dt},$$

Using alebra, we find that \(dh/dt\) is given by

$$\frac{dh}{dt}=\frac{108}{3πh^2}\frac{ft}{minute}.\tag{3}$$

As we can see from Equation (3), the rate-of-change of the height of water depends upon \(h\) or, in other words, at what height the water is being poured in at. If the height \(h\) of the water inside of the cone is known, then Equation (3) can be used to determine—right at that exact moment in time—how quickly the height of the water is changing with time.


This article is licensed under a CC BY-NC-SA 4.0 license.

Sources: Khan Academy & Mathalino