The volume of a cylinder varies jointly with the square of the radius and the height. If the volume of the cylinder is \begin{align*}64 \ \text{units}^3\end{align*}

### Joint Variation

The last type of variation is called **joint variation**. This type of variation involves three variables, usually \begin{align*}x, y\end{align*} and \begin{align*}z\end{align*}. For example, in geometry, the volume of a cylinder varies jointly with the square of the radius and the height. In this equation the constant of variation is \begin{align*}\pi\end{align*}, so we have \begin{align*}V= \pi r^2h\end{align*}. In general, the joint variation equation is \begin{align*}z=kxy\end{align*}. Solving for \begin{align*}k\end{align*}, we also have \begin{align*}k=\frac{z}{xy}\end{align*}.

Let's write an equation for the following relationships.

- \begin{align*}y\end{align*} varies inversely with the square of \begin{align*}x\end{align*}.

\begin{align*}y=\frac{k}{x^2}\end{align*}

- \begin{align*}z\end{align*} varies jointly with \begin{align*}x\end{align*} and the square root of \begin{align*}y\end{align*}.

\begin{align*}z=kx \sqrt{y}\end{align*}

- \begin{align*}z\end{align*} varies directly with \begin{align*}x\end{align*} and inversely with \begin{align*}y\end{align*}.

\begin{align*}z=\frac{kx}{y}\end{align*}

Now, let's solve the following problems.

- \begin{align*}z\end{align*} varies jointly with \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. If \begin{align*}x = 3, y = 8,\end{align*} and \begin{align*}z = 6\end{align*}, find the variation equation. Then, find \begin{align*}z\end{align*} when \begin{align*}x = -2\end{align*} and \begin{align*}y = 10\end{align*}.

Using the equation when it is solved for \begin{align*}k\end{align*}, we have:

\begin{align*}k=\frac{z}{xy}=\frac{6}{3 \cdot 8}=\frac{1}{4}\end{align*}, so the equation is \begin{align*}z=\frac{1}{4}xy\end{align*}.

When \begin{align*}x = -2\end{align*} and \begin{align*}y = 10\end{align*}, then \begin{align*}z=\frac{1}{4} \cdot -2 \cdot 10=-5\end{align*}.

**Geometry Connection**The volume of a pyramid varies jointly with the area of the base and the height with a constant of variation of \begin{align*}\frac{1}{3}\end{align*}. If the volume is \begin{align*}162 \ units^3\end{align*} and the area of the base is \begin{align*}81 \ units^2\end{align*}, find the height.

Find the joint variation equation first.

\begin{align*}V=\frac{1}{3} \ Bh\end{align*}

Now, substitute in what you know to solve for the height.

\begin{align*}162&=\frac{1}{3} \cdot 81 \cdot h \\ 162&=27 \ h \\ 6&=h\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find the height of the cylinder.

The formula for the volume of a cylinder, \begin{align*}V= \pi r^2h\end{align*}, is a joint variation equation in which the constant \begin{align*}k = \pi\end{align*}.

We can therefore plug in the given values and solve for *h*, the height.

\begin{align*}V = \pi r^2h\\ 64\pi = \pi 4^2(h)\\ 64\pi = 16\pi(h)\\ 4 = h\end{align*}

Therefore, the cylinder has a height of 4 units.

#### Example 2

Write the equation for \begin{align*}z\end{align*}, that varies jointly with \begin{align*}x\end{align*} and the cube of \begin{align*}y\end{align*} and inversely with the square root of \begin{align*}w\end{align*}.

\begin{align*}z=\frac{kxy^3}{\sqrt{w}}\end{align*}

#### Example 3

\begin{align*}z\end{align*} varies jointly with \begin{align*}y\end{align*} and \begin{align*}x\end{align*}. If \begin{align*}x = 25, z = 10,\end{align*} and \begin{align*}k=\frac{1}{5}\end{align*}, find \begin{align*}y\end{align*}.

The equation would be \begin{align*}z=\frac{1}{5}xy\end{align*}. Solving for \begin{align*}y\end{align*}, we have:

\begin{align*}10&=\frac{1}{5} \cdot 25 \cdot y \\ 10&=5y \\ 2&=y\end{align*}

#### Example 4

Kinetic energy \begin{align*}P\end{align*} (the energy something possesses due to being in motion) varies jointly with the mass \begin{align*}m\end{align*} (in kilograms) of that object and the square of the velocity \begin{align*}v\end{align*} (in meters per seconds). The constant of variation is \begin{align*}\frac{1}{2}\end{align*}.

Write the equation for kinetic energy.

\begin{align*}P=\frac{1}{2} \ mv^2\end{align*}

If a car is travelling 104 km/hr and weighs 8800 kg, what is its kinetic energy?

The second portion of this problem isn’t so easy because we have to convert the km/hr into meters per second.

\begin{align*}\frac{104 \ \cancel{km}}{\cancel{hr}} \cdot \frac{\cancel{hr}}{3600 \ s} \cdot \frac{1000 \ m}{\cancel{km}}=0.44 \ \frac{m}{s}\end{align*}

Now, plug this into the equation from part a.

\begin{align*}P&=\frac{1}{2} \cdot 8800 \ kg \cdot \left(0.44 \ \frac{m}{s}\right)^2 \\ &=1955.56 \ \frac{kg \cdot m^2}{s^2}\end{align*}

Typically, the unit of measurement of kinetic energy is called a joule. A joule is \begin{align*}\frac{kg \cdot m^2}{s^2}\end{align*}.

### Review

For questions 1-5, write an equation that represents relationship between the variables.

- \begin{align*}w\end{align*} varies inversely with respect to \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
- \begin{align*}r\end{align*} varies inversely with the square of \begin{align*}q\end{align*}.
- \begin{align*}z\end{align*} varies jointly with \begin{align*}x\end{align*} and \begin{align*}y\end{align*} and inversely with \begin{align*}w\end{align*}.
- \begin{align*}a\end{align*} varies directly with \begin{align*}b\end{align*} and inversely with \begin{align*}c\end{align*} and the square root of \begin{align*}d\end{align*}.
- \begin{align*}l\end{align*} varies directly with \begin{align*}m\end{align*}, and inversely with \begin{align*}p\end{align*}.

Write the variation equation and answer the given question in each problem.

- \begin{align*}z\end{align*} varies jointly with \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. If \begin{align*}x=2,y=3\end{align*} and \begin{align*}z=4\end{align*}, write the variation equation and find \begin{align*}z\end{align*} when \begin{align*}x=-6\end{align*} and \begin{align*}y=2\end{align*}.
- \begin{align*}z\end{align*} varies jointly with \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. If \begin{align*}x=5,y=-1\end{align*} and \begin{align*}z=10\end{align*}, write the variation equation and find \begin{align*}z\end{align*} when \begin{align*}x=- \frac{1}{2}\end{align*} and \begin{align*}y=7\end{align*}.
- \begin{align*}z\end{align*} varies jointly with \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. If \begin{align*}x=7,y=3\end{align*} and \begin{align*}z=-14\end{align*}, write the variation equation and find \begin{align*}y\end{align*} when \begin{align*}z=-8\end{align*} and \begin{align*}x=3\end{align*}.
- \begin{align*}z\end{align*} varies jointly with \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. If \begin{align*}x=8,y=-3\end{align*} and \begin{align*}z=-6\end{align*}, write the variation equation and find \begin{align*}x\end{align*} when \begin{align*}z=12\end{align*} and \begin{align*}y=-16\end{align*}.
- \begin{align*}z\end{align*} varies inversely with \begin{align*}x\end{align*} and directly \begin{align*}y\end{align*}. If \begin{align*}x=4,y=48\end{align*} and \begin{align*}z=-2\end{align*}, write the variation equation and find \begin{align*}x\end{align*} when \begin{align*}z=8\end{align*} and \begin{align*}y=96\end{align*}.
- \begin{align*}z\end{align*} varies inversely with \begin{align*}x\end{align*} and directly \begin{align*}y\end{align*}. If \begin{align*}x=\frac{1}{2},y=5\end{align*} and \begin{align*}z=20\end{align*}, write the variation equation and find \begin{align*}x\end{align*} when \begin{align*}z=-4\end{align*} and \begin{align*}y=8\end{align*}.

Solve the following word problems using a variation equation.

- If 20 volunteers can wash 100 cars in 2.5 hours, find the constant of variation and find out how many cars 30 volunteers can wash in 3 hours.
- If 10 students from the environmental club can clean up trash on a 2 mile stretch of road in 1 hour, find the constant of variation and determine how low it will take to clean the same stretch of road if only 8 students show up to help.
- The work \begin{align*}W\end{align*} (in joules) done when lifting an object varies jointly with the mass \begin{align*}m\end{align*} (in kilograms) of the object and the height \begin{align*}h\end{align*} (in meters) that the object is lifted. The work done when a 100 kilogram object is lifted 1.5 meters is 1470 joules. Write an equation that relates \begin{align*}W, m,\end{align*} and \begin{align*}h\end{align*}. How much work is done when lifting a 150 kilogram object 2 meters?
- The intensity \begin{align*}I\end{align*} of a sound (in watts per square meter) varies inversely with the square of the distance \begin{align*}d\end{align*} (in meters) from the sound’s source. At a distance of 1.5 meters from the stage, the intensity of the sound at a rock concert is about 9 watts per square meter. Write an equation relating \begin{align*}I\end{align*} and \begin{align*}d\end{align*}. If you are sitting 10 meters back from the stage, what is the intensity of the sound you hear?

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 9.3.