# Concept of Joint and Combined Variation

Often the value of a variable quantity depends on the values of several other quantities, for instance the amount of simple interest on an investment depends on the interest rate, the amount invested, and the period of time involved. For compound interest the amount of interest depends on an additional variable: how often the compounding takes place. A situation in which one variable depends on several others is called combined variation. An important type of **combined variation** is defined below.

__Joint Variation__

If a variable quantity $$y$$ is proportional to the product of two or more variable quantities we say that $$y$$ **is jointly proportional** to these quantities, or $$y$$ **varies jointly as** (or with) these quantities.

For instance, if $$y$$ is jointly proportional to $$u$$ and $$v$$, then $$y$$ is related to $$u$$ and $$v$$ by the formula

\[y = kuv\]

where $$k$$ is a constant. Sometimes the word “jointly” is omitted and we simply say that $$y$$ is proportional to $$u$$ and $$v$$.

__Example__:

In chemistry, the absolute temperature $$T$$ of a perfect gas varies jointly as its pressure $$P$$ and its volume $$V$$. Given that $$T = 500$$ Kelvin when $$P = 50$$ pounds per square inch and $$V = 100$$ cubic inches, find a formula for $$T$$ in terms of $$P$$ and $$V$$ find $$T$$ when $$P = 100$$ pounds per square inch and $$V = 75$$ cubic inches.

__Solution__:

Since $$T$$ varies jointly as $$P$$ and $$V$$, there is a constant $$k$$ such that $$T = kPV$$. Putting $$T = 500$$, $$P = 50$$ and $$V = 100$$, we find that $$500 = k(50)(100)$$ or $$k = \frac{{500}}{{50(100)}} = \frac{1}{{10}}$$

Thus, the desired formula is

\[T = \frac{1}{{10}}PV\]

When $$P = 100$$ and $$V = 75$$, we have $$T = \frac{1}{{10}}(100)(75) = 750$$ Kelvin.

We sometimes have joint variation together with inverse variation. Perhaps the most important historical discovery of this type of combined variation is *Newton’s law of universal gravitation*, which states that the gravitation force of attraction $$F$$ between two particles is jointly proportional to their masses $$m$$ and $$M$$, and inversely proportional to the square of the distance $$d$$ between them. In other words, $$F$$ is related to $$m$$, $$M$$ and $$d$$ by the formula

\[F = G\frac{{mM}}{{{d^2}}}\]

where $$G$$ is constant of proportionality.

__Example__:

Careful measurements show that two $$1 – $$ kilogram masses $$1$$ meter apart exact a mutual gravitational attraction of $$6.67 \times {10^{ – 11}}$$ Newton (one pound of force is approximately $$4.45$$ Newton). The Earth has a mass of $$5.98 \times {10^{24}}$$ kilograms. Find the Earth’s gravitational force on a space capsule that has a mass of $$1000$$ kilograms and that is $${10^8}$$ meters from the center of the Earth.

__Solution__:

Putting $$F = 6.67 \times {10^{ – 11}}$$ Newton, $$m = 1$$ kilogram, $$M = 1$$ kilogram, and $$d = 1$$ meter in the formula $$F = G(mM/{d^2})$$, we find that

\[6.67 \times {10^{ – 11}} = G\frac{{1(1)}}{{{1^2}}} = G\]

So that, for arbitrary values of $$m$$, $$M$$ and $$d$$,

\[F = (6.67 \times {10^{ – 11}})\frac{{mM}}{{{d^2}}}\]

Now we substitute $$M = 5.98 \times {10^{24}}$$,$$m = {10^3}$$, and $$d = {10^8}$$ to obtain

\[F = (6.67 \times {10^{ – 11}})\frac{{{{10}^3}(5.98 \times {{10}^{24}})}}{{{{({{10}^8})}^2}}} \approx 39.9\]

(or less than $$9$$ pounds of gravitational force).