Related rates

Many of you are familiar with the Ideal Gas Equation:

 

PV = nRT,

where P is pressure (in atmospheres), V is volume (in liters), n is the number of moles of the gas, R is the ideal gas constant: 0.08206 liter-atmospheres/mole/degree, and T is the temperature (in Kelvins). Excepting R of course, anywhere from two to all four of the quantities related by this formula may vary with time. Thus the rate of change of the left-hand side of the Ideal Gas Equation with respect to time will always equal the rate of change of the right-hand side of the Ideal Gas Equation with respect to time, and we have

 

PV' + P'V = R(nT' + n'T)

Thus if we know at some particular time that:

• The pressure is 2 atmospheres: P = 2; and
• The pressure is increasing at rate of 0.01 atmospheres per hour: P' = 0/.01;
• The volume is 30 liters: V = 3; and
• The volume is increasing at a rate of 0.02 liters per hour: V' = 0.01;
• There are 2 moles of gas: n=2;
• The amount of gas is constant: n' = 0;
• The temperature is 300 degrees Kelvin: T = 300.

Then the rate of change of the temperature at that time is found by solving

(2)(0.02) + (0.01)(30) = (0.08206)(2T' + (0)(300))

for T'.

Similar calculations may be carried out in a variety of situations. A seemingly boring version is that one in which one side, S, and the hypotenuse, H, of a right triangle vary with time while the third side, F, remains fixed. Since the Pythagorean Theorem asserts that

H² = F² + S²

we have

$$2HH' = 2F\cdot 0 + 2SS'$$

or more simply,

HH' = SS'.

If this seems irrelevant to your life, think of S' as your highway speed and H' as your speed measured by a police officer off to the side of the road. Since H > S, your speed as measured on the radar gun is always less than your actual speed, making it clear why people argue defective speedometer in traffic court rather than defective radar units.

One last example. Suppose that a rubber ball of is being inflated at a rate of 5 cubic centimeters per second. When it reached a radius of 30 centimeters, how fast is the surface area increasing? Here, we have to look at two equations, $V = 4\pi R^3/3$ and $SA = 4\pi R^2$. These give us

$$\begin{array} {rcl} V' & = & 4\pi R^2 R'\\ SA' & = & 8\pi R R'\end{array}$$

The given information along with the first of these rate equations gives a value for R' which when substituted into the second equation produces a value for SA': $SA' = (1/3){\rm cm}^2$per second. In fact, one can eliminate R' from both equations and show that 2V'/R = SA'. Again, this may look like a silly problem, but the same principle is needed when looking at stresses on a surface due to expansion and contraction of the volume it contains. This is the basis for mechanical de-icers on air craft, for example.