Page 76 - Calc for the Clueless

Finally, sometimes you need two different related rates equations to solve the problem.

Example 23—

Find the rate at which the volume changes with respect to time when the change of surface area with respect to
time is 600 square inches per hour and the volume is 1000 cubic inches.

Now V = x . So dV/dt = 3x dx/dt. We need to know both x and dx/dt, but we are given neither directly.
3
2
However, V = x = 1000. Sooo × = 10 inches. We are given dS/dt. So the surface area of a cube S = 6x .
2
3

Finally ...

The Gravity of the Situation

The last type of word problem we will deal with is throwing an object into the air. There are three variables that
determine how high an object goes. One is gravity. On this planet, gravity is -32 feet per second squared (or -
9.8 meters per second squared); the minus sign indicates down. The second is the initial velocity (positive if the
object is thrown upwards, negative if the object is thrown down, and zero if the object is dropped). The third is
the initial height (it goes higher if I throw something from the Empire State Building than from the ground).
Interestingly, the weight has no bearing. If I give a rock or a refrigerator the same initial velocity (which I can't,
of course ... or can I?), both will go just as high. Friction, winds, etc., are not included.

The following symbols are usually used:

v 0: initial velocity—the velocity at time t = 0

y 0: initial height—the height at time t = 0
The initial height is either zero, the ground, or positive. It is very difficult to throw something upward from
under the ground. The acceleration is dv/dt. So the velocity is the antiderivative of the acceleration.

The velocity is dy/dt. The height y is the antiderivative of the velocity.

Example 24—

On the planet Calculi, gravity is 20 feet per second squared. A ball is thrown upward—initial velocity 40 feet
per second, initial height 50 feet.

A. Write the equation for the height.

B. Find the ball's maximum height.