RELATED RATES PROBLEMS |

Pay attention to the words because, as always in math, they say exactly what they mean. In these questions, we're dealing with **rates of change that are related to one another** through a well defined function. (As with all word problems, **we should make a diagram whenever possible!**)

For instance, when a perfectly spherical snowball melts at a perfectly constant rate (we're in Calculusland here -- where there are such things), **the rate at which the volume decreases is directly related to the rate at which the radius decreases**. The function that relates these two changing variables is the formula for the volume of a sphere:

Since both the volume and the radius are decreasing with respect to time, we can say that the volume is a function of time and so is the radius. That is:

V = f(t) and r = g(t)

**Example 1**

A spherical snowball is melting at a constant rate.

The radius changes from 12 inches to 8 inches in 45 minutes.

How fast is the volume changing when the radius is 10 inches?

We know , since the radius decreased by 4 inches in 45 minutes.

This means that . . and we need at r = 10 inches.

Since , when we differentiate both sides with respect to time we get:

Now we substitute values to find .

Notice how the units indicate a change in volume per unit time.

**Example 2**

This one's an exam classic.

The length of the legs of a right triangle are increasing at 3 cm/sec and 5 cm/sec.

How fast is the length of the hypotenuse increasing when the first leg is 6 cm.

and the second is 8 cm.?

now we substitute values for x, y and z.

we know x = 6 cm and y = 8 cm so we find z by Pythagoras. z = 10 cm.

**** **We Never** **substitute** values for the variables **until we've differentiated** the relation between the variables and expressed the unknown in terms of the other variables. We always put in the units of measure to determine whether we've found the right quantities.

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** Hint: **label the quantities given**. If told that gas is being pumped into a balloon at

**10 cm ^{ 3} / sec** , label it

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**Example 3:** Gas is being pumped into a spherical balloon at a rate of 5 ft^{ 3} / min.

If the pressure is constant, find the **rate at which the radius is changing** when the **diameter** reaches 18 inches.

First we change the 9 inch **radius** to feet since the volume is being measured in ft^{ 3} .

r = 9 in = 0.75 ft., | dV/dt = 5 ft^{ 3} / min, |
we want dr/dt |

the relation between Volume and radius of a sphere is:

= 0.71 ft/min

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**Example 4:** A circular metal disk is being heated and the **diameter** increases at a rate of

0.01 cm/min. At what rate is the **Area** of one side of the disk increasing when the diameter reaches 5 cm. ?

Since dd/dt = 0.01 cm/min, and the radius is half the diameter, dr/dt = 0.005 cm/min.

When the diameter reaches 5 cm, the radius r will be 2.5cm.

For a circle, A = o r^{ 2} .

This is the relation then that links the changing area of the disk to the changing radius.

So

1) A weather balloon is rising vertically at a rate of 2 ft/sec. An observer is situated 100 yds. from the point on the ground directly below the balloon. At what rate is the distance between the balloon and the observer changing when the altitude of the balloon is 500 ft. ? (watch the mixed units!!)

2) A point P(x, y) moves along the graph of y^{ 2} = x^{ 2} - 9 such that dx/dt = 1/x units/sec.

Find dy/dt at the point (5, 4).

3) A point P(x, y) moves along the graph of y = x^{ 3} + x^{ 2} + 1 with the x value changing at a rate of 2 units / sec. How fast is y changing at the point (1, 3)?

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1) A weather balloon is rising vertically at a rate of 2 ft/sec. An observer is situated 100 yds. from the point on the ground directly below the balloon. At what rate is the distance between the balloon and the observer changing when the altitude of the balloon is 500 ft. ? (watch the mixed units!!)

2) A point P(x, y) moves along the graph of y^{ 2} = x^{ 2} - 9 such that dx/dt = 1/x units/sec.

Find dy/dt at the point (5, 4).

y^{ 2} = x^{ 2} - 9 so 2y dy/dt = 2x dx/dt

at (5, 4), x = 5 and y = 4. so units/sec.

3) A point P(x, y) moves along the graph of y = x^{ 3} + x^{ 2} + 1 with the x value changing at a rate of 2 units / sec. How fast is y changing at the point (1, 3)?

y = x^{ 3} + x^{ 2} + 1, so dy/dt = (3 x^{ 2} + 2x) dx/dt

set x = 1

dy/dt = (3 + 2)2 units/sec = **10 units / sec**

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Back to Calculus 1 Lessons Links List

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