**Introduction to Related Rates**

If a tree is growing in a forest, then both its height and its radius will be increasing. These two growths will depend in turn on (i) the amount of sunlight that hits the tree, (ii) the amount of nutrients in the soil, (iii) the proximity of other trees. We may wish to study the relationship among these various parameters. For example, if we know that the amount of sunlight and nutrients are increasing at a certain rate then we may wish to know how that affects the rate of change of the radius. This consideration gives rise to *related rates* problems.

**Examples**

**Example 1**

A toy balloon is in the shape of a sphere. It is being inflated at the rate of 20 in. ^{3} /min. At the moment that the sphere has volume 64 cubic inches, what is the rate of change of the radius?

**Solution 1**

We know that volume and radius of a sphere are related by the formula

The free variable in this problem is time, so we differentiate equation (*) with respect to time *t* . It is important that we keep the chain rule in mind as we do so. ^{ 1} The result is

^{1} The point is that we are *not* differentiating with respect to *r* .

Now we are given that *dV/dt* = 20. Our question is posed at the moment that *V* = 64. But, according to (*), this means that Substituting these values into equation (**) yields

Solving for *dr/dt* yields

**Example 2**

A 13-foot ladder leans against a wall ( Fig. 3.15 ). The foot of the ladder begins to slide away from the wall at the rate of 1 foot per minute. When the foot is 5 feet from the wall, at what rate is the top of the ladder falling?

**Solution 2**

Let *h* ( *t* ) be the height of the ladder at time *t* and *b* ( *t* ) be the distance of the base of the ladder to the wall at time *t* . Then the Pythagorean theorem tells us that

*h* ( *t* ) ^{2} + *b* ( *t* ) ^{2} = 13 ^{2} .

We may differentiate both sides of this equation with respect to the variable *t* (which is time in minutes) to obtain

2 · *h* ( *t* ) · *h′* ( *t* ) + 2 · *b* ( *t* ) · *b′* ( *t* ) = 0.

Solving for *h′* ( *t* ) yields

At the instant the problem is posed, *b* ( *t* ) = 5, *h* ( *t* ) = 12 (by the Pythagorean theorem), and *b′* ( *t* ) = 1. Substituting these values into the equation yields

Observe that the answer is negative, which is appropriate since the top of the ladder is falling.

**You Try It:** Suppose that a square sheet of aluminum is placed in the hot sun. It begins to expand very slowly so that its diagonal is increasing at the rate of 1 millimeter per minute. At the moment that the diagonal is 100 millimeters, at what rate is the area increasing?

**Example 3**

A sponge is in the shape of a right circular cone ( Fig. 3.16 ). As it soaks up water, it grows in size. At a certain moment, the height equals 6 inches, and is increasing at the rate of 0.3 inches per second. At that same moment, the radius is 4 inches, and is increasing at the rate of 0.2 inches per second. How is the volume changing at that time?

**Solution 3**

We know that the volume *V* of a right circular cone is related to the height *h* and the radius *r* by the formula

Differentiating both sides with respect to the variable *t* (for time in seconds) yields

Substituting the values for *r* , *dr/dt* , *h* , and *dh/dt* into the right-hand side yields

**You Try It:** In the heat of the sun, a sheet of aluminum in the shape of an equilateral triangle is expanding so that its side length increases by 1 millimeter per hour. When the side length is 100 millimeters, how is the area increasing?

Find practice problems and solutions for these concepts at: Applications of the Derivative Practice Test.

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