# Related Rates

Suppose we have two quantities, which are connected to each other and both changing with time. A related rates problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

Let the two variables be x and y. The relationship between them is expressed by a function y = f (x). The rates of change of the variables x and y are defined in terms of their derivatives dx/dt and dy/dt. If dx/dt is known, we can determine dy/dt (and vice versa).

Any related rates problem can be solved as follows:

1. Decide what are the two variables describing your system or process.
2. Drawing a picture can often be useful.
3. Write an equation that relates the variables. To derive the equation you may use a geometric fact (like the Pythagorean theorem or similar triangles), a trigonometric identity, or a physical law.
4. Take the derivative $$\frac{d}{{dt}}$$ of both sides of the equation.
5. Solve for the unknown rate of change.
6. Substitute all known values to get the final answer.

As an example, let's consider the well-known sliding ladder problem.

A ladder $$13$$ feet long leans against a wall. The top of the ladder slides down at a constant rate of $$12\,\frac{\text{ft}}{\text{sec}}.$$ Find the velocity of the ladder when $$h = 5\,\text{ft}.$$

By the Pythagorean equation:

${x^2} + {h^2} = {13^2}.$

Differentiate both sides with respect to time $$t.$$

$\frac{d}{{dt}}\left( {{x^2} + {h^2}} \right) = \frac{d}{{dt}}\left( {{{13}^2}} \right),\;\; \Rightarrow 2x\frac{{dx}}{{dt}} + 2h\frac{{dh}}{{dt}} = 0,\;\; \Rightarrow x\frac{{dx}}{{dt}} = - h\frac{{dh}}{{dt}},\;\; \Rightarrow \frac{{dx}}{{dt}} = - \frac{h}{x}\frac{{dh}}{{dt}}.$

As $$x = \sqrt {{{13}^2} - {h^2}} ,$$ we have

$\frac{{dx}}{{dt}} = - \frac{h}{{\sqrt {{{13}^2} - {h^2}} }}\frac{{dh}}{{dt}}.$

Substitute the known values:

$\frac{{dh}}{{dt}} = - 12\,\frac{\text{ft}}{\text{sec}}$

(the minus sign denotes that the ladder is sliding down),

$h = 5\,\text{ft}.$

This yields:

$\frac{{dx}}{{dt}} = - \frac{h}{{\sqrt {{{13}^2} - {h^2}} }}\frac{{dh}}{{dt}} = - \frac{5}{{\sqrt {{{13}^2} - {5^2}} }} \cdot \left( { - 12} \right) = \frac{5}{{12}} \cdot 12 = 5\,\frac{\text{ft}}{\text{sec}}.$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Oil from an uncapped well is radiating outward in the form of a circular film on the surface of the water. If the radius of the circle is increasing at the rate of $$0.5$$ meters per minute, how fast is the area of the oil film growing at the instant when the radius is $$100\,\text{m}?$$

### Example 2

A triangle has two sides $$a = 1\,\text{cm}$$ and $$b = 2\,\text{cm}.$$ How fast is the third side $$c$$ increasing when the angle $$\alpha$$ between the given sides is $$60^\circ$$ and is increasing at the rate of $$3^\circ$$ per second (Figure $$2$$)?

### Example 3

A light is at the top of a $$16$$ ft pole. A man $$6$$ ft tall walks away from the pole at a rate of $$5\,\frac{\text{ft}}{\text{sec}}$$ (Figure $$3$$). How fast is the tip of his shadow moving when he is $$20$$ ft from the pole?

### Example 4

The volume of a cube is increasing at the rate of $$2$$ cubic inches per minute. How fast is the surface of the cube increasing when the side is $$8$$ inches?

### Example 1.

Oil from an uncapped well is radiating outward in the form of a circular film on the surface of the water. If the radius of the circle is increasing at the rate of $$0.5$$ meters per minute, how fast is the area of the oil film growing at the instant when the radius is $$100\,\text{m}?$$

Solution.

Suppose that t is time in minutes, R and A are the radius and area of the circle, respectively.

The rate of change of the area is given by the derivative $$\frac{{dA}}{{dt}},$$ where

$A = \pi {R^2}.$

Differentiating the right-hand side of the relation by the chain rule, we get

$\frac{{dA}}{{dt}} = \frac{d}{{dt}}\left( {\pi {R^2}} \right) = 2\pi R\frac{{dR}}{{dt}}.$

It is known that $$\frac{{dR}}{{dt}} = 0.5\,\frac{\text{m}}{\text{min}}.$$ Therefore, the oil spot is growing at the rate

$\frac{{dA}}{{dt}} = 2\pi R\frac{{dR}}{{dt}} = 2\pi R \cdot 0.5 = \pi R.$

For $$R = 100\,\text{m},$$ we have

$\frac{{dA}}{{dt}} = 100\pi \approx 314\,\frac{{{\text{m}^2}}}{\text{min}}.$

### Example 2.

A triangle has two sides $$a = 1\,\text{cm}$$ and $$b = 2\,\text{cm}.$$ How fast is the third side $$c$$ increasing when the angle $$\alpha$$ between the given sides is $$60^\circ$$ and is increasing at the rate of $$3^\circ$$ per second (Figure $$2$$)?

Solution.

According to the law of cosines,

${c^2} = {a^2} + {b^2} - 2ab\cos \alpha .$

We differentiate both sides of this equation with respect to time t:

$\frac{d}{{dt}}\left( {{c^2}} \right) = \frac{d}{{dt}}\left( {{a^2} + {b^2} - 2ab\cos \alpha } \right),$
$2c\frac{{dc}}{{dt}} = - 2ab\left( { - \sin \alpha } \right)\frac{{d\alpha }}{{dt}},$

or

$\frac{{dc}}{{dt}} = \frac{{ab\sin \alpha }}{c}\frac{{d\alpha }}{{dt}}.$

Calculate the length of the side $$c:$$

$c = \sqrt {{a^2} + {b^2} - 2ab\cos \alpha } = \sqrt {{1^2} + {2^2} - 2 \cdot 1 \cdot 2 \cdot \cos 60^\circ } = \sqrt {1 + 4 - 2} = \sqrt 3 .$

Now we know all quantities to determine the rate of change $$\frac{{dc}}{{dt}}:$$

$\frac{{dc}}{{dt}} = \frac{{ab\sin \alpha }}{c}\frac{{d\alpha }}{{dt}} = \frac{{1 \cdot 2 \cdot \sin 60^\circ }}{{\sqrt 3 }}\frac{{d\alpha }}{{dt}} = \frac{{2\frac{{\sqrt 3 }}{2}}}{{\sqrt 3 }} \cdot 3 = 3\,\frac{\text{cm}}{\text{sec}}.$

### Example 3.

A light is at the top of a $$16$$ ft pole. A man $$6$$ ft tall walks away from the pole at a rate of $$5\,\frac{\text{ft}}{\text{sec}}$$ (Figure $$3$$). How fast is the tip of his shadow moving when he is $$20$$ ft from the pole?

Solution.

By similar triangles,

$\frac{L}{{16}} = \frac{{L - x}}{6}.$

Solving for $$L,$$ we have:

$6L = 16\left( {L - x} \right),\;\; \Rightarrow 6L = 16L - 16x,\;\; \Rightarrow L = \frac{8}{5}x.$

Differentiating both sides with respect to time $$t$$ yields:

$\frac{{dL}}{{dt}} = \frac{8}{5}\frac{{dx}}{{dt}}.$

We are told that the rate $$\frac{{dx}}{{dt}} = 5\,\frac{\text{ft}}{\text{sec}}.$$ Therefore

$\frac{{dL}}{{dt}} = \frac{8}{5} \cdot 5 = 8\,\frac{\text{ft}}{\text{sec}}.$

Thus, the tip of the shadow is moving at the rate of $$8\,\frac{\text{ft}}{\text{sec}}.$$

### Example 4.

The volume of a cube is increasing at the rate of $$2$$ cubic inches per minute. How fast is the surface of the cube increasing when the side is $$8$$ inches?

Solution.

The volume of the cube of side $$a$$ is given by

$V = {a^3}.$

We differentiate both sides of the equation to find the relation between the rates of change:

$\frac{{dV}}{{dt}} = 3{a^2}\frac{{da}}{{dt}}.$

Solve for $$\frac{{da}}{{dt}}:$$

$\frac{{da}}{{dt}} = \frac{1}{{3{a^2}}}\frac{{dV}}{{dt}}.$

Recall that the surface area of the cube is

$A = 6{a^2}.$

Similarly, take the derivatives of both sides:

$\frac{{dA}}{{dt}} = 12a\frac{{da}}{{dt}}.$

Now we substitute the expression for $$\frac{{da}}{{dt}}$$ from the first equation. This yields

$\frac{{dA}}{{dt}} = 12a\frac{{da}}{{dt}} = 12a \cdot \frac{1}{{3{a^2}}}\frac{{dV}}{{dt}} = \frac{4}{a}\frac{{dV}}{{dt}}.$

As the volume of the cube is increasing at the rate of $$2\,\frac{{{\text{in}^3}}}{\text{min}},$$ we can write

$\frac{{dA}}{{dt}} = \frac{4}{a}\frac{{dV}}{{dt}} = \frac{4}{a} \cdot 2 = \frac{8}{a}.$

Finally, we substitute the side length of the cube, which is equal to $$8$$ inches. As a result, we find the rate of change of the area:

$\frac{{dA}}{{dt}} = \frac{8}{a} = \frac{8}{8} = 1\,\frac{{{\text{in}^2}}}{\text{min}}.$

See more problems on Page 2.