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# Analyzing problems involving related rates

Related rates problems are word problems where we reason about the rate of change of a quantity by using information we have about the rate of change of another quantity that's related to it. Let's get acquainted with this sort of problem.
Related rates problems are applied problems where we find the rate at which one quantity is changing by relating it to other quantities whose rates are known.

## Worked example of solving a related rates problem

Imagine we are given the following problem:
The radius r, left parenthesis, t, right parenthesis of a circle is increasing at a rate of 3 centimeters per second. At a certain instant t, start subscript, 0, end subscript, the radius is 8 centimeters.
What is the rate of change of the area A, left parenthesis, t, right parenthesis of the circle at that instant?

#### Making sense of the quantities and their rates

In general, we are dealing here with a circle whose size is changing over time. There are two quantities referenced in the problem:
r, left parenthesis, t, right parenthesis is the radius of the circle after t seconds. It is measured in centimeters.
A, left parenthesis, t, right parenthesis is the area of the circle after t seconds. It is measured in square centimeters.
A circle has a radius labeled r of t and an area labeled A of t.
The problem also refers to the rates of those quantities. The rate of change of each quantity is given by its derivative:
r, prime, left parenthesis, t, right parenthesis is the instantaneous rate at which the radius changes at time t. It is measured in centimeters per second.
A, prime, left parenthesis, t, right parenthesis is the instantaneous rate at which the area changes at time t. It is measured in square centimeters per second.

#### Making sense of the given information

We are given that the radius is increasing at a rate of 3 centimeters per second. This means that start color #1fab54, r, prime, left parenthesis, t, right parenthesis, equals, 3, end color #1fab54 for any value of t.
We are also given that at a certain instant t, start subscript, 0, end subscript, the radius is 8 centimeters. This means that start color #11accd, r, left parenthesis, t, start subscript, 0, end subscript, right parenthesis, equals, 8, end color #11accd. Notice that this is only the case for t, start subscript, 0, end subscript, and not for just any value of t.
Finally, we are asked to find the rate of change of A, left parenthesis, t, right parenthesis at the instant t, start subscript, 0, end subscript. Mathematically, we are looking for start color #e07d10, A, prime, left parenthesis, t, start subscript, 0, end subscript, right parenthesis, end color #e07d10.

#### Relating the area and the radius

After we've made sense of the relevant quantities, we should look for an equation, or a formula, that relates them. The quantities in our case are the area and the radius of a circle. These quantities are related using the formula for the area of a circle:
A, equals, pi, r, squared

#### Differentiating

To find start color #e07d10, A, prime, left parenthesis, t, start subscript, 0, end subscript, right parenthesis, end color #e07d10 we need to take the derivative of each side of the equation. Once we do that, we will be able to relate start color #e07d10, A, prime, left parenthesis, t, start subscript, 0, end subscript, right parenthesis, end color #e07d10 with other known values, like start color #1fab54, r, prime, left parenthesis, t, start subscript, 0, end subscript, right parenthesis, end color #1fab54, which will allow us to solve for start color #e07d10, A, prime, left parenthesis, t, start subscript, 0, end subscript, right parenthesis, end color #e07d10.
Since we don't have the explicit formulas for A, left parenthesis, t, right parenthesis and r, left parenthesis, t, right parenthesis, we will use implicit differentiation:
\begin{aligned} A(t)&=\pi [r(t)]^2 \\\\ \dfrac{d}{dt}[A(t)]&=\dfrac{d}{dt}\Bigl[\pi [r(t)]^2\Bigr] \\\\ \goldD{A'(t)}&=2\pi\blueD{r(t)}\greenD{r'(t)} \end{aligned}
This is the core of our solution: by relating the quantities (i.e. A and r) we were able to relate their rates (i.e. A, prime and r, prime) through differentiation. This is why these problems are called "related rates"!

#### Solving

Note that the equation we got is true for any value of t and specifically for t, start subscript, 0, end subscript. We can substitute start color #11accd, r, left parenthesis, t, start subscript, 0, end subscript, right parenthesis, equals, 8, end color #11accd and start color #1fab54, r, prime, left parenthesis, t, start subscript, 0, end subscript, right parenthesis, equals, 3, end color #1fab54 into that equation:
\begin{aligned} \goldD{A'(t_0)}&=2\pi\blueD{r(t_0)}\greenD{r'(t_0)} \\\\ &=2\pi(\blueD 8)(\greenD 3) \\\\ &=48\pi \end{aligned}
In conclusion, we found that at t, start subscript, 0, end subscript, the area is increasing at a rate of 48, pi square centimeters per second.
Problem 1.A
Problem set 1 will walk you through the steps of analyzing the following problem:
The base b, left parenthesis, t, right parenthesis of a triangle is decreasing at a rate of 13 start text, m, slash, h, end text and the height h, left parenthesis, t, right parenthesis of the triangle is increasing at a rate of 6 start text, m, slash, h, end text. At a certain instant t, start subscript, 0, end subscript, the base is 5 start text, m, end text and the height is 1 start text, m, end text. What is the rate of change of the area A, left parenthesis, t, right parenthesis of the triangle at that instant?
Match each expression with its units.

start text, m, end text
start text, m, slash, h, end text
start text, m, end text, squared
start text, m, end text, squared, start text, slash, h, end text
b, prime, left parenthesis, t, right parenthesis
A, left parenthesis, t, start subscript, 0, end subscript, right parenthesis
h, left parenthesis, t, start subscript, 0, end subscript, right parenthesis
start fraction, d, A, divided by, d, t, end fraction

Want more practice? Try this exercise.

### Common mistake: Confusing which expressions are variables and which are constants

As you've seen, related rates problems involve multiple expressions. Some represent quantities and some represent their rates. Some are changing, some are constants.
It's important to make sure you understand the meaning of all expressions and are able to assign their appropriate values (when given).
We recommend performing an analysis similar to those shown in the example and in Problem set 1: what are all the relevant quantities? What are their rates? What are their units? What are their values?
Problem 2
Consider this problem:
Two cars are driving towards an intersection from perpendicular directions. The first car's velocity is 50, start text, space, k, m, slash, h, end text and the second car's velocity is 90, start text, space, k, m, slash, h, end text. At a certain instant t, start subscript, 0, end subscript, the first car is a distance x, left parenthesis, t, start subscript, 0, end subscript, right parenthesis of 0, point, 5, start text, space, k, m, end text from the intersection and the second car is a distance y, left parenthesis, t, start subscript, 0, end subscript, right parenthesis of 1, point, 2, start text, space, k, m, end text from the intersection. What is the rate of change of the distance d, left parenthesis, t, right parenthesis between the cars at that instant?
Which equation should be used to solve the problem?

### Common mistake: Selecting an equation that misrepresents the given problem

As you've seen, the equation that relates all the quantities plays a crucial role in the solution of the problem. It's usually helpful to have some kind of diagram that describes the situation with all the relevant quantities. Let's take Problem 2 for example. The problem describes a right triangle.
A right triangle is formed between the intersection, first car, and second car. The right angle is at the intersection. The leg to the first car is labeled x of t. The leg to the second car is labeled y of t. The hypotenuse, between the cars, measures d of t.
The diagram makes it clearer that the equation we're looking for relates all three sides of the triangle, which can be done using the Pythagoream theorem:
open bracket, d, left parenthesis, t, right parenthesis, close bracket, squared, equals, open bracket, x, left parenthesis, t, right parenthesis, close bracket, squared, plus, open bracket, y, left parenthesis, t, right parenthesis, close bracket, squared
Without the diagram, we might accidentally treat d, left parenthesis, t, right parenthesis as if it's the triangle's area...
d, left parenthesis, t, right parenthesis, equals, start fraction, x, left parenthesis, t, right parenthesis, dot, y, left parenthesis, t, right parenthesis, divided by, 2, end fraction
...or treat x, left parenthesis, t, right parenthesis, y, left parenthesis, t, right parenthesis, and d, left parenthesis, t, right parenthesis as if they are the three angles of the triangle...
d, left parenthesis, t, right parenthesis, plus, x, left parenthesis, t, right parenthesis, plus, y, left parenthesis, t, right parenthesis, equals, 180
...or maybe treat d, left parenthesis, t, right parenthesis as if it's an angle and form some trig equation
tangent, open bracket, d, left parenthesis, t, right parenthesis, close bracket, equals, start fraction, y, left parenthesis, t, right parenthesis, divided by, x, left parenthesis, t, right parenthesis, end fraction.
All of these equations might be useful in other related rates problems, but not in the one from Problem 2.
Problem 3
Consider this problem:
A 20-meter ladder is leaning against a wall. The distance x, left parenthesis, t, right parenthesis between the bottom of the ladder and the wall is increasing at a rate of 3 meters per minute. At a certain instant t, start subscript, 0, end subscript, the top of the ladder is a distance y, left parenthesis, t, start subscript, 0, end subscript, right parenthesis of 15 meters from the ground. What is the rate of change of the angle theta, left parenthesis, t, right parenthesis between the ground and the ladder at that instant?
Which equation should be used to solve the problem?