Unit 5

Applications of Derivatives

B. Related Rates Problems

Lesson 4: Trigonometric Functions

1. Draw and label a diagram. Include the values that never change (constants) and include variables for the values that do change.
2. List all given and required rates as derivatives with respect to time.
3. Write an equation that relates the variables from Step 2, and take the derivative. This equation may stem from a well-known geometric formula for area or volume, the Pythagorean Theorem, the properties of similar triangles, or trigonometry. The chain rule is then used to differentiate the equation with respect to time.
4. Substitute the given information into the differentiated equation, and solve for the unknown rate.
   Note: A common error is made when the given numerical information is substituted too soon. If a value is true only for a particular instant, it can only be used after differentiation.
   

Example 1

An airplane, in level flight, is approaching the spot above where you are standing. The speed of the airplane is $100\mathrm{} \mathrm{m}/\mathrm{s}$ and it is flying at an altitude of $1 000 \mathrm{m}$. What is the rate of change of the angle of elevation, $\theta$, when the distance from where you are standing to a point directly below the plane is $2 000 \mathrm{m}$?

Solution

Step 1:

Draw and label a diagram.

Let $\theta$ be the angle of elevation.

Let $x$ be the distance from your position to the point directly below the airplane.

Let $z$ be the distance from your position to the airplane.

Step 2:

State the given and required related rates.

${\left.\frac{dx}{dt}\right|}_{x=2 000}=-100 \mathrm{}\mathrm{m}/\mathrm{s}$

${\left.\frac{d\theta }{dt}\right|}_{x=2 000}=\mathrm{?}$

Note that $\frac{dx}{dt}$ is negative since the distance $x$ is decreasing.

Step 3:

Write an equation.

$\begin{array}{rcl}\tan \theta & =& \frac{1 000}{x}\\ \frac{d}{dt}\left(\tan \theta \right)& =& \frac{d}{dt}\left(\frac{1 000}{x}\right)\\ {\sec }^2\theta \bullet \frac{d\theta }{dt}& =& -\frac{1 000}{x^2}\bullet \frac{dx}{dt}\\ \frac{1}{{\cos }^2\theta }\bullet \frac{d\theta }{dt}& =& -\frac{1 000}{x^2}\bullet \frac{dx}{dt}\\ \frac{d\theta }{dt}& =& -\frac{1 000}{x^2}\bullet \frac{dx}{dt}\bullet {\cos }^2\theta \end{array}$

Find $\cos \theta$ before proceeding.

Solve for $z$ when $x=2 000$.

$\begin{array}{rcl}{z}^2& =& x^2+1 {000}^2\\ z^2& =& 2 {000}^2+1 {000}^2\\ z& =& \sqrt{4\mathrm{}000\mathrm{}000+1\mathrm{}000\mathrm{}000}\\ & =& \sqrt{5\mathrm{}000\mathrm{}000}\\ & & \\ \cos \theta & =& \frac{2 000}{\sqrt{5\mathrm{}000\mathrm{}000}}\end{array}$

Note: Sometimes it takes until the derivative has been computed to determine what additional information is required. In this case, the derivative contains ${\cos }^2\theta$, so some additional information must be determined from the given information.

Step 4:

Substitute the values of $x$, $\frac{dx}{dt}$, and $\cos \theta$ into the derivative and solve for $\frac{d\theta }{dt}$.

$\begin{array}{rcl}\frac{d\theta }{dt}& =& -\frac{1 000}{x^2}\bullet \frac{dx}{dt}\bullet {\cos }^2\theta \\ & =& -\frac{1 000}{2 {000}^2}\bullet \left(-100\right)\bullet {\left(\frac{2 000}{\sqrt{5\mathrm{}000\mathrm{}000}}\right)}^2\\ & =& \left(\frac{-1 000}{4\mathrm{}000\mathrm{}000}\right)\left(-100\right)\left(\frac{4\mathrm{}000\mathrm{}000}{5\mathrm{}000\mathrm{}000}\right)\\ & =& \frac{100\mathrm{}000}{5\mathrm{}000\mathrm{}000}\\ & =& \frac{1}{50}\end{array}$

    The angle of elevation is increasing at a rate of $\frac{1}{50}\mathrm{} \mathrm{rad}/\mathrm{s}$ .