Related Rates
In a related rates problem, the idea is to compute the rates of change of two or more related variables with respect to time. The procedure involves finding an equation that relates the variables and using the Chain Rule to differentiate both sides with respect to time.
Guidelines For Solving Related Rate Problems
1. Read the problem carefully and draw a diagram if possible.
2. Name the variables and constants. Use $t$ for time.
3. Write an equation that relates the variables whose rates of change are given. If necessary, use the geometry or trigonometry of the situation and write an equation whose rates of change are to be determined.
4. Combine two or more equations to get a single equation that relates the variable.
5. Use the Chain Rule to differentiate both sides of the equation with respect to $t$.
6. Substitute the given numerical information into the resulting equation and solve for the unknown rate.
Example
- Water runs into a conical tank at a rate of $0.5 \mathrm{~m}^3 / \mathrm{min}$. The tank stands point down and has a height of $4 \mathrm{~m}$ and a base radius of $2 \mathrm{~m}$. How fast is the water level rising when the water is $2.5 \mathrm{~m}$ deep?
▶️Answer/Explanation
Solution
Draw a picture and name the variables and constants. Let $V=$ volume of water in the tank, $h=$ depth of water in the tank, and $r=$ radius of the surface of the water at time $t$.
Volume of a cone is $V=\frac{1}{3} \pi r^2 h$.
In order to eliminate $r$ we use the similar triangle in the figure.
$
\frac{2}{4}=\frac{r}{h} \Rightarrow r=\frac{h}{2}
$
$
V=\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi\left(\frac{h}{2}\right)^2 h=\frac{\pi}{12} h^3
$ Substitute $r=\frac{h}{2}$.
$
\frac{d V}{d t}=\frac{\pi}{12}\left(3 h^2 \frac{d h}{d t}\right)=\frac{\pi h^2}{4} \frac{d h}{d t}
$ Differentiate $V$ and $h$ with respect to $t$.
$
0.5=\frac{\pi(2.5)^2}{4} \frac{d h}{d t}
$$\frac{d V}{d t}=0.5$ and $h=2.5$.
$
\frac{d h}{d t}=\frac{(0.5) 4}{\pi(2.5)^2} \approx 0.102
$
So, when $h=2.5$, the water level is rising at a rate of 0.102 meters per minute.
Example 2
- Car $A$ is traveling due west toward the intersection at a speed of 45 miles per hour. $\operatorname{Car} B$ is traveling due north away from the intersection at a speed of $30 \mathrm{mph}$ Let $x$ be the distance between $\operatorname{Car} A$ and the intersection at time $t$, and let $y$ be the distance between $\operatorname{Car} B$ and the intersection at time $t$ as shown in the figure at the right.
(a) Find the rate of change, in miles per hour, of the distance between the two cars when $x=32$ miles and $y=24$ miles.
(b) Let $\theta$ be the angle shown in the figure. Find the rate of change of $\theta$, in radians per hour, when $x=32$ miles and $y=24$ miles.
▶️Answer/Explanation
Solution
- (a) Let s = the distance between the two cars.
$
s^2=x^2+y^2
$Pythagorean Theorem
(b) $\tan \theta=\frac{y}{x}$
$
\frac{d}{d t}(\tan \theta)=\frac{d}{d t}\left(\frac{y}{x}\right)
$ Differentiate both sides with respect to $t$.
$\begin{aligned} & \sec ^2 \theta \frac{d \theta}{d t}=\frac{x \frac{d y}{d t}-y \frac{d x}{d t}}{x^2} \\ & \left(\frac{40}{32}\right)^2 \frac{d \theta}{d t}=\frac{(32)(30)-(24)(-45)}{(32)^2} \\ & \frac{d \theta}{d t}=1.275 \mathrm{rad} / \mathrm{hr}\end{aligned}$
Example3
- A man 6 feet tall walks at a rate of 3 feet per second away from a light that is 20 feet above the ground.
(a) At what rate is the tip of his shadow moving when he is 12 feet from the base of the light.
(b) At what rate is the length of his shadow changing when is 12 feet from the base of the light.
▶️Answer/Explanation
Solution
Draw a picture and name the variables and constants. Let $x=$ the distance from the base to the man and $s=$ the distance from the man to the tip of his shadow.
By similar triangles, $\frac{20}{x+s}=\frac{6}{s}$. $\Rightarrow 20 s=6(x+s) \Rightarrow 14 s=6 x$ $\Rightarrow s=\frac{3}{7} x$
(a) The tip of the shadow moves at a rate of
$
\frac{d}{d t}(x+s)
$ $x+s$ is the distance from the base to the tip of shadow.
$
\begin{array}{ll}
=\frac{d}{d t}\left(x+\frac{3}{7} x\right)=\frac{10}{7} \frac{d x}{d t} & s=\frac{3}{7} x \\
=\frac{10}{7}(3)=\frac{30}{7} \mathrm{ft} / \mathrm{sec} & \frac{d x}{d t}=3
\end{array}
$
(b) The length of the shadow changing at a rate of
$
\begin{aligned}
& \frac{d}{d t}(s)=\frac{d}{d t}\left(\frac{3}{7} x\right)=\frac{3}{7} \frac{d x}{d t} \\
& =\frac{3}{7}(3)=\frac{9}{7} \mathrm{ft} / \mathrm{sec}
\end{aligned}
$
$s$ is the length of shadow.
Exercises – Related Rates
Multiple Choice Questions
- The radius of a circle is changing at the rate of $1 / \pi$ inches per second. At what rate, in square inches per second, is the circle’s area changing when $r=5$ in ?
(A) $\frac{5}{\pi}$
(B) 10
(C) $\frac{10}{\pi}$
(D) 15
▶️Answer/Explanation
Ans:B
Example
- The volume of a cube is increasing at the rate of $12 \mathrm{in}^3 / \mathrm{min}$. How fast is the surface area increasing, in square inches per minute, when the length of an edge is 20 in?
(A) 1
(B) $\frac{6}{5}$
(C) $\frac{4}{3}$
(D) $\frac{12}{5}$
▶️Answer/Explanation
Ans:D
Example
- In the figure shown above, a hot air balloon rising straight up from the ground is tracked by a television camera $300 \mathrm{ft}$ from the liftoff point. At the moment the camera’s elevation angle is $\pi / 6$, the balloon is rising at the rate of $80 \mathrm{ft} / \mathrm{min}$. At what rate is the angle of elevation changing at that moment?
(A) 0.12 radian per minute
(B) 0.16 radian per minute
(C) 0.2 radian per minute
(D) 0.4 radian per minute
▶️Answer/Explanation
Ans:C
Example
A car is approaching a right-angled intersection from the north at $70 \mathrm{mph}$ and a truck is traveling to the east at $60 \mathrm{mph}$. When the car is 1.5 miles north of the intersection and the truck is 2 miles to the east, at what rate, in miles per hour, is the distance between the car and truck is changing?
(A) Decreasing 15 miles per hour
(B) Decreasing 9 miles per hour
(C) Increasing 6 miles per hour
(D) Increasing 12 miles per hour
▶️Answer/Explanation
Ans:C
Example
- The radius $r$ of a sphere is increasing at a constant rate. At the time when the surface area and the radius of sphere are increasing at the same numerical rate, what is the radius of the sphere? (The surface area of a sphere is $S=4 \pi r^2$.)
(A) $\frac{1}{8 \pi}$
(B) $\frac{1}{4 \pi}$
(C) $\frac{1}{3 \pi}$
(D) $\frac{\pi}{8}$
▶️Answer/Explanation
Ans:A