Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 4

Differentiation

Related Rates

Related rate problems deal with quantities that change with respect to time and are related to each other. We will be looking for the rate of change of one quantity in terms of the related rates of the others. The aim is to find the equation that relates these quantities, and to use implicit differentiation with respect to time to find the rate we are looking for.

A possible strategy for approaching such problems is given on page 199 of the textbook. We repeat it here, with additional notes.

1. Read the problem carefully. Identify the quantities that change with respect to time and the quantities that are constant. You will be able to do this only if you understand the problem. Pay attention to the units.
2. Draw a diagram if possible. It is not always possible to draw a diagram, but try to draw one whenever possible to help you visualize and understand the problem.
3. Introduce notation. Assign symbols to all quantities that are functions of time, and indicate the constants in the diagram. Do not include in your diagram the known rates or the rate to be found under the conditions given.
4. Express the given information and the required rate in terms of the derivatives. Identify the known rates and the ones you are looking for.
5. Write an equation that relates the various quantities of the problem. If necessary, use the geometry of the situation to eliminate one of the variables by substitution. Use the quantities of your diagram to find the corresponding equation.
6. Use the chain rule to differentiate both sides of the equation with respect to time. It is important that you identify the quantities that change with respect to time, because you must differentiate the equation implicitly with respect to $t$.
7. Substitute the given information into the resulting equation, and solve for the rate you are looking for.

Example 4.62. A ladder $13$ feet long is leaning against the side of a building. If the foot of the ladder is pulled away from the building at a constant rate of $0\text{.}2$ feet per second, how fast is the angle formed by the ladder and the ground changing at the instant when the top of the ladder is $12$ feet above the ground?

We identify the constants and variables that change with respect to time.

$x\text{:}$ distance of the ladder to the wall (variable)

$y\text{:}$ distance from the top of the ladder to the ground (variable)

$\theta \text{:}$ angle formed by the ladder and the ground (variable)

$13$ ft: the length of the ladder (constant)

A picture for this problem is given in Figure 4.20, below.

Figure 4.20. Diagram for Example 4.62

The known rate is ${\displaystyle {\displaystyle \frac{dx}{dt}}=0\text{.}2}$ ft/s.

The rate we want to know is ${\displaystyle {\displaystyle \frac{d\theta }{dt}}}$, when $y=12$ ft.

We must relate the variables $\theta$ and $x$ and the constant $13$. From the diagram we have $\mathrm{cos }\theta =x/13\text{.}$ This is the equation that we differentiate with respect to time:

$\left(-\sin \theta \right){\displaystyle \frac{d\theta }{dt}}={\displaystyle \frac{1}{13}}{\displaystyle \frac{dx}{dt}}\text{.}$

Solving for the rate we are looking for,

${\displaystyle \frac{d\theta }{dt}}=-{\displaystyle \frac{\csc \theta }{13}}{\displaystyle \frac{dx}{dt}}\text{.}$

We have $dx/dt=0.2$, and we need to know $\csc \theta$ when $y=12\text{.}$

We refer to the Figure 4.20, above, and we see, by Pythagoras’ Theorem, that $x=5\text{.}$ Therefore, $\csc \theta =13/12$.

Hence,

${\displaystyle \frac{d\theta }{dt}}=-{\displaystyle \frac{13}{12}}{\displaystyle \frac{1}{13}}\left(0.2\right)=-{\displaystyle \frac{1}{60}}$.

The angle is decreasing at a rate of $1∕60$ rad/sec.

Example 4.63. A water trough with vertical cross section in the shape of an equilateral triangle is being filled at a rate of $4$ cubic feet per minute. Given that the trough is $12$ feet long, how fast is the level of the water rising at the instant the water reaches a depth of $1\text{.}5$ feet?

The constants and variables are given below.

length of the trough: 12 ft,

volume of water: $V$, and

depth of water: $h$.

We have ${\displaystyle {\displaystyle \frac{dV}{dt}}=4}$ ft${}^3$/min, and we want to know ${\displaystyle {\displaystyle \frac{dh}{dt}}}$ when $h=1\text{.}5$ ft.

Figure 4.21. Diagram for Example 4.63

Figure 4.21, above, shows the trough as a whole, and in cross-section.

The volume of the water is the product of the area of the lower isosceles triangle shown in the cross-section times the length of the trough.

The area of the isosceles triangle is equal to half the product of the base times the height.

Figure 4.22. Right-angle triangle, one-half of the lower triangle (cross-section), shown in Figure 4.21

Figure 4.22, above, shows half of the lower triangle, with $b$ being the base. Hence,

${\displaystyle \cot {\displaystyle \frac{\pi }{3}}={\displaystyle \frac{b}{h}}\text{,}\text{ and so, }b=h\cot {\displaystyle \frac{\pi }{3}}={\displaystyle \frac{h}{\sqrt{3}}}\text{.}}$

The area of half of the lower triangle is

${\displaystyle \left({\displaystyle \frac{h}{2\sqrt{3}}}\right)h={\displaystyle \frac{h^2}{2\sqrt{3}}}\text{.}}$

The volume of the water is

${\displaystyle V={\displaystyle \frac{12h^2}{\sqrt{3}}}\text{.}}$

This is the equation that relates the variables and the constants.

Implicitly differentiating with respect to time, we obtain

${\displaystyle {\displaystyle \frac{dV}{dt}}=\left({\displaystyle \frac{24h}{\sqrt{3}}}\right){\displaystyle \frac{dh}{dt}}\text{.}}$

We solve for ${\displaystyle {\displaystyle \frac{dh}{dt}}}$ and get

${\displaystyle {\displaystyle \frac{dh}{dt}}=\left({\displaystyle \frac{\sqrt{3}}{24h}}\right){\displaystyle \frac{dV}{dt}}\text{.}}$

When $h=1\text{.}5$ ft the rate of change of the depth $h$ is

${\displaystyle \frac{dh}{dt}}={\displaystyle \frac{\sqrt{3}}{24\left(1.5\right)}}\left(4\right)={\displaystyle \frac{\sqrt{3}}{9}}\text{ft}\text{/min}\text{.}$

When the depth is $1\text{.}5$ ft, the level of the water is rising at a rate of $\sqrt{3}∕9$ ft/min.

Tutorial 4: Related Rates

Tutorial 5: Related Rates