3.2: Related Rates

In a related rates problem, we are given the rate of change of one quantity and wish to find the rate of change of another. Such problems can often be solved by implicit differentiation.

Example \(\PageIndex{1}\)

The point of a fountain pen is placed on an ink blotter, forming a circle of ink whose area increases at the constant rate of \(0.03 \ \text{in.}^{2}/\text{sec}\). Find the rate at which the radius of the circle is changing when the circle has a radius of \(\frac{1}{2}\) inch. We solve the problem in four steps.

Solution

Step 1
Label all quantities involved and draw a diagram. \[t = \text{time} \quad A = \text{area} \quad r = \text{radius of circle} \nonumber\]

Both \(A\) and \(r\) are functions of \(t\). The diagram is shown in Figure \(\PageIndex{1}\).

Figure \(\PageIndex{1}\).

Step 2
Write the given information in the form of equations. \[dA/dt = 0.03, \quad A = \pi r^{2}. \nonumber\]

The problem is to find \(dr/dt\) when \(r = 1/2\).

Step 3
Differentiate both sides of the equation \(A = \pi r^{2}\) with respect to \(t\). \[\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}, \text{ whence } 0.03 = 2 \pi r \frac{dr}{dt}. \nonumber\]

Step 4
Set \(r = \frac{1}{2}\) and solve for \(dr/dt\). \[0.03 = 2\pi \frac{1}{2} \cdot \frac{dr}{dt}, \text{ so } \frac{dr}{dt} = \frac{0.03}{\pi} \text{in./sec.} \nonumber\]

Example \(\PageIndex{2}\)

A 10-foot ladder is propped against a wall. The bottom end is being pulled along the floor away from the wall at the constant rate of 2 ft/sec. Find the rate at which the top of the ladder is sliding down the wall when the bottom end is 5 ft from the wall. Warning: although the bottom end of the ladder is being moved at a constant rate, the rate at which the top end moves will vary with time.

Solution

Step 1
\(\begin{align*} t &= \text{time}, \\ x &= \text{distance of the bottom end from the wall}, \\ y &= \text{height of the top end above the floor.} \end{align*}\)

The diagram is shown in Figure \(\PageIndex{2}\).

Figure \(\PageIndex{2}\).

Step 2
\(dx/dt = 2, \quad x^{2} + y^{2} = 10^{2} = 100\).

Step 3
We differentiate both sides of \(x^{2} + y^{2} = 100\) with respect to \(t\). \[2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0, \text{ whence } 4x + 2y \frac{dy}{dt} = 0. \nonumber\]

Step 4
Set \(x = 5 \ \text{ft}\) and solve for \(dy/dt\). We first find the value of \(y\) when \(x = 5\). \[x^{2} + y^{2} = 100, \quad y = \sqrt{100 - x^{2}} = \sqrt{100 - 5^{2}} = \sqrt{75}. \nonumber\]

Then we can solve for \(dy/dt\), \[\begin{align*} 4x + 2y \frac{dy}{dt} &= 0, \\ 4 \cdot 5 + 2 \sqrt{75} \frac{dy}{dt} = 0, \\ \frac{dy}{dt} &= - \frac{4 \cdot 5}{2 \sqrt{75}} = - \frac{2}{\sqrt{3}} \ \text{ft/sec} \end{align*}\]

The sign of \(dy/dt\) is negative because \(y\) is decreasing.

Related rates problems have the following form.

Given:
(a) Two quantities which depend on time, say \(x\) and \(y\).
(b) The rate of change of one of them, say \(dx/dt\).
(c) An equation showing the relationship between \(x\) and \(y\).

(Usually this information is given in the form of a verbal description of a physical situation and part of the problem is to express it in the form of an equation.)

The problem:
Find the rate of change of \(y\), \(dy/dt\), at a certain time \(t_{0}\). (The time \(t_{0}\) is sometimes specified by giving the value which \(x\), or \(y\), has at that time.) Related rates problems can frequently be solved in four steps as we did in the examples.

Step 1: Label all quantities in the problem and draw a picture. If the labels are \(x\), \(y\), and \(t\) (time). The remaining steps are as follows:

Step 2: Write an equation for the given rate of change \(dx/dt\). Write another equation for the given relation between \(x\) and \(y\).

Step 3: Differentiate both sides of the equation relating \(x\) and \(y\) with respect to \(t\). We choose the time \(t\) as the independent variable. The result is a new equation involving \(x\), \(y\), \(dx/dt\), and \(dy/dt\).

Step 4: Set \(t = t_{0}\) and solve for \(dy/dt\). It may be necessary to find the values of \(x\), \(y\), and \(dx/dt\) at \(t = t_{0}\) first.

The hardest step is usually Step 2, because one has to start with the given verbal description of the problem and set it up as a system of formulas. Sometimes more than two quantities that depend on time are given. Here is an example with three.

Example \(\PageIndex{3}\)

One car moves north at 40 mph (miles per hour) and passes a point \(P\) at time 1:00. Another car moves east at 60 mph and passes the same point \(P\)
at time 2:30. How fast is the distance between the two cars changing at the time 2:00?

Solution

It is not even easy to tell whether the two cars are getting closer or farther away at time 2:00. This is part of the problem.

Step 1
\(\begin{align*} t &= \text{time}, \\ y &= \text{position of the first car tra veiling north,} \\ x &= \text{position of the second car travelling east,} \\ z &= \text{distance between the two cars.} \end{align*}\)

In the diagram in Figure \(\PageIndex{3}\), the point \(P\) is placed at the origin.

Figure \(\PageIndex{3}\).

Step 2
\[\frac{dy}{dt} = 40, \quad \frac{dx}{dt} = 60, \quad x^{2} + y^{2} = z^{2}. \nonumber\]

Step 3
\[2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt}, \text{  whence  } 60x + 40y = z \frac{dz}{dt}. \nonumber\]

Step 4
We first find the values of \(x\), \(y\), and \(z\) at the time \(t = 2\) hrs. We are given that when \(t = 1\), \(y = 0\). In the next hour the car goes 40 miles, so at \(t = 2\), \(y = 40\). We are given that at time \(t = 2 \frac{1}{2}\), \(x = 0\). One-half hour before that the car was 30 miles to the left of \(P\), so at \(t = 2\), \(x = -30\). To sum up, \[\text{at} \ t = 2, \ y = 40 \text{ and } x = -30. \nonumber\]

We can now find the value of \(z\) at \(t = 2\), \[z = \sqrt{x^{2} + y^{2}} = \sqrt{(-30)^{2} + 40^{2}} = 50. \nonumber\]

Finally, we solve for \(dz/dt\) at \(t = 2\). \[60 \cdot (-30) + 40 \cdot 40 = 50 \frac{dz}{dt}, \quad \frac{dz}{dt} = \frac{-1800 + 1600}{50} = -4. \nonumber\]

The negative sign shows that \(z\) is decreasing. Therefore at 2:00 the cars are getting closer to each other at the rate of 4 mph.

Example \(\PageIndex{4}\)

The population of a country is growing at the rate of one million people per year, while gasoline consumption is decreasing by one billion gallons per year. Find the rate of change of the per capita gasoline consumption when the population is 30 million and total gasoline consumption is 15 billion gallons per year.

By the per capita gasoline consumption we mean the total consumption divided by the population.

Solution

Step 1
\(\begin{align*} t &= \text{time} \\ x &= \text{population} \\ y &= \text{gasoline consumption} \\ z &= \text{per capita gasoline consumption.} \end{align*}\)

Step 2
At \(t = t_{0}\), \[\begin{align*} dx/dt &= 1 \ \text{million} = 10^{6} \\ dy/dt &= -1 \ \text{billion} = -10^{9} \\ z &= y/x. \end{align*}\]

Step 3
\[\begin{align*} \frac{dz}{dt} &= \frac{x (dy/dt) - y (dx/dt)}{x^{2}}, \\[4pt] \frac{dz}{dt} &= \frac{-10^{9} x - 10^{6}y}{x^{2}}. \end{align*}\]

Step 4
At \(t = t_{0}\), we are given \[\begin{align*} x &= 30 \ \text{million} = 30 \times 10^{6}. \\ y &= 15 \ \text{billion} = 15 \times 10^{9}. \end{align*}\]

Thus \[\begin{align*} \frac{dz}{dt} &= \frac{-10^{9} \cdot 30 \cdot 10^{6} - 10^{6} \cdot 15 \cdot 10^{9}}{(30 \cdot 10^{6})^{2}} \\ &= - \frac{45 \cdot 10^{15}}{900 \cdot 10^{12}} = -50. \end{align*}\]

The per capita gasoline consumption is decreasing at the annual rate of 50 gallons per person.

We conclude with another example from economics. In this example the independent variable is the quantity \(x\) of a commodity. The quantity \(x\) which can be sold at price \(p\) is called the demand function \(D(p)\), \[x = D(p). \nonumber\]

When a quantity \(x\) is sold at price \(p\), the revenue is the product \[R = px. \nonumber\]

The additional revenue from the sale of an additional unit of the commodity is called the marginal revenue and is given by the derivative \[\text{marginal revenue} = dR/dx. \nonumber\]

Example \(\PageIndex{5}\)

Suppose the demand for a product is equal to the inverse of the square of the price. Find the marginal revenue when the price is $10 per unit. 

Solution

Step 1
\(p = \text{price}, \quad x = \text{demand}, \quad R = \text{revenue}.\)

Step 2
\(x = 1/p^{2}, \quad R = px.\)

Step 3
\[\begin{align*} \frac{dR}{dx} &= p \frac{dx}{dx} + x \frac{dp}{dx} = p + x \frac{dp}{dx}, \\ \frac{dx}{dp} &= -2p^{-3}, \end{align*}\]

so by the Inverse Function Rule, \[\frac{dp}{dx} = \frac{1}{dx/dp} = - \frac{1}{2p^{-3}} = -\frac{1}{2} p^{3}. \nonumber\]

Substituting, \[\frac{dR}{dx} = p + \left(\frac{1}{p^{2}}\right) \left(-\frac{1}{2} p^{3}\right) = \frac{1}{2} p. \nonumber\]

Step 4
We are given p = $10. Therefore the marginal revenue is \[dR/dx = $5. \nonumber\]

An additional unit sold would bring in an additional revenue of $5.

Here is a list of formulas from plane and solid geometry which will be useful in related rates problems. We always let \(A\) = area and \(V\) = volume.

Rectangle with sides \(a\) and \(b\): \(A = ab\), \(\text{perimeter} = 2a + 2b\)
Triangle with base \(b\) and height \(h\): \(A = \frac{1}{2} bh\)
Circle of radius \(r\): \(A = \pi r^{2}\), \(\text{circumference} = 2 \pi r\)
Sector (pie slice) of a circle of radius \(r\) and central angle \(\theta\) (measured in radians): \(A = \frac{1}{2} r^{2} \theta\)
Rectangular solid with sides \(a\), \(b\), \(c\): \(V = abc\)
Sphere of radius \(r\): \(V = \frac{4}{3} \pi r^{3}\), \(A = 4 \pi r^{2}\)
Right circular cylinder, base of radius \(r\), height of \(h\): \(V = \pi r^{2}h\), \(A = 2\pi rh\)
Prism with base of area \(B\) and height \(h\): \(V = Bh\)
Right circular cone, base of radius \(r\), height \(h\): \(V = \pi r^{2} h/3\), \(A = \pi r \sqrt{r^{2} + h^{2}}\)

Problems for Section 3.2

1.	Each side of a square is expanding at the rate of 5 cm/sec. How fast is the area changing when the length of each side is 10 cm?
2.	The area of a square is decreasing at the constant rate of 2 sq cm/sec. How fast is the length of each side decreasing when the area is 1 sq cm?
3.	The vertical side of a rectangle is expanding at the rate of 1 in./sec, while the horizontal side is contracting at the rate of 1 in./sec. At time \(t = 1\) sec the rectangle is a square whose sides are 2 in. long. How fast is the area of the rectangle changing at time \(t = 2\) sec?
4.	Each edge of a cube is expanding at the rate of 1 in./sec. How fast is the volume of the cube changing when the volume is 27 in.^3?
5.	Two cars pass point \(P\) at approximately the same time, one traveling north at 50 mph, the other traveling west at 60 mph. Find the rate of change of the distance between the two cars one hour after they pass the point \(P\).
6.	A cup in the form of a right circular cone with radius \(r\) and height \(h\) is being filled with water at the rate of 5 in.^3/sec. How fast is the level of the water rising when the volume of the water is equal to one half the volume of the cup?
7.	A spherical balloon is being inflated at the rate of 10 in.^3/sec. Find the rate of change of the area when the balloon has radius 6 in.
8.	A snowball melts at the rate equal to twice its surface area, with area in square inches and melting measured in cubic inches per hour. How fast is the radius shrinking?
9.	A ball is dropped from a height of 100 ft, at which time its shadow is 500 ft from the ball. How fast is the shadow moving when the ball hits the ground? The ball falls with velocity \(gt\) ft/sec, and the shadow is cast by the sun. Here \(g = 32\) and \(t\) = time after drop.
10.	A 6-foot-tall man walks away from a 10-foot-high lamp at the rate of 3 ft/sec. How fast is the tip of his shadow moving?
11.	A car is moving along a road at 60 mph. To the right of the road is a bush 10 ft away and a parallel wall 30 ft away. Find the rate of motion of the shadow of the bush on the wall cast by the car headlights.
12.	A car moves along a road at 60 mph. There is a bush 10 ft to the right of the road, and a wall 30 ft behind the bush is perpendicular to the road. Find the rate of motion of the shadow of the bush on the wall when the car is 26 ft from the bush.
13.	An airplane passes directly above a train at an altitude of 6 miles. If the airplane moves north at 500 mph and the train moves north at 100 mph, find the rate at which the distance between them is increasing two hours after the airplane passes over the train.
14.	A rectangle has constant area, but its length is growing at the rate of 10 ft/sec. Find the rate at which the width is decreasing when the rectangle is 3 ft long and 1 ft wide.
15.	A cylinder has constant volume. but its radius is growing at the rate of 1 ft/sec. Find the rate of change of its height when the radius and height are both 1 ft.
16.	A country has constant national income, but its population is growing at the rate of one million people per year. Find the rate of change of the per capita income (national income divided by population) when the population is 20 million and the national income is 20 billion dollars.
17.	If at time \(t\) a country has a birth rate of \(1,000,000 t\) births per year and a death rate of \(300,000 \sqrt{t}\) deaths per year, how fast is the population growing?
18.	The population of a country is 10 million and is increasing at the rate of 500,000 people per year. The national income is $10 billion and is increasing at the rate of $100 million per year. Find the rate of change of the per capita income.
19.	Work Problem 18 assuming that the population is decreasing by 500,000 per year.
20.	Sand is poured at the rate of 4 in.^3/sec and forms a conical pile whose height is equal to the radius of its base. Find the rate of increase of the height when the pile is 12 in. high.
21.	A circular clock has radius 5 in. At time \(t\) minutes past noon, how fast is the area of the sector of the circle between the hour and minute hand increasing? \((t \leq 60)\).
22.	The demand \(x\) for a commodity at price \(p\) is \(x = 1/\left(1 + \sqrt{p}\right)\). Find the marginal revenue, that is, the change in revenue per unit change in \(x\), when the price is $100 per unit.
23.	\(x\) units of a commodity can be produced at a total cost of \(y = 100 + 5x\). The average cost is defined as the total cost divided by \(x\). Find the change in average cost per unit change in \(x\) (the marginal average cost) when \(x = 100\).
24.	The demand for a commodity at price \(p\) is \(x = 1/\left(p + p^{3}\right)\). Find the change of the price per unit change in \(x\), \(dp/dx\), when the price is 3 dollars per unit.
25.	In one day a company can produce \(x\) items at a total cost of \(200 + 3x\) dollars and can sell \(x\) items at a price of \(5 - x/1000\) dollars per item. Profit is defined as revenue minus cost. Find the change in profit per unit change in the number of items x (marginal profit).
26.	In one day a company can produce \(x\) items at a total cost of \(200 + 3x\) dollars and can sell \(x = 1000/\sqrt{y}\) items at a price of \(y\) dollars per item. (a) Find the change in profit per dollar change in the price \(y\) (the marginal profit with respect to price). (b) Find the change in profit per unit change in \(x\) (the marginal profit).
27.	An airplane \(P\) flies at 400 mph one mile above a line \(L\) on the surface. An observer is at the point \(O\) on \(L\). Find the rate of change (in radians per hour) of the angle \(\theta\) between the line \(L) and the line \(OP\) from the observer to the airplane when \(\theta = \pi/6\).
28.	A train 20 ft wide is approaching an observer standing in the middle of the track at 100 ft/sec. Find the rate of increase of the angle subtended by the train (in radians per second) when the train is 20 ft from the observer.
29.	Find the rate of increase of \(e^{2x+ 3y}\) when \(x = 0\), \(y = 0\), \(dx/dt = 5\), and \(dy/dt = 4\).
30.	Find the rate of change of \(\ln A\) where \(A\) is the area of a rectangle of sides \(x\) and \(y\) when \(x = 1\), \(y = 2\), \(dx/dt = 3\), \(dy/dt = -2\).