1. Related Rates
In general, when two or more quantities are related to each
other, their rates of change with respect to time (speeds)
are also related to each other. In this section, we'll solve problems of finding a rate of change with respect to time by
searching for how it's related to one or more other rates of change with respect to time that are known or easily
determined. That is, we'll solve problems of related rates.
For the remainder of this section, when we say rate of change
without specifying what it's with respect to, we mean
rate of change with respect to time.
2. Rates Of Change That Are Constant
The radius of a circle is increasing at a rate of 3 cm/sec. How fast is the circumference of the circle changing?
The given rate of change dr/dt of the radius is constant.
The wanted rate of change dC/dt of the circumference is
constant. As the radius is increasing at a constant speed, the circumference is increasing at also a constant speed.
i. We define the symbols r and C
to assign to the radius and circumference respectively.
ii. We are to find the rate of change dC/dt of C.
iii. The rate of change dr/dt of r is known. So we determine an equation relating C to r.
iv. We differentiate that equation with respect to time. The Leibniz notation is used.
v. We substitute the value of dr/dt and obtain the value of dC/dt.
vi. We conclude with a statement answering the question asked. We specify that C is increasing, not just changing. It's
increasing because dC/dt > 0.
Using The Leibniz Notation
It's a common practice to use the Leibniz notation in
related-rates problems. The reason is that it's suggestive of the fact
that the derivative is the rate of change.
3. Rates Of Change That Are Changing
The radius of a circle is increasing at a rate of 3 cm/sec.
How fast is the area of the circle changing when the radius is
5 cm long?
A circle of radius r and area A.
i. A figure is drawn.
ii. After differentiation, there's an additional factor r.
iii. The value of r is also substituted.
4. Rates Of Change As Signed Quantities
One side of a rectangle is increasing at a rate of 3 cm/sec
and the other side is decreasing at a rate of 4 cm/sec. How
fast is the area of the rectangle changing when the increasing side is 12 cm long and the decreasing side is 10 cm long?
Let x, y, and A be the increasing side, decreasing side, and area of the rectangle at time t respectively. We have A = xy.
When the increasing side is 12 cm long and the decreasing
one is 10 cm long, the area is decreasing at a rate of
The side y
is decreasing, so dy/dt < 0. That's why in the
solution we have dy/dt = 4, although in the
no minus () sign. If we didn't take care of this aspect, our answer would be wrong.
The resulting value of dA/dt is negative. So in the concluding statement,
we specify that the area is decreasing at a
of 18 cm2/sec. Note that we use the positive 18 cm2/sec instead of the negative 18 cm2/sec in the statement.
Remember, rates of change are signed quantities. Rates of
increase are positive and rates of decrease are negative. And
in calculations don't forget the minus sign () for negative quantities.
5. Handling Constants Of Proportionality
A water tank is in the shape of a cylinder with radius 5 m
and height greater that 8 m. Water leaks out of the tank at a
rate proportional to the depth of the water in the tank. When the water in the tank is 8 m deep, it's leaking at 0.2 m3/min.
How fast is the water level in the tank changing at that time?
A cylindrical water tank of radius 5 m and height H m, where H > 8.
and V be the
depth and volume of the water in the tank at time t respectively. Refer to Fig. 5.1. We are given
dV/dt = kh, where k is a constant. Since dV/dt = 0.2 when h = 8, we get 0.2 = k(8), so that k = ( 0.2)/8 =
0.025. This yields dV/dt = (0.025)h.
Hence, the water level in the tank is dropping at
approximately 2.5 mm/min when the water in the tank is 8 m deep.
The problem statement states that water leaks out of the
tank, meaning that the volume of the water in the tank
decreases, at a rate proportional to the depth of the water in the tank. This means that dV/dt = kh, where k is the
constant of proportionality. (Because dV/dt < 0 and h > 0, we must have had k < 0. As expected, we did.) The problem
statement gives us a particular value of h and the corresponding value of dV/dt. We utilize this gift and the equation
dV/dt = kh to find the value of k.
6. Using The Radian Measure For Angles
A lighthouse is located on a tiny island 3 km north of a
point A on a straight
shoreline. The lighthouse lamp rotates at 5
revolutions per minute. How fast is the illuminated spot on the shoreline moving along the shoreline when it is 6 km from
Lighthouse L is 3 km north of straight shoreline AI.
We have to convert revolutions/min to rad/min. In general,
angles must be measured in radians when employing
differentiation formulas. This is because trigonometric differentiation formulas are obtained by measuring angles in
radians. See Section 6.1.4. For example, we could use formulas such as (d/du) tan u = sec2 u only if the angle u is
measured in radians.
7. When To Differentiate
Air is being pumped into a spherical balloon. Suppose the volume
of the balloon is increasing at a rate of 400 cm3/sec
when the radius is 30 cm. How fast is the radius increasing at that time?
Let r and V be the radius and volume of the balloon at time t respectively. So:
The radius is increasing at about 0.035 cm/sec when it's 30
8. Procedure Of Solving Related-Rates Problems
When solving related-rates problems we adopt the suggested procedure with the following steps:
i. Draw a figure when appropriate.
ii. Define letters and symbols to assign to quantities and variables.
iii. Determine an equation relating the quantity whose rate is to find to the quantity whose rate is known.
iv. Differentiate that equation with respect to time.
Use the Leibniz notation. Angles are in radians.
v. Substitute the values of given and known quantities. Angles must be in radians. Rates of change are signed quantities.
Rates of increase are positive and rates of decrease are negative.
vi. Solve for the required rate.
vii. Conclude with a statement answering the question asked.
Problems & Solutions
1. Each side of a square is increasing at a rate of 3 cm/sec. How fast is the perimeter changing?
and P be the
side and perimeter of the square respectively. We have P = 4a,
so that dP/dt = 4 da/dt
= 4 x 3 =
12 cm/sec. The perimeter is increasing at 12 cm/sec.
2. A ladder
is 10 m long. Its top is slipping down along a vertical wall while its base is
being pulled away from the base
of the wall at a speed of 1/4 m/sec. How fast is the top of the ladder slipping down when it is 6 m above the base of
be the distance from the base of the wall to the base of the ladder and h the height of the top of the
the base of the wall, both at time t. We have b2 + h2 = 102 = 100. So:
The top of the ladder is slipping down at 1/3 m/sec when it's 6 m above the base of the wall.
that when a hard candy ball is dropped in a glass of water, it dissolves at a
rate proportional to its surface
area. Prove that the radius of the ball decreases at a constant rate.
airplane is flying due east at an altitude of 2 km at a speed of 410 km/h. At some
instant it passes directly above a
car travelling due southeast on a straight level road at 100 km/h. How fast is the distance between the airplane and
the car increasing 30 seconds later?
and y be the
distances travelled by the airplane and the car respectively, and s the distance separating
them, all at
t hours after the airplane passes directly above the car at time t = 0. We have:
Hence, 30 seconds later the distance between the airplane and the car is increasing at approximately 285 km/h.