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Calculus 1 Problems & Solutions – Chapter 5 – Section 5.2 |
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5.2
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Contents
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Review |
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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”.
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2. Rates Of Change That Are Constant |
Example 2.1
The radius of a circle is increasing at a rate of 3 cm/sec. How fast is the circumference of the circle changing?
Solution
EOS
The given rate of change dr/dt of the radius is constant. The
wanted rate of change dC/dt of the
circumference is also
constant. As the radius is increasing at a constant speed, the circumference is
increasing at also a constant speed.
For the solution procedure we remark that:
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.
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3. Rates Of Change That Are Changing |
Example 3.1
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?
Solution
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Fig. 3.1
A circle of radius r and area A. |
EOS
For the solution procedure we remark that there are the following additional events:
i. A figure is drawn.
ii. After differentiation, there's an additional factor r.
iii. The value of r
is also substituted.
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4. Rates Of Change As Signed Quantities |
Example 4.1
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?
Solution
Let x,
y,
and A
be the increasing side, decreasing side, and area of the rectangle at time t
respectively. We have A
= xy.
This yields:
When the increasing side is 12 cm long and the decreasing
one is 10 cm long, the area is decreasing at a rate of
18 cm2/sec.
EOS
The side y
is decreasing, so dy/dt < 0.
That's why in the solution we have dy/dt = – 4, although in the question there's
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 rate
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.
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5. Handling Constants Of Proportionality |
Example 5.1
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?
Solution
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Fig. 5.1
A cylindrical water tank of radius 5 m and height H m, where H > 8. |
Let h
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 that
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.
EOS
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.
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6. Using The Radian Measure For Angles |
Example 6.1
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
A?
Solution
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Fig. 6.1
Lighthouse L is 3 km north of straight shoreline AI. |
EOS
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
4.1.1.4. For example, we could use formulas such as (d/du) tan u = sec2 u only if the angle u is
measured in radians.
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7. When To Differentiate |
Example 7.1
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?
Solution
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
cm long.
EOS
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8. Steps In Solving Related-Rates Problems |
When solving related-rates problems we adopt the following suggested 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.
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Problems & Solutions |
1. Each side of a square is increasing at a rate of 3 cm/sec. How fast is the perimeter changing?
Solution
Let a
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
the wall?
Solution
Let b
be the distance from the base of the wall to the base of the ladder and h the height
of the top of the ladder above
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.
Solution
4. Suppose 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.
Solution
5. An 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?
Solution
Let x
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.
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