Download - Related Rates Examples
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Related Rate Problems
Problem 1: (Depth) A conical (cone-shaped) tank (with vertex down) is 10 feet across the top
and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the
rate of change of the depth of the water when the water is 8 feet deep.
12 ’
h
Figure 1: Conical Tank
Step 1: Identify what is given in the problem.
Top of conical tank has a diameter of 10 feet (radius = 5 feet).
Height of conical tank is 12 feet.
Rate at which the water is flowing into the tank :dV
dt= 10 ft3/min
Want to find the rate of change of the depth of the water when water is
8 feet deep:dh
dt= ?
Step 2: Develop a model or formula
In this problem we have a conical (cone) shaped tank.
V =13πr2h (Volume of a cone where, V = volume, h = height, r = radius)
Step 3: Use the model and the given information to solve the problem
V =13πr2h (1)
Substitute in r =512
h (see similar triangles section below) and simplifying we end up with
V =25432
πh3 (2)
Differentiate both sides of equation(3) with respect to t (time)
dV
dt=
75432
πh2 dh
dt(3)
Equation (3) is called a related rate problem, because there is a relation between the rate at which
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the water is flowing into the tank and the rate of change of the depth of the water.
Solve fordh
dtwhen height of the water is 8 feet (h = 8). Substitute in
dV
dt= 10 and h = 8 into
equation (3),
10 =75432
π(8)2dh
dt(4)
Therefore,dh
dt=
910π
feet/minute.
Similar Triangles:
r
5
12
h
r
5=
h
12
Solving for r, r =512
h
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Problem 2: (Air Traffic Control) An airplane is flying at an altitude of 5 miles and passes directly
over a radar antenna. When the plane is 10 miles away (s = 10), the radar detects that the distance
s is changing at a rate of 240 miles per hour. What is the speed of the plane ?
y
x
s
Figure 2: Airplane
Step 1: Identify what is given in the problem.
Plane is 10 (s = 10) miles away from the radar.
Plane is flying at an altitude of 5 miles. (Let y = 5 )
Rate at which the distance between the plane and radar is changing:ds
dt= 240 miles / hour.
Want to find the the speed of the plane when s = 10 ?
Note: speed = | velocity |.So we really want to find the velocity when s = 10 ?
dx
dt= ?
Step 2: Develop a model or formula
We must apply the Pythagorean Theorem: s2 = x2 + y2
Step 3: Use the model and the given information to solve the problem
Using implicit differentiation,
2sds
dt= 2x
dx
dt+ 2y
dy
dt(5)
Since the plane is flying in the ”x-direction” then there is no change in y, hencedy
dt= 0.
Equation (5) becomes
2sds
dt= 2x
dx
dt(6)
This further reduces to
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sds
dt= x
dx
dt(7)
Looking at the above figure, when s = 10 and y = 5, implies that x =√
102 − 52 = 5√
3.
Substitutingds
dt= 240, s = 10, and x = 5
√3 into equation (7), we get that
dx
dt= 480/
√3 ≈ 277.13
miles / hour. Therefore, the speed is 277.13 miles / hour.
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Problem 3: (Electricity) The combined electrical resistance R of R1 and R2, connected in parallel,
is given by1R
=1
R1+
1R2
(8)
where R, R1, and R2 are measured in ohms. R1 and R2 are increasing at rates of 1 and 1.5 ohms
per second, respectively. At what rate is R changing when R1 = 50 ohms and R2 = 75 ohms ?
Step 1: Identify what is given in the problem.
Increasing rate at resistor 1:dR1
dt= +1 ohms/sec
Increasing rate at resistor 2:dR2
dt= +1.5 ohms/sec
Want to find the the rate of R when R1 = 50 ohms and R2 = 75 ohms:dR
dt= ?
Step 2: Develop a model or formula
In this case we are provided a model to work with.
1R
=1
R1+
1R2
(9)
Step 3: Use the model and the given information to solve the problem
We can re-write equation (9) as
R−1 = R−11 + R−1
2 (10)
Using implicit differentiation on equation we get the following related rate equation,
1R2
dR
dt=
1R2
1
dR1
dt+
1R2
2
dR2
dt(11)
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Solving fordR
dt
dR
dt= R2
(1
R21
dR1
dt+
1R2
2
dR2
dt
)(12)
We are givendR1
dt,
dR2
dt, R1, and R2 but not provided with R. Using equation
(9) we can solve for R:1R
=150
+175⇒ R = 30
Solving fordR
dt:
dR
dt= 302
(1
502(1) +
1752
(1.5))
(13)
Therefore,dR
dt= 0.6 ohms/sec.