Related Rates
Problem 1:
Water is leaking out of an inverted conical tank at a rate of 10,000 cm3/min at the
same time that water is being pumped into the tank at a constant rate. The tank has a
height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of
20cm/min when the height of the water is 2 m, find the rate at which water is being
pumped into the tank.
Tank:
Givens:
Height = h = 2 m or 200 cm
Diameter= 4 m
Radius= 2m
C= rate of water being pumped in= dV/dt= C – 10,000
V= (1/3)πr2h
dh/dt= 20 cm3/min
To solve for one variable use similar triangles to solve for r in
terms of h.
(r/2)= (h/6)
r = (h/6)•2
r = (1/3)h
Now
V= (1/3) [(1/3)h]2 hπ
= (1/3) [(1/9)h2]hπ
= (1/27) h3π
Now take the derivative of the new and improved equation
dV/dt = (π/9)h2 dh/dt
Plug in the known variables
h, dh/dt, dV/dt
C – 10,000 = (π/9) (200)2(20)
C – 10,000 = 800,000π/9
C = (800,000π/9) + 10,000 ≈ 289,253 cm3/ min
Problem 2:
A water trough is 10 m long and a cross-section has the shape of an isosceles
trapezoid that is 30 cm wide at the bottom, 80 cm wide at the top and has a height 50 cm.
If the trapezoid is being filled with water at the rate of 0.2 m3/min, how fast is the water
level rising when the water is 30 cm deep?
Trapezoid:
0.25
0.3
Givens:
0.25
a
0.5
Area of a trapezoid = (1/2) (b1+ b2) h
then the V of the trough = 10A
h
dV/dt = 20 cm3/ min
dh/dt = ?
a= 0.25 m
h = 0.5 m
0.3
Solution:
V = (1/2) [.03 + (0.3 + 2a)]h (10)
V = 5 (0.6 + h)h
= 3h + 5h2
use the equation (a/h) = (0.25/.05) =(1/2)
So 2a = 2(0.5h) = h
0.2 = dV/dt = (3 + 10h)(dh/dt)
then separate the variables dh/dt and h
dh/dt = 0.2/ (3 +10h)
Next, allow h = .3 m as stated in the problem and then solve.
dh/dt = 0.2/ (3 + 10(0.3))
= 0.2/6 m/min
= 10/3 cm/min
Problem 3:
A man walks along a straight path at a speed of 4 m/s. A searchlight is located on
the ground 20 m from the path and is kept focused on the map. At what rate is the
searchlight rotating when the man is 15 m from the point on the path closest to the
searchlight?
Picture:
x
20
θ
searchlight
Let x be distance from point of
path closest to the searchlight
to the man. θ is the angle between
the beam of searchlight and the
perpendicular to the path.
Givens:
dx/dt = 4 m/s
Find d θ /dt = ? , when x = 15
Use the following equation to relate x and θ :
x/20 = tan θ
x = 20 tan θ
Differentiate both sides of the equation with respect to t to find d θ /dt.
dx/dt = 20 sec2 θ d θ /dt
d θ /dt = (1/20) cos2 θ dx/dt
= (1/20) cos2 θ (4)
= (1/5) cos2 θ
When x = 15 then the length of the beam is 25, so cos θ = (4/5) and
d θ /dt = (1/5)(4/5)2
= 16/125
= 0.128
The searchlight is rotating at 0.128 rad/s.
Problem 4:
Two people start from the same point. One walks east at 3 mi/h and the other
walks northeast at 2 mi/h. How fast is the distance between the people changing after 15
minutes?
Picture:
y
We are given that
dx/dt = 3 mi/h
dy/dt = 2 mi/h
z
45°
Use the Law of Cosines to find and equation to find dz/dt.
x
z2 = x2 + y2 – 2xycos 45°
z2 = x2 + y2 - √2xy
Then differentiate
2z(dz/dt) = 2x(dx/dt) + 2y(dy/dt) -√2x(dy/dt) - √2y(dx/dt)
After 15 minutes,
x = ¾ and y = ½ so z = [√(13 - 6√2)]/4
Lastly insert all the known variables and slove for dz/dt.
2[√(13 - 6√2)]/4 (dz/dt) = 2(3/4)(3) + 2(1/2)(2) - √2(3/4)(2) - √2(1/2)(3)
[√(13 - 6√2)]/2 (dz/dt) = (18/4) + 2 – (2/3)√2 – (3/2)√2
dz/dt ≈ 2.125 mi/h
Problem 5:
Boyle’s Law states that when a sample of gas is compressed at a constant
temperature, the pressure P and volume V satisfy the equation PV = C, where C is a
constant. Suppose that at a certain instant the volume is 600 cm3, the pressure is 150 kPa,
and the pressure is increasing at a rate of 20 kPa/min at what rate is the volume
decreasing at this instant?
Givens:
V= 600 cm3
P= 150 kPa
dP/dt = 20 kPa/min
PV = C
Differentiate the given equation.
P(dV/dt) + V(dP/dt) = 0
dV/dt = -[V(dP/dt)]/P
Plug in the givens and solve.
dV/dt = -[600(20)]/150
dV/dt = -80 cm3/min