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Related Rates

Computing related rates requires use of the chain
rule:
dy/dx = (dy/du)(du/dx)
Example 1.
A swimming pool is 20m long, 10m wide, 4m deep at the deep
end and 2m deep at the shallow end. If water is pumped into
the pool at the rate of 2 m^{3}/min, how fast is the
water level rising when it is 1.5m deep at the deep end?
^ < L >
   D'
  .
D  L' .
 .
 h  .
v .
D = 4
D' = 2
L = 20
W = 10
dV/dt = 2
Using the property of similar triangles:
D/L = h/L' ∴ 4/20 = h/L' or L' = 5h
Volume of water, V = (1/2)hL' * 10
= 25h^{2}
dV/dt = (dV/dh)(dh/dt) = 50hdh/dt
dh/dt = (dV/dt)/50h = 2/(50*1.5) =
Example 2.
A water tank in the shape of a right circular cone has
a height of 10 feet. The top rim of the tank is a circle
with a radius of 4 feet. If water is being pumped into
the tank at the rate of 2 ft^{3}/min, what is the rate of
change of the water depth, in feet per minute, when the
depth is 5 feet?
<R>
\ r / ^
\/ 
\ / ^ H
\ / h 
\/ v v
H = 10
R = 4
dV/dt = 2
Using the property of similar triangles:
H/R = h/r
∴ r = hR/H = 4h/10
Volume of water, V = (π/3)r^{2}h = (4π/75)h^{3}
dV/dt = (dV/dh)(dh/dt)
= (4π/25)h^{2}(dh/dt)
∴ dh/dt = (dV/dt)/(4π/25)h^{2} = 1/2π
Example 3.
A ladder 20 feet long leans against a building. If the
bottom of the ladder slides away from the building
horizontally at a rate of 4 ft/sec, how fast is the
ladder sliding down the house when the bottom of the
ladder is 8 feet from the wall.
y

 
v \
 \
 \h
 \
 \
 \
> x
>
h = 17
dx/dt = 4
x = 8
h^{2} = x^{2} + y^{2} ... Pythagoras
y^{2} = h^{2}  x^{2}
∴ y = √(h^{2}  x^{2})
dy/dt = (dy/dx)(dx/dt)
= x/√(h^{2}  x^{2})(dx/dt)
Note: we could have also used implicit differentiation
to get:
2ydy/dt = 2hdh/dt  2xdx/dt
dh/dt = 0 (the ladder is not changing in length!)
∴ dy/dt = (x/y)dx/dt
= (x/√(h^{2}  x^{2})(dx/dt)
= (8/√(20^{2}  8^{2})4
= 32/√(336)