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Contents
Rates of Change
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Rates of Change
The application of the derivative is not just limited to geometric representation on the cartesian plane. It can also represent the rates of change of variables with respect to another variable, usually time. An example is the rate of change of area A with respect to time t. Say A and t are related by the expression A = 4t2 + 1. The rate of change of area would be dA/dt = 8t. Chain Rule
dr/dt = 3 cm/s (given) A = pr2 formula for area of circle
dA/dr = 2pr
when r = 2, dA/dr = 4p
Rate of change of area,
= 4p
.3 = 12p
cm2/s
 A
cylindrical jar of radius 4 cm contains water to a depth of 5 cm. The water is
then poured at a steady rate into a hemispherical bowl. After t seconds, the
depth of water in the bowl is x cm and the volume, V cm3, of the
water that has been transferred is given by V = 1/3px2
(18 - x) cm3.
Given
that all the water is transferred in 10 seconds, find
(i) dV/dt
(ii) the rate at which x is increasing when x = 2.
Solution:
The
quantities required to calculate volume of water are given.
Volume of water in cylinder = p(4)25
= 80p
It
is also given that all the water is transferred in 10 seconds.
dV/dt = 80p/10
= 8p
We
are given 2 variables, V and x. Hence the chain rule must come in.
Since
we are required to find dx/dt, we need to find dV/dx. We are given the
equation relating V and x.
V = 1/3px2
(18 - x) = 6px2
- 1/3px3
dV/dx = 12px
- px2
Substitute
the expressions into the chain rule,
8p
= (12px
- px2)
dx/dt
when x = 2,
dx/dt = 2/5 cm/s
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Usually
in problems involving rates of change, two variables are employed. The chain
rule can be used to link the two variables.