|
1. Rates of Change
If y is a function of t, then the rate of change of y with respect to t is denoted by \(dy \over dt\)
Differentiation is applied to calculate rate of a quantity that cannot be measured directly whereby an equation relating the involved variables is differentiated to get another equation relating the rate to be calculated from the known rates.
The derivative with respect to time represent the rates of change whereby a positive derivative indicates an increment whilst a negative derivative indicates a decrement in the quantity.
2. Steps in finding rate of change
Step 1
Sketch a diagram and name all the variables and constants. All variables are differentiable functions of time, t.
Step 2
Write down the numerical information using mathematical symbols.
(Step 2 is usually combined with Step 1)
Step 3
Form an equation that relates the variables.
Step 4
Differentiate with respect to t. Express the rate to be calculated in terms of the rate and variables known.
Step 5
Evaluate using the given values.
Example
The radius of a circular plate of metal being heated in an oven increases at the rate of 0.01cm/min. At what rate is the plate’s area increasing when the radius is 50 cm?
Step 1 & Step 2
Step 3
Step 4
| \({dA \over dt}= {2 \pi r}{dr \over dt}\) |
Step 5
| \({dA \over dt}=2π(50)(0.01) = \pi \text { }cm^2/min \) |
Example
A cylindrical shaped cup is 12 cm in height and 5 cm in radius. If the cup is being filled up at the rate of 1 cm3/min, how fast is the water level in the cup rising when the cup has been filled 3 cm deep?
Step 1 & Step 2
Step 3
Note that radius is constant throughtout the cup that is r=5
Step 4
|
\(1=25π {dh \over dt} \)
\({dh \over dt}={1 \over 25π} cm/min\)
|
In this case, the water level in the cup is rising at a constant rate.
Example
A cone shaped tank with a height of 10 ft and a base radius of 5 ft is placed on the ground point down. If water is poured into the tank at a rate of 9 ft3/min, how fast is the water level rising when the water is 6 ft deep?
Step 1 & Step 2
Step 3
| \(V={4 \over 3} \pi r^3 \) |
Using the ratio of the radius and height of the cup, \({r \over h}={5 \over 10}\) that is \(r={h \over 2}\).
Therefore, volume in terms of the height is given by
|
\(V={4 \over 3} \pi ({h \over 2})^3 \)
\(={\pi h^3 \over 6}\)
|
Step 4
| \({dV \over dt}= {\pi h^2 \over 2}{dh \over dt}\) |
Step 5
|
\(9= \pi {(6)^2 \over 2}{dh \over dt} \)
\( {dh \over dt}={1 \over 2π} ft/min \)
|
The above questions can be solved without drawing the diagrams. The diagram helps you to visualize the question better but is not absolutely necessary. In some instances, a diagram cannot be drawn.
Example
Variables x and y are connected by the equation 2xy=195. If x is increasing at a rate of 1.5 units/sec, find the rate of change of y at the instant when x=15 units.
|
\(y={195 \over 2x}\)
\({dy \over dt}=-{97.5 \over x^2}{dx \over dt} \)
\( =-{97.5 \over 15^2} (1.5) \)
\( =-0.65 units/sec \)
|
Example
A study of a certain community shows that there will be
|
\(Q(p)=p^2+4p+900 \text { }units \)
|
of a harmful pollutant in the air when the population is p people. If the population is currently 20, 000 and is increasing at the rate of 15 per year, at what rate is the level of pollution increasing?
|
\({dQ \over dt}=(2p+4){dp \over dt} \)
\( =2×20000+415 \)
\( =600060 \text { } people/year \)
|
|