The derivative of a function is simply referred to as the rate of change of the function. If a quantity y depends on and varies with a quantity x, then the rate of change of y with respect to x is \( \frac{dy}{dx} \)
For instance, the rate of change of pressure p with respect to height h is
\( \frac{dp}{dh} \)Example:
1. Suppose the amount of water in a holding tank at time t is given by
v(t) = 2t2 – 16t + 35
Determine;
a. if the volume of water in the tank is increasing or decreasing at time t = 1m
b. if the volume of water of in the tank is increasing or decreasing at time t = 5m
c. time at which the volume of water in the tank is not changing.
Solution
a. v(t) = 2t2 – 16t + 35
\( \frac{dv}{dt} \) = 4t – 16
\( \frac{dv}{dt} \) = 4(1) – 16
\( \frac{dv}{dt} \) = -12
At t = 1, \( \frac{dv}{dt} \) = -12
Rate of change is negative, therefore the volume is decreasing.
b. \( \frac{dv}{dt} \) = 4t – 16
At t = 5
\( \frac{dv}{dt} \) = 4(5) – 16
\( \frac{dv}{dt} \) = 20 – 16
\( \frac{dv}{dt} \) = 4
Rate of change is positive, therefore the volume of water is decreasing.
c. When the volume of water is constant,
\( \frac{dv}{dt} \) = 4t – 16 = 0
4t – 16 = 0
4t = 16
t = \( \frac {16}{4} \)
t = 4m
At time 4m there was no change in volume of water in the tank.
Example
2. The distance x meters moved by a car in a time t seconds is given by
x = 3t3 – 2t2 + 4t – 1
Determine the velocity and acceleration when;
i. t = 0
ii. t = 1.5s
Solution:
x = 3t3 – 2t2 + 4t – 1
Velocity, v = \( \frac {dx}{dt} \scriptsize = 9t^2 \; – \; 4t + 4 \)
Acceleration, a = \( \frac {dv}{dt} \scriptsize = 18t \; – \; 4\)
i. when t = 0
velocity, v = 9(0) – 4(0) + 4 = 4m/s
acceleration, a = 18(0) – 4 = -4ms-2
ii. when t = 1.5s
velocity, v = 9(1.5)2 – 4(1.5) + 4 = 18.25m/s
acceleration, a = 18(1.5) – 4 = 23ms-2
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