Differentiation - Application of Differentiation, Rate of Change

Differentiation – Application of Differentiation, Rate of Change

January 26, 2012 in Additional Mathematics, Application of Derivation, Differentiation by Shivakkumar Vadiveyl

Chapter 3 Section K Differentiation – Application of Differentiation, Rate of Change

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One of the basic application of rate of change is with regards to motion of an object.

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When an object moves along a line, the distance travelled by the object changes with respect to time. As the time increases the distance will also increase.

For example, the equation for the distance travelled by an object with respect to time could be written in this form, $s=t^{2}+4t$

At time t = 0, the distance is 0. At time t = 1, the distance, s = 5. At time t = 2, the distance, s = 12 and so on.

We know that speed is distance travelled divided by time it took to travelled that distance. If we derive the rate of change of distance with respect to time, $dfrac{ds}{dt}$ , that would give us the speed, v, of the object at the given point in time. The rate at which the distance changes with respect to time is called speed of the object.

$begin{array}{ccccc}& s & = & t^{2}+4tv= & dfrac{ds}{dt} & = & 2t+4end{array}$

At time t = 0, the speed, v, is 4. At time t = 1, v = 6.

We also know that the rate of change of speed, $dfrac{dv}{dt}$ , is acceleration, a.

$begin{array}{ccccc}& s & = & t^{2}+4tvspace{2mm}v= & dfrac{ds}{dt} & = & 2t+4vspace{2mm}a= & dfrac{dv}{dt} & = & 2end{array}$

This means that the acceleration, a, is a constant indicating that the object is continously increasing in speed.

In this lesson, a simple concept of rate of change was introduced by referring to an example of distance travelled by an object with respect to time, the speed and the acceleration of the object.

Exercise K – Differentiation – Rate of Change

A property of an object, y, is given as $y=2x^{3}-2x^{2}+5x+1$ . It is also given that the rate of change of x with respect to time, t, is 5 units per second. Express the rate of change of y with respect to time. Using the expression of the rate of change of y with respect to time, find the value of y when x is 6.

$begin{array}{ccccc}& y & = & 2x^{3}-2x^{2}+5x+1 & rightarrow A\& dfrac{dx}{dt} & = & 5, units/s & rightarrow Bend{array}$

We are requested to express the rate of change of y with respect to time. The rate of change of y with respect to time is $dfrac{dy}{dt}$ . We already know $dfrac{dx}{dt}$ and since we know the expression of y in terms of x, we can also find $dfrac{dy}{dx}$

Using the chain rule,

$begin{array}{ccccc}& dfrac{dy}{dt} & = & dfrac{dy}{dx}timesdfrac{dx}{dt}end{array}$

$begin{array}{ccccc}& y & = & 2x^{3}-2x^{2}+5x+1vspace{2mm}& dfrac{dy}{dx} & = & 6x^{2}-4x+5vspace{2mm}& dfrac{dy}{dt} & = & dfrac{dy}{dx}timesdfrac{dx}{dt}vspace{2mm}& & = & (6x^{2}-4x+5)5vspace{2mm}& dfrac{dy}{dt} & = & 5(6x^{2}-4x+5)end{array}$

When x = 6 , the rate of change of y with respect to x is,

$begin{array}{ccccc}& dfrac{dy}{dt} & = & 5(6x^{2}-4x+5)& & = & 5(6(6)^{2}-4(6)+5)& & = & 5(216-24+5)& dfrac{dy}{dt} & = & 985end{array}$

In this example, we used expressions x, y and t. t as given, is time. x and y are not given. It could be any property such as temperature and weight. For example, if y = weight of an object, the $dfrac{dy}{dt}$ is the rate of change of weight with respect to time. If x = temperature, then, we can say that the $dfrac{dy}{dx}$ is the rate of change of weight with respect to temperature. The important thing to remember is that the rate of change can be between any properties such as weight against time or weight against temperature. It does not necessarily be the rate of change against time.

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