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The average rate of change between two points on a curve is determined by finding the gradient of the straight line joining these two points. We often need to find the rate of change at a particular instant. In such cases, the method used for finding the average rate of change is no longer appropriate.

However, it does provide the foundation that leads to obtaining the instantaneous rate of change. We now refine the definition of the average rate of change to incorporate the notion of the instantaneous rate of change.

The basic argument revolves around the notion of magnifying near the point where we wish to find the instantaneous rate of change, that is, by repeatedly ‘closing in’ on a section of the curve. This will give the impression that over a very small section, the curve can be approximated by a straight line.

Finding the gradient of that straight line will provide us with a very good approximation of the rate of change of the curve.

To obtain the exact rate of change at a particular point on the curve we will then need to use a limiting approach.

Since an interval centered at x = a always has the form [k – k, k + h] (with length 2h), this can be written:

$\frac{dy}{dx}$ = $\lim_{_{h\rightarrow 0}}$$\frac{f(k+h)-f(k-h)}{2h}$In a geometrical sense, it is the slope of the tangent to the graph of ‘ f ‘ at x = k.

If a variable quantity ‘y’ is some function of time ‘t’, that is y = f (t) 0, then a small change in time $\Delta$ 't' will have a corresponding change $\Delta$ y in ‘ y ‘.

Thus, the average rate of change = $(\frac{\Delta y}{\Delta t})$

When the limit $\Delta t\rightarrow 0$ is applied, the rate of change becomes instantaneous and we get the rate of change at the instant ‘ x ‘

$\lim_{_{\Delta t\rightarrow 0}}$($\frac{\Delta y}{\Delta t}$) = $\frac{dy}{dt}$

The differential coefficient of y with respect to ‘x’, that is $(\frac{dy}{dx})$ , is nothing but the rate of increase of y relative to ‘x’.

The velocity of a moving particle is defined as the rate of change of its displacement with respect to time and acceleration is defined as the rate of change of its velocity with respect to time.

Let a particle 'A' move rectilinear. Let 'S' be the displacement from a fixed point 'O' along the path at time ‘t’, ‘s’ is considered to be positive on the right of the point 'O' and negative on the left of the point 'O'.

Also,$\Delta$ s is positive when ‘s’ increases, that is when the particle moves towards the right.

Thus if $\Delta$s is the increment in s in time $\Delta$t. The average velocity on this interval is

$(\frac{\Delta s}{\Delta t})$

And the instantaneous velocity that is velocity at time ‘t’ is

$v=\lim_{\Delta t\rightarrow 0}$($\frac{\Delta s}{\Delta t}$) = $\frac{ds}{dt}$

If the velocity varies, then there is a change of velocity $\Delta$ v in time $\Delta$ t.

Hence, the acceleration at time:

$a=\lim_{_{\Delta t\rightarrow 0}}$($\frac{\Delta v}{\Delta t}$) = $\frac{dv}{dt}$

Function y=x

a)Find the average rate of change of y with respect to x over the interval [3,5].

b)Find the instantaneous rate of change of y with respect to x at the point x=2

Given y = x

=>f(3) = 3

=>f(5) = 5

Average rate of change over the interval [3,5]

=> $\frac{[f(5)-f(3)]}{[5-3]}$

=> $\frac{[21-5]}{2}$

=> $\frac{16}{2}$

=>8

=> f(2) = 2

=> f(x1) = x1

Instantaneous rate of change at the point x=2

6(2)

=>$ \lim_{x1 \to 2} $ $\frac{x1^{2}-4-0}{x1-2}$

=>$ \lim_{x1 \to 2} $ $\frac{x1^{2}-4}{x1-2}$

=>$ \lim_{x1 \to 2} $ $\frac{(x1-2)(x1+2)}{x1-2}$

=>$ \lim_{x1 \to 2} $ (x1+2)

=>(2+2)

=>4

Example 2.

Find the instantaneous rate of change of 4x

Let f(x)=4x

at x = 2

=>f(2)=4(2)

=>f(2)=20

=>f(x1)=4(x1)

=> $ \lim_{x1 \to 2} $ $\frac{4(x1)^{2} + 3(x1) -2 - 20}{x1-2}$

=> $ \lim_{x1 \to 2} $ $\frac{(x1-2)(4x1+11)}{x1-2}$

=> $ \lim_{x1 \to 2} $ [4x1 + 11]

= >19