**Applications of derivatives: rate of change of bodies, increasing/decreasing functions, tangents and normal, use of derivatives in approximation, maxima and minima (first derivative test motivated geometrically and second derivative test is given as a provable tool)**

To solve practical problems such as engineering optimization, the greatest challenge is often to convert the word problem into a mathematical optimization problem—by setting up the function that is to be maximized or minimized. Similarly, we can also find the rate of change of a body using differentiation.

Recall that dy/dx is positive if y increases as x increases and is negative if y decreases as x increases.Also, the slope of the tangent of a curve f(x) can be found out by deriving the function of the curve at the given point. Because the normal line is perpendicular to the curve f(x) at a given point, the gradient of the normal is -1/f’(x).

Example: A ship can reach its top speed in 5 minutes.During that time its distance from the start can be calculated using the formula *D* = *t* + 50*t*^{2} where *t* is the time in seconds* and D is measured in metres. *How fast is it accelerating?

Speed, v m/s, is the rate of change of distance with respect to time.

*v* = 1 + (100 5) = 501 m/sec.

Acceleration, *a* m/s^{2}, is the rate of change of *speed *with respect to time or second derivative of distance with respect to time.

Now consider the graph below:

The signs indicate where the gradient of the curve is =, – or 0. In each case:

A function is strictly increasing in a region where *f*´(*x*) > 0. A function is strictly increasing in a region where *f*´(*x*) < 0. A function is stationary where *f*´(*x*) = 0

Example: *f*(*x*) = 2*x*^{3} – 3*x*^{2} – 12*x* + 1.

Identify where the function is (i) increasing (ii) decreasing (iii) stationery

Solution: f’(x) = 6x^{2} – 6x – 12

A sketch of the derivative shows us that

*f*´(*x*) < 0 for 0 <*x*< 2 … *f*(*x*) decreasing

*f*´(*x*) > 0 for *x*< 0 or *x*> 2 … *f*(*x*) increasing

*f*´(*x*) = 0 for *x* = 0 or *x* = 2 … *f*(*x*) stationary

When a function is defined on a closed interval, *a* ≤ *x* ≤ *b*, then it must have a maximum and a minimum value in that interval.

These values can be found either at

• a stationary point [where *f*´(*x*) = 0]

• an end-point of the closed interval. [*f*(*a*) and *f*(*b*)]

All you need do is find these values and pick out the greatest and least values.

Example: A manufacturer is making a can for holding 250 ml of juice.The cost of the can is dependent on its radius, *x* cm.For practical reasons, the radius must be between 2.5 cm and 4.5 cm.The cost can be calculated from the formula

C = *x*^{3} –5*x*^{2} + 3*x* + 15, 2.5 ≤ *x* ≤ 4.5.

Calculate the maximum and minimum values of the cost function.

Solution

which equals zero at stationary points.

3*x*^{2} – 10*x* + 3 = 0

- (3
*x*– 1)(*x*– 3) = 0 *x*=^{1}/_{3}or*x*= 3- Working to 1 d.p.
*f*(^{1}/_{3}) = 15.5*f*(3) = 6*f*(2.5) = 6.9*f*(4·5) = 18.4- By inspection
*f*= 18.4 (when_{max}*x*= 4.5) and*f*= 6 (when_{min}*x*= 3).

Let f : D → R, D ⊂R, be a given function and let y = f (x). Let ∆x denote a small increment in x. Recall that the increment in y is corresponding to the increase in x, denoted by ∆y, is given by ∆y = f (x + ∆x) – f (x). We define the following:

- The differential of x, denoted by dx, is defined by dx = ∆x.
- The differential of y, denoted by dy, is defined by dy = f′( x) dx

In case dx = ∆x is relatively small when compared with x, and for y, we denote dy ≈ ∆y.

Example: Use differential to approximate $\sqrt{36.5}$

Solution:

Take $y= \sqrt{x}$, let x = 36 and ?x=0.5

Therefore, $\Delta y$ = $\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{36.5}-\sqrt{36}=\sqrt{36.5}-6$

Hence, $\sqrt{36.5}=6+\Delta y$

Since dy is approximate = $\Delta y$, it is given by:

$dy=(\frac{dy}{dx})\Delta x=\frac{1}{2\sqrt{x}}(0.5)=\frac{1}{2\sqrt{36}}(0.5)=\frac{1}{24}=0.04$, since $y= \sqrt{x}$

Therefore, the approx. value of $\sqrt{36.5}$ is 6 + 0.04=6.04