Application Of Derivatives - I

APPLICATION OF DERIVATIVES -I The derivatives of functions have many applications in various disciplines e.g. engineering, science, social science and many other fields. Here we shall discuss a few important applications among them, as example, a) finding the equations of tangent and normal lines on a curve at a point, b) finding the rate of change of quantities, c) solving distance-velocity-acceleration problems, d) approximation of some function values, e) finding the intervals in which a function is increasing or decreasing, f) finding the turning points on the graph of a function which help us to locate points where largest or smallest values (local extrema) of a function occurs.

TANGENTS & NORMAL LINES

We know that,

slope of the tangent drawn on the curve at a point =

the derivative of the function at that point =

This means, if we take a point

b

`

x 0 ,f x 0

ac

or `

`

` a

f. x

a

x 0 , y 0 on the graph of the curve y = f(x),

a

the slope of the tangent at that point is f. x 0 . We can then write the equation of the tangent at that point by the slope-point form of a straight line.

The equation of the tangent is

y @ y 0 = f. x 0 x @ x 0 `

a`

a

The normal line to the curve at a point is the line perpendicular to the tangent line at that point. Because the product of slopes of two perpendicular lines is -1, the slopes of the tangent line and the normal line are negative reciprocals of one another.

Slope of the normal drawn on the curve at a point derivative of the function at that point

=

=

the negative reciprocal of the

1f f f f f f f f f f f f f f f f f ` a @ f f. x

The equation of the normal is y @ y0 = @

` a 1f f f f f f f f f f f f f f f f f f f f f ` a x @ x0

f. x 0

Example : Find the equation of the tangent and normal lines to the curve f x = 3x 2 + 4 ` a

at the point where x=3. Solution : f 3 = 3.3 + 4 = 31 ` a

2

b

c

So the point is 3,31 f x = 3x 2 + 4 ` a f. x = 6x ` a

b

c

The slope of the tangent at the point 3,31 is m = f. 3 = 6.3 = 18 Using the point @ slope form of a staright line, the equation of the tangent is ` a y @ y1 = m x @ x1 , ` a y @ 31 = 18 x @ 3 18x @ y @ 23 = 0 ` a

b

c

The slope of the normal at 3,31 is 1f 1f f f f f f f f f f f f f f f f f f f f f f f ` a =@ m =@ f f. 3 18 ` a The equation of the normal is y @ y1 = m x @ x1 , ` a 1f f f f f f x @3 y @ 31 = @ f 18 b c ` a 18 y @ 31 = @ x @ 3 18y @ 558 = @ x + 3 x + 18y @ 561 = 0

Example : Find the points at which the tangent the curve parallel to the x-axis.

f x = x 3 @ 3x 2 @ 9x + 7 is ` a

Solution : f x = x 3 @ 3x 2 @ 9x + 7 ` a

f . x = 3x 2 @ 6x @ 9 As the slope of any line parallel to the x @ axis is 0, and the slope of the tangent ` a at any point on the curve is f . x , ` a # at that point f . x = 0 ` a

3x 2 @ 6x @ 9 = 0 x 2 @ 2x @ 3 = 0 x = @ 1 and x = 3 ` a ` a3 ` a2 ` a f @ 1 = @ 1 @ 3 @ 1 @ 9 @ 1 + 7 = @ 1 @ 3 + 9 + 7 = 12 f 3 = 3 @ 3 3 @ 9 3 + 7 = 27 @ 27 @ 27 + 7 = @ 20 ` a ` a3

` a2

b

` a

c

b

c

So, the points are @ 1,12 and 3, @ 20 A

FINDING THE CRITICAL POINTS The points on the graph of the function where the derivative is zero or the derivative does not exist are important in curves of different functions and many applications. If at the point x = x 0 in the domain of the function f , b

`

exist, then x 0 , f x 0

ac

is called a critical point of f(x).

f . x = 0 or f . x does not ` a

The geometrical interpretation is that in such cases the tangent at the point either horizontal, vertical or does not exist. Example : Find all the critical points of f x = x 3 @ x 2 ` a

Solution : The domain of

` a

f x

are all real numbers A

f . x = 3x @ 2x ` a So f . x exists at all values of x in the domain A ` a

2

When f . x = 0, ` a

3x 2 @ 2x = 0, ` a x 3x @ 2 = 0 2f f x = 0, x = f 3 ` a ` a3 ` a2 f 0 = 0 @ 0 =0

f

f g f g3 f g2 2f 2f f f 2f f f f f

=

3

b

c

The critical points are 0,0

3

and

@

f

3

8f 4f f f f f f 4f f f f f f f f = f @ f =@ f 27 9 27

2f f f 4 f f f f f f ,@ f A 27 3 g

` a

b

`

x0 , f x0

ac

is

Example : Find all the critical points of f x = sin 2x on 0,2π B

` a

Solution : The function

` a

f x

C

B

C

exists for all x values in the interval 0,2π A

f. x = 2 cos 2x ` a So f. x exists at all values of x in the given intervalA ` a 2 cos 2x = 0, When f. x = 0, cos 2x = 0 ` a πf f f f f where n is any integer, 2x = 2n + 1 2 ` a πf f f f f x = 2n + 1 4 puting different values of n we get the following values of x in the given interval πf f f f f 3π f f f f f f f f 5π f f f f f f f 7π f f f f f f f f , f , f x= , 4 4 4 4 d e f g πf 3π 3π πf f f f f f f f f f f f f f f f f f f f f f f f = sin = 1, f f = sin f = @ 1, 2 4 2 4 ` a

5π 5π f f f f f f f f f f f f f f f f = sin f = 1, 2 4 f

7π 7π f f f f f f f f f f f f f f f f f f = sin f = @ 1, 4 2

g

The critical points are

f

f

g πf f f f f f f f f

4,1

,

f

g

3π f f f f f f f f ,@1 , 4 g

f

5π f f f f f f f f f , 4,1 g

f

7π f f f f f f f f f ,@1 A 4 g

THE DERIVATIVE AS A RATE OF CHANGE Using calculus, we can find the rate of change of one variable with respect to another variable. Whenever one quantity y varies with another quantity x, satisfying some rule y = f(x), then

` a dy f f f f f f f f or f. x represents the rate of change of y with respect to x. That is, in a dx

differentiable function, y = f(x), the rate of change of y with respect to x at x = x 0 is ` a dy f f f f f f f f at x 0 = f. x 0 . dx

Further, if two variables x and y are varying with respect to another variable t, i.e. if x = dy f f f f f f f f f

f(t) and y = g(t),

then by chain rule

dy dx f f f f f f f f dt f f f f f f f f f f f f f f f f = if ≠ 0 .Usually the variable t f f f f f f f f f dt dx dx dt

represents time. The sign of the derivative say

dy f f f f f f f f is positive or negative, depending on whether y dx

dx f f f f f f f increases or decreases with increase in x. For example, f is positive if x increases with dt time, and negative if x decreases with time.

Example : Radius of a circle is increasing at the rate of 2 in/sec. Find the rate of change of its area when the radius is 4 inches.

Solution : If the area of the circle = A and the radius = r , then A = πr 2 differentiating with respect to time t, b c b c dA df df f f f f f f f f f f f f f f 2 f f f f f f 2 dr f f f f f f f = πr = πr A dt dt dr dt dr f f f f f f f = 2πr dt dr f f f f f f f As the radius is increasing with time the sign of will be positive dt dr f f f f f f f = 2 inch / sec dt ` a` a dA f f f f f f f f f = 2π 4 inch A 2 inch / sec dt 2 = 16π inch / sec 2 Hence, the area is increasing at the rate of 16π inch / sec when the radius is 4 inches A

Example : A ladder 10 ft long is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of ½ ft/sec. How fast is its height on the wall changing when the foot of the ladder is 6 ft away from the wall?

Solution :

w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w q100 @ x 2 ft

10 ft

x ft

Let the foot of the ladder be at a distance x ft from the wall, and the height on the wall is h A dx f f f f f f f f 1f f = f ft / sec It is given dt 2 2 2 Using Pythagorus relation , h + x 2 =10 b c d` a df f f f f f 2 f f f f f h + x2 = f 100 dt dt D E w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w w dh dx 2 2 f f f f f f f f f f f f f f f f q 2h A + 2x A = 0, when x = 6 ft, h = 10 @6 = 8 ft dt dt g b c dh b c f1 f f f f f f f f f f f 2 A 8 ft A + 2 6 ft A ft / sec = 0 dt 2 B C 3f dh f f f f f f f f f =@ f ft / sec the negativce sign here implies decrease in h with time 8 dt 3f f So, the height is decreasing at the rate of f ft / sec A 8