#### • Class 11 Physics Demo

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# Applications of Derivatives in Real Life

Application of Derivative:
Application of Derivative is a part of mathematics relates the change of  one function with respect to another function. Generally the application of derivative is used to measure changes of the system with respect to time. The maximum and minimum range of a function can be determined which helps in graphing of a function without any difficulty.

Application of Derivative in real life :
In everyday day to day we come with application of  derivative.  Some of the application are given below
• Rate of change of displacement
• Rate of change of velocity
• Rate of change of area and volume
• Angle between two curves
• Find the point of curve  at which tangents are parallel to the axis
• To determine tangents and subnormal
Angle between two curves relates angle between  their respective tangents at the point of intersection  (x1,y1)

$\Theta =\tan ^{-1}\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}$

Application of Derivative Problem with Answer:

1. A particle is thrown from a fixed point whose distance is given by S = 2t- 4t -3 units. Find the velocity of the given particle at
t = 3 seconds.

Solution: S = 2t- 4t - 3
Differentiate wit respect to t .

$\frac{dS}{dt}$= 4t - 4

Velocity of the particle  is 4t - 4
Velocity at t = 3
= 4*3 - 4
= 12 - 4
= 8m/sec

2. The volume of the sphere is increasing at 3 c.c per second. Find  the rate of increases of the radius when the radius is 2 cm.

Solution: Given the rate of change of volume of sphere as 3cc/sec

The volume of Sphere = $\frac{4}{3}$ πr3

To find the rate of increases of radius, differentiate volume with respect to t

$\frac{dv}{dt}=\frac{4}{3}3\pi r^{2}\frac{dr}{dt}$

$\frac{dv}{dt}=4\pi r^{2}\frac{dr}{dt}$

3 = $4\pi r^{2}\frac{dr}{dt}$

3 = $4\pi (2)^{2}\frac{dr}{dt}$

$\frac{3}{16\pi }=\frac{dr}{dt}$

The rate of increases of the radius is $\frac{3}{16\pi}$ cm/sec.

3. Find the equation of the tangent to the curve x = et and y = e-t at (1,1)

Solution: Differentiate both the term with respect to t.
x = et     y = e-t

$\frac{dx}{dt}=e^{t}$                $\frac{dy}{dt}=-e^{-t}$

Slope of the tangent is given by $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-e^{t}}{e^{t}}=\frac{-y}{x}=-1$

Equation of the tangent :
y -1  = (-1) (x-1)
x + y -2 = 0