Derivatives are used to find rate of change of a function with respect to variables. To find rate of change of function with respect to a variable differentiating it with respect to that variable is required. Rate of change of function y = f(x) with respect to x is defined by dy/dx or f'(x). For example y = x2 + x then, dy/dx = f'(x) = d(x2+ x)/dx = 2x + 1.
When two variables (let x = f(t) and y = g(t)) varying with respect to another variable t then the rate of change is calculated by the chain rule. Rate of change of y with respect to x is: dy/dx = (dy/dt) × (dt/dx). For example, x = t and y = 2t , then dy/dx is calculated as: dx/dt = 1 or dt/dx = 1 and dy/dt = 2. Hence, dy/dx = (dy/dt) × (dt/dx) = 2 × 1 = 2. Therefore, the rate of change of y with respect to x is calculated as the rate of change of y with respect to t and the rate of change of t with respect to x.
Application of derivatives
Derivatives are majorly used in mathematics to find any variable’s value change w.r.t another variable. In real life also, derivatives are seen and discussed. For instance, the word speed is mentioned a lot, but very few know that speed is also a derivative, speed is the change in the distance w.r.t time. Let’s look at some important applications of derivatives,
- To find the rate of change of a function with respect to variables.
- To find tangent and normal curves.
- To find the maximum and minimum value of a function.
- To find whether a function is increasing or decreasing and interval where the function is increasing or decreasing.
- To find the concavity (whether a function is concave up or concave down) of a function.
- To calculate velocity from displacement and acceleration from velocity (velocity is a change of displacement with respect to time and acceleration is the change of velocity with respect to time).
- To find the approximate value of certain quantities (for example square root, cube root of a number).
Tangent and Normal
Tangent to a curve at a given point is a straight line that touches the curve at a given point (does not intersect the curve, only touches the curve at a given point). Whereas Normal is a straight line that is perpendicular to the tangent.
Hence, the product of slope of tangent and normal is -1. Let denoted slope of tangent and normal be mTand mNrespectively. Then,
mT × mN = -1.
If tangent makes angle Θ with x-axis then slope of tangent = mT = tan Θ.
Since, mT× mN = -1 So, tan Θ × mN = -1.
Hence, the slope of normal is -1/tan Θ or -cot Θ.
Equation of Tangent and Normal
Slope of tangent to a curve whose equation is y = f(x) at a point a is f'(a) (derivative of f(x) at point a). Hence, equation of tangent in point-slope form is
(y – f(a))/(x – a) = f'(a),
and using equation mT × mN = -1,
Equation of normal is:
(y – f(a))/(x – a) = -1/f'(a).
Sample Problems
Question 1: y = x2 is an equation of a curve, find the equation of tangent, and normal at point (1, 1).
Solution:
First differentiate y = x2, y’ = 2x .
Slope of tangent at point (1, 1) is 2 × 1 = 2 and slope of normal is -1/2 .
Hence, equation of tangent at point (1, 1) is: (y – 1)/(x – 1) = 2 => y = 2x – 1
And equation of normal at point (1, 1) is: (y – 1)/(x – 1) = -1/2 = 2y = -x + 3 .
Question 2: y = x4 + x2 is an equation of a curve, find the equation of tangent and normal at point (1, 2)
Solution:
First differentiate y = x4 + x2 , we get y’ = 4x3 + 2x .
Slope of tangent at point (1, 2) is 4 × (1)3 + 2 × 1 = 6 and slope of normal is -1/6.
Hence, equation of tangent at point (1, 2) is: (y – 2)/(x – 1) = 6 => y = 6x – 4
And equation of normal at point (1, 2) is: (y – 2)/(x – 1) = -1/6 => 6y = -x + 13.
Question 3: y = x is an equation of a curve, find the equation of tangent and normal at point (1, 1).
Solution:
First differentiate y = x , we get y’ = 1 .
The slope of the tangent at point (1, 1) is 1, and the slope of normal is -1.
Hence, equation of tangent at point (1, 1) is: (y – 1)/(x – 1) = 1 => y = x
And equation of normal at point (1, 1) is: (y – 1)/(x – 1) = -1 => 2y = -x + 2.
Question 4: x = t and y = t is an equation of curve, find equation of tangent and normal at point t = 2.
Solution:
First differentiate x = t, we get dx/dt = 1 or dt/dx = 1.
And differentiate y = t, dy/dt = 1.
Hence, dy/dx = (dy/dt) × (dt/dx) = 1 × 1 = 1.
The slope of the tangent at point t = 2 is 1 and slope of normal is -1.
Hence, equation of tangent at point at point t = 2 is: (y – 2)/(x – 2) = 1 => y = x
And equation of normal at point at point t = 2 is (y – 2)/(x – 2) = -1 => y = -x + 4.
Question 5: x = t and y = t2 is an equation of curve and, find equation of tangent and normal at point t = 1.
Solution:
First, differentiate x = t, dx/dt = 1 or dt/dx = 1.
And differentiate y = t2, dy/dt = 2t = 2 × 1 = 2.
Hence, dy/dx = (dy/dt) × (dt/dx) = 2 × 1 = 2.
The slope of the tangent at point t = 1 is 2 and the slope of normal is -1/2.
Hence, equation of tangent at point at point t = 1 is (y – 1)/(x – 1) = 2 => y = 2x – 1.
And equation of normal at point at point t = 1 is (y – 1)/(x – 1) = -1/2 => 2y = -x + 3.