The applications of derivatives are:

- determining rate of change of quantities,
- finding the
**equations of tangent and normal**to a curve at a point, - finding turning points on the graph of a function which in turn will help us to locate points at which largest or smallest value (locally) of a function occurs.

In this article, we will learn to use differentiation to find the equation of the tangent line and the normal line to a curve at a given point.

**Also, read:**

**Tangent Line to a Curve**

Tangent is a straight line which touches the curve at a given point and it has the same slope of the curve at that point.

Consider the function y = f(x) given in the figure,

The first derivative of a function at a point gives its slope of the tangent at that point.

For the function *f*, first derivative i.e. *f’*at the point where x =a is *f*‘(a).

*f*‘(a) = slope of the tangent at the point where x =a

Equation of a straight line **\( \frac {(y-y_1)}{(x-x_1)}\) = m**, where “m” is slope of the line.

**\( \frac{(y – f(a))}{(x-a)}\)= f’(a)** is the equation of tangent of the function at x=a.

**Example:**

Consider the function *f*(x) = x^{3}-3x^{2}+2x+1. Find equation of tangent of f(x) when x = 2.

*f*(x) = x^{3}-3x^{2}+2x+1

*f*(2) = (2)^{3}-3(2)^{2}+2(2)+1

First derivative of the function *f*(x).*f*‘(x) = d/dx(x^{3}-3x^{2}+2x+1)

= 3x^{2} – 6x + 2

*f*‘(2)= 3 × 2^{2} – 6 × 2 + 2 = 12 – 12 + 2 = 2

Equation of the tangent = \( \frac {(y-1)}{(x-2)}\) = 2.

y – 1 = 2(x – 2)

y – 1 = 2x – 4

y = 2x – 3 is the equation of tangent.

## Normal to a curve

Normal is a line which is perpendicular to the tangent to a curve. Tangent and normal of *f*(x) is drawn in the figure below.

Since tangent and normal are perpendicular to each other, product of slope of the tangent and slope of the normal will be equal to -1.

**(y – f(a))/(x-a)} = f‘(a); **is the equation of tangent of the function y =

*f*(x) at x = a.

Then, equation of the normal will be,\( \frac{(y – f(a))}{(x-a)}\) = \( \frac {-1}{f'(a)} \)

Example:

Consider the function, f(x) = x^{2} – 2x + 5. Find the equation of tangent and equation of normal at x = 3.

f(x) = x^{2} – 2x + 5

f(3) = 3^{2} – 2 × 3 + 5 = 9 – 6 + 5 = 8

First derivative, *f*‘(x) = d/dx(x^{2} – 2x + 5) = 2x – 2

*f*(‘x) when x = 3,*f*(3) = 2 × 3 – 2 = 6 – 2 = 4

Equation of the tangent is, (y-8)/(x-3) = 4

y – 8 = 4(x – 3)

y – 8 = 4x – 12

y = 4x – 4, is equation of tangent.

Slope of normal = – 1/slope of tangent = -1/4

Equation of normal is, (y-8)/(x-3) = -1/4

4(y – 8) = (3 – x)

4y – 32 = 3 – x

y = -x/4+35/4

### Points to Remember

- If a tangent line to the curve y = f (x) makes an angle θ with x-axis in the positive direction, then dy/dx = slope of the tangent tan = θ.
- If the slope of the tangent line is zero, then tan θ = 0 and so θ = 0 which means the tangent line is parallel to the x-axis. In this case, the equation of the tangent at the point (x
_{0}, y_{0}) is given by y = y_{0} - If θ →π/2, then tan θ → ∞, which means the tangent line is perpendicular to the x-axis, i.e., parallel to the y-axis. In this case, the equation of the tangent at (x
_{0}, y_{0}) is given by x = x_{0}

## Tangent and Normal lines Problems with Solutions

**Q.1: Find the slope of the tangent to the curve y = x ^{2}+25 at x = 5**

Solution: Given, y = x

^{2}+25

On differentiating the given expression, we get;

dy/dx = 2x

dy/dx = y’ = y'(5) = 2 x 5 = 10

Hence, the slope of the tangent to the curve is 10.

**Q.2: Find the equation of lines having slope 2 and being tangent to the curve y + 2/(x-1) ^{2}**

Solution: Slope of the tangent to the given curve at any point (x, y) is given by:

dy/dx = 2/(x-1)^{3}

But the slope is given here as 3. Hence,

2 = 2/(x-1)^{3}

(x-1)^{3} = 1

x – 1 = 1

x = 2

If x = 2, then y = -2.

Hence, the equation of tangent passing through the point (2, -2), is given by;

y + 2 = 2(x-2)

y – 2x + 4 = 0