Tangent - Equation Of Tangent And Normal

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,

Tangent Line to a Curve

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) = x3-3x2+2x+1, . Find equation of tangent of f(x)when . x = 2.

f(x) = x3-3x2+2x+1

f(2) = (2)3-3(2)2+2(2)+1

First derivative of the function f(x).f‘(x) = d/dx(x3-3x2+2x+1)

= 3x2 – 6x + 2 

f‘(2)= 3 × 22 – 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.

Tangent - Normal to a curve

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) = x2 – 2x + 5 . Find the equation of tangent and equation of normal at x = 3.

f(x) = x2 – 2x + 5

f(3) = 32 – 2 × 3 + 5 = 9 – 6 + 5 = 8

First derivative, f‘(x) = d/dx(x2 – 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 (x0, y0) is given by y = y0
  • 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 (x0, y0) is given by x = x0

Tangent and Normal lines Problems with Solutions

Q.1: Find the slope of the tangent to the curve y = x2+25 at x = 5
Solution: Given, y = x2+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

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