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Tangents and Normal to a Curve

A tangent is a line that touches a curve. A tangent meets or touches a circle only at one point, whereas the tangent line can meet a curve at more than one point, as the diagrams below illustrate.

Tangent to a circle      Tangent to a curve

On the other hand, a line may meet the curve once, but still not be a tangent.

Curve_line

As we noticed in the geometrical representation of differentiation of a function, a secant PQ – as Q approaches P – becomes a tangent to the curve. Hence a tangent to a curve is best described as a limiting position of a secant.

The gradient or slope of the tangent at a point ‘x = a’ is given by dy/dx at ‘x = a’.

Note: When you are asked to find the gradient of a tangent at x = a, we find dy/dx or f'(x) or y’, and substitute it in the gradient or differential function with x = a.

 

Example 1: Find the gradient of the tangent to a curve y = x3 + 2x +4 at x =1

dy/dx = y’ = 3x2 + 2

So y’ at x=2

= m = 3 x 22 + 2  =  14.

So the gradient of the tangent at x = 1 is m = 14.

Using Coordinate Geometry, we know two lines are said to be perpendicular if the products of the slopes of lines is equal to -1. i.e. m1m2 = -1.

In other words, if two lines with gradients m1 and m2 respectively are perpendicular to each other, then m1m2  =  -1.

A normal is a straight line that is perpendicular to the tangent at the same point of contact with the curve i.e. the tangent and normal will have the same point of contact on the curve, as the diagram below illustrates.

Tangent & normal to a curve

 

Example 2: Find the equation of the tangent and the normal to the curve y = x4 – 3x3 + 6x + 2 at the point (2, 6)

Step 1: First find the gradient of the tangent at the point (2, 6)

y’ = dy/dx = 4x3 – 9x2 + 6

dy/dx at x = 2 = mT = 4 x 23 – 9 x 22 + 6

mT = 2  (mT is the gradient or slope of the tangent)

Step 2: Using the equation of line with point-slope form, we get

Equation of tangent: y – y1 = mT(x – x1)

y – 6 = 2(x – 2),  or

-2x + y – 6 + 4 = 0

-2x + y – 2 = 0, or

2x – y + 2 = 0

Step 3: Using m1m2 = -1 for gradients of tangent and normal which are perpendicular at (2, 6) we get

mT x mN = -1;   2 x mN = -1  (here mN is the gradient or slope of the normal)

Therefore mN = -1/2 is the slope of normal.

Step 4: Equation of normal at (2, 6) and mN = -1/2, we get

(y – 6) =  -1/2(x – 2)

2y – 12  = -x + 2

x + 2y – 12 – 2 = 0

x + 2y – 14 = 0

Equation of tangent: 2x – y + 2 = 0, and

Equation of normal:  x + 2y – 14 = 0

 

Example 3: Find the coordinate of point Q where the tangent to the curve y = x2 + 3x +2 is parallel to the line 2x + y + 2 = 0

The gradient of the tangent to  y = x2 + 3x +2 which is parallel to 2x + y + 2 = 0 is the same as the line 2x + y + 2 = 0.

Therefore y = -2x – 2

doubleright m = -2

Gradient of tangent to y = x2 + 3x +2 is -2.

dy/dx = 2x + 3

dy/dx = 2x + 3  = m = -2

2x + 3 = -2

x = -5/2

The point at which the tangent is parallel to the given line is x = -5/2. So

y = (-5/2)^2 + 3 x (-5/2) + 2

25/4 – 15/2 + 2

-5/4 + 2  =  3/4

= (-5/2, 3/4) is the coordinate for Q.