Everything Maths Grade 12

CHAPTER 7. DIFFERENTIAL CALCULUS 7.5

Example 9: Finding the Equation of a Tangent to a Curve

QUESTION

Find the equation of the tangent to the curve y = x^2 at the point (1; 1) and draw both

functions.

SOLUTION

Step 1 : Determine what is required

We are required to determine the equation of the tangent to the curve defined

by y = x^2 at the point (1; 1). The tangent is a straight line and we can findthe

equation by using derivatives to find the gradient of the straight line. Then we

will have the gradient and one point on the line, so we can find the equation

using:

y−y 1 = m(x−x 1 )

from Grade 11 Coordinate Geometry.

Step 2 : Differentiate the function

Using our rules of differentiation we get:

y�= 2x

Step 3 : Find the gradient at thepoint (1; 1)

In order to determine the gradient at the point (1; 1), we substitute the x-value

into the equation for thederivative. So, y�at x = 1 is:

m = 2(1) = 2

Step 4 : Find the equation of the tangent

y−y 1 = m(x−x 1 )

y− 1 = (2)(x− 1)

y = 2x− 2 + 1

y = 2x− 1

Step 5 : Write the final answer

The equation of the tangent to the curve definedby y = x^2 at the point (1; 1) is

y = 2x− 1.

Step 6 : Sketch both functions