Past Question and Solution
Obafemi Awolowo University, Ile-Ife, Osun, Nigeria
MTH 106: Elementary Mathematics
Topic: Equation of tangent and normal to a curve at a given point.
Question:
1. Find the equation of the tangent and normal to the curve: ๐ฆ = ๐ฅ3โ 3๐ฅ2+ 2 at ๐ฅ = 1 and
๐ฆ = 0.
Solution:
- Since the tangent and normal are lines, their equations will be linear.
- The point-slope form of the equation of a line is given as:
๐ฆ โ ๐ฆ1= ๐(๐ฅ โ ๐ฅ1)
Where:
(๐ฅ1, ๐ฆ1) is a point on the line.
๐ is the slope or gradient of the line.
- The slope or gradient in the given task can be determined by differentiating the
equation of the curve with respect to ๐ฅ and substituting in the given value of ๐ฅ.
- Use the rule ๐
๐๐ฅ (๐๐ฅ๐)=(๐ ร ๐)๐ฅ๐โ1 for differentiation.
- Differentiating a constant results in 0.
- If the gradient of the tangent is ๐1, then the gradient of the normal is ๐2= โ 1
๐1
Differentiate the expression:
๐
๐๐ฅ (๐ฅ3โ 3๐ฅ2+ 2)=๐
๐๐ฅ (๐ฅ3)โ๐
๐๐ฅ (3๐ฅ2)+๐
๐๐ฅ (2)
๐
๐๐ฅ (๐ฅ3โ 3๐ฅ2+ 2)= 3๐ฅ2โ 6๐ฅ + 0
๐
๐๐ฅ (๐ฅ3โ 3๐ฅ2+ 2)= 3๐ฅ2โ 6๐ฅ
Substitute ๐ฅ = 1 into the expression on the right-hand side:
๐
๐๐ฅ (๐ฅ3โ 3๐ฅ2+ 2)= 3(1)2โ 6(1)
Simplify the expression on the right-hand side:
๐
๐๐ฅ (๐ฅ3โ 3๐ฅ2+ 2)= 3 โ 6 = โ3
