- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
Implicit differentiation and inverse trigonometric functions practice question 17
The point lies on the curve below. Find an equation of the normal line to the curve there: the line through perpendicular to the tangent line.
The normal is perpendicular to the tangent, so you need the tangent slope first. How do you get it from an equation with and in it?
Differentiate implicitly; needs the product rule. Solve for and substitute to get the slope of the tangent.
Perpendicular slopes multiply to : the normal’s slope is the negative reciprocal of the tangent’s.
An equation of a line. Use ^ for powers and brackets for every input: sin(2x), e^(x^2), log(x+1). Give exact values: 1/6, not 0.167.
Theory UBC Math 100 §4.5CLP-1 §2.11Key results
Worked solution
Strategy
Find the tangent slope by implicit differentiation, turn it into the perpendicular slope, and write the line through the point.
Working
The chain rule on y³; the product rule on xy.
Collect the dy/dx terms, divide, then substitute the point.
The negative reciprocal, so the two slopes multiply to −1.
Point-slope form through (2, 2).
Answer
The answer is y equals 7x minus 12.
Check
The curve passes through (2.01, 1.99857), so the tangent slope is about −0.143 ≈ −1/7, and the normal’s slope 7 is its negative reciprocal.
Where this goes wrong
Giving the tangent line, or changing only the sign of its slope. The normal is perpendicular to the tangent, so its slope is the negative reciprocal of −1/7, which is 7.
Leans on
Equation of a tangent line, from The definition of the derivative
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