- Topic
- Linear approximation
- Difficulty
- Challenging
- Suggested time
- 7 min
- Marks
- 4
Linear approximation practice question 10
The equation defines implicitly as a function of near the point . Use the tangent-line approximation at that point to estimate when .
Differentiate implicitly to find at .
Differentiating both sides gives ; the term needs the product rule. Solve for and put in .
Then .
An exact fraction or expression.
Theory UBC Math 100 §9.2CLP-1 §3.4Key results
Worked solution
Strategy
A linear approximation only needs the value and the slope at the base point. The value is given; the slope comes from implicit differentiation, since there is no formula for y.
Working
Product rule on xy; solve for y′.
Check first that (1, 2) is on the curve: 8 + 2 = 10.
Tangent-line estimate. The true value is 1.98462 to five places as well, so give the estimate exactly.
Answer
The answer is 2 minus 1 over 65, which is 129 over 65, about 1.9846.
Check
Putting y = 129/65 ≈ 1.9846 back in, y³ + 1.1y ≈ 9.9999, within 0.0002 of 10.
Where this goes wrong
Trying to solve the cubic for y at x = 1.1. That is exactly the calculation linear approximation is meant to avoid, and it is not what was asked.
Leans on
Implicit differentiation, from Implicit differentiation and inverse trigonometric functions
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