- Topic
- Optimization
- Difficulty
- Challenging
- Suggested time
- 7 min
- Marks
- 4
Optimization practice question 13
Find the -coordinate of the point on the curve closest to the point .
Minimising the distance and minimising the squared distance give the same point, and the squared distance is far easier to differentiate.
A point on the curve is , so .
A decimal or a fraction.
Theory UBC Math 100 §8CLP-1 §3.5Key results
Worked solution
Strategy
Write the squared distance from a general point on the curve to the fixed point. Because the square root is increasing, whatever minimises the square also minimises the distance, so the root can be dropped.
Working
Parametrise by x so there is only one variable from the start.
Squaring the root collapses the second term, which is what makes this approach worth taking.
Differentiate the squared distance.
Solve for the critical point.
The second derivative is 2 for every x ≥ 0, so D² is concave up on the whole domain and x = 7/2 gives its absolute minimum. The closest point is (3.5, √3.5).
Answer
The answer is x = 3.5, that is, 7/2.
Check
The segment from (3.5, √3.5) to (4, 0) has slope −2√3.5 and the tangent there has slope 1/(2√3.5): perpendicular, as the closest point needs.
Where this goes wrong
Taking the point straight above (4, 0), at x = 4. It is 2 away, but the curve comes closer a little to the left: minimise the squared distance (x − 4)² + x rather than guessing.
Leans on
Distance between two points, from Functions, graphs and inverses
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