How do you find the point on the curve y=2x^2 closest to (2,1)?

Lokubovumn

Lokubovumn

Answered question

2022-08-13

How do you find the point on the curve y = 2 x 2 closest to (2,1)?

Answer & Explanation

merneh7

merneh7

Beginner2022-08-14Added 13 answers

Every point on the curve has the form: ( x , 2 x 2 )
The closest point is the one whose distance is minimum.
The distance between ( x , 2 x 2 ) and (2,1) is ( x - 2 ) 2 + ( 2 x 2 - 1 ) 2
Because the square root is an increasing function, we can minimize the distance by minimizing:
f ( x ) = ( x - 2 ) 2 + ( 2 x 2 - 1 ) 2
f ( x ) = 4 x 4 - 3 x 2 - 4 x + 5
To minimize f
f ( x ) = 16 x 3 - 6 x - 4
Now use whatever tools you have available to solve this cubic equation. (Formula, graphing technology, successive approximation, whatever you have.)
The critical number is approximately 0.824.
Clearly f'(0) is negative and f'(1) is positive, so f(0.824) is a local minimum.
f is a polynomial with only one critical number, so "local" implies "global"
The minimum value of distance occurs at x 0.824 and f ( 0.824 ) 1.358
The closest point is about (0.824,1.358).

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Multivariable calculus

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?