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

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Answer 1 · 2015-09-02T17:59:14

See the explanation.

Every point on the curve has the form: #(x,2x^2)#

The closest point is the one whose distance is minimum.

The distance between #(x, 2x^2)# and #(2,1)# is #sqrt((x-2)^2+(2x^2-1)^2)#.

Because the square root is an increasing function, we can minimize the distance by minimizing:

#f(x) = (x-2)^2+(2x^2-1)^2#

#f(x) = 4x^4-3x^2-4x+5#

To minimize #f#,

#f'(x) = 16x^3-6x-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)#.