Question

Find the point in the first quadrant on the graph of y=x^2 that is closest to the point (0,4)

Answer 1

use
d( (x,y) , (0,4) ) = sqrt ( (x-0)^2 + (y-4)^2)
and substitute y = x^2

Answer 2

distance between a point on graph y=x^2 and (0,4) is
d = sqrt ( x^2 + (x^2 - 4)^2 )

Answer 3

d = sqrt ( x^2 + x^4 - 8x^2 + 16 )

Answer 4

d = sqrt ( x^4 -7x^2 + 16)

Answer 5

so a trick is actually to square both sides first
d^2 = x^4 - 7x^2 + 16

Answer 6

Okay, got that. Do I need to plug in the points now?

Answer 7

you need to solve d'(x)

Answer 8

2*d *d'(x) = 4x^3 -14x

d'(x) = (4x^3 -14x)/ (2d)

Answer 9

set that equal to zero
(4x^3 -14x) / (2d) = 0
4x^3 - 14x = 0 the denominator =0 is also a critical pt, but the denominator is strictly greater than zero

Answer 10

2x(2x^2 -7) = 0
x^2 = 7/2
x = + - sqrt (7/2)

Answer 11

And how could I solve for y?

Answer 12

we know y = x^2

Answer 13

( sqrt (7/2), 7/2)

Answer 14

So square x right?

Answer 15

Ha! Okay got it! Thanks!!

Answer 16

did that make sense

Answer 17

Yes a lot actually! You factored the numerator and equaled it to zero to solve for x, and then plugged it into the given equation.

Answer 18

right, and we don't care about the denominator 2d , because it is always greater than zero

Answer 19

technically we skipped the part which shows this is the minimum .
you show that d'(x) > 0 for 0< x < sqrt(7/2)
and d'(x) < 0 for x < sqrt(7/2)