Question

Check it please: Find the (shortest) distance from the origin to the parabola y=4-x^2

Answer 1

I got z= sqrt( (x-0)^2 + (y-0)^2 )

Answer 2

Then z= sqrt( (x-0)^2 + (4-x^2-0)^2 )

Answer 3

Then z= sqrt( (x)^2 + (4-x^2)^2 )

Answer 4

Then what?

Answer 5

Find the interecepts and then take the square root of the sum of their squares

Answer 6

The next step would be to derive z with respect to x

Then you find the roots of the derivative function you found

Answer 7

dz/dx = (x(2x^2-7))/(sqrt(x^4-7x^2+16)) ??

Answer 8

close, but not quite

Answer 9

oh wait, i see how you got that

Answer 10

yes you are correct

Answer 11

now set this equal to zero and solve for x

Answer 12

so x=sqrt(7/2)

Answer 13

x = 0 is also a root

Answer 14

those are your critical values

now you have to use the first or second derivative test to see which one of those critical values are minimums

Answer 15

the sqrt(7/2) is

Answer 16

and so is -sqrt(7/2)

Answer 17

and this makes sense because you have symmetry with respect to the y-axis

Answer 18

we plug x into sqrt(x^4-7x^2+16)

Answer 19

yes

Answer 20

to find the distance

Answer 21

and get + or - 1.936494154 ?

Answer 22

I'm getting the same

Answer 23

Woot thanks a bunch for the help!

Answer 24

you're welcome