Question

Find the point P on the parabola y=x^2 closest to the point (3,0).

Answer 1

Do you know any calculus?

Answer 2

lol, yes I figured out to use the distance formula and have taken the derivative of that, but I dont know where to go from there.

Answer 3

Ok, great. You take the derivative of the distance formula to minimize it ("Find closest point")
Set first derivative =0 to get critical points.

Answer 4

thats where I'm stuck, what I got for the derivative is 4x^3+2x-6, and it comes out to some nasty number lol.

Answer 5

It looks like there's some stuff missing from that derivative. Can you show me how you got that?

Answer 6

Or is that all you used to =0 to find the critical points?

Answer 7

Yeah, thats what I used to try and find the critical points.

but this is the equation I got first \[(x^2)^2+(x-3)^2\] and I derived that.

Answer 8

Alright, it'll work out in the end, but remember that distance formula is the pythagorean theorem, so there should a square root happening in there.

Answer 9

When you set it equal to zero, the denominator won't matter, so you don't really need it.

Answer 10

What numbers did you find for critical points?

Answer 11

hmmm, not quite sure if I did it right but on one of my tries I got 1/3?

Answer 12

I don't think 1/3 is an option. Did you use rational root theorem?

Answer 13

No I didn't use the rational root theorem.

Answer 14

The distance between P(x,y) and (3,0) is \( z=\sqrt{x^4+x^2-6x+9}\)

The derivative of that is \(\large \frac{4x^3+2x-6}{2\sqrt{x^4+x^2-6x+9}}\).

That is equal to zero when the numerator equals zero, so solve \( 4x^3+2x-6=0\).

Conveniently enough, x=1 is a root. Verify: \(4(1)^3+2(1)-6=0\)
Turns out, that is the only critical point, so that must be your minimizing value.

Answer 15

hmmm I guess when I solved for zero I did something wrong.

Answer 16

I think I remember another way to solve it using the fact that the shortest distance line is perpendicular to the tangent line at the point on the parabola.

This you can also see because the line from (1,1) on the parabola to (3,0) has a slope of -1/2, and the slope of the tangent line at x=1 is 2.

Answer 17

I think I know where I made my mistake now, had to do with the cube root.

Answer 18

cube root? Where did that show up?

Answer 19

In the derivative when I was solving for zero.

Answer 20

Oh, you mean solving the cubic polynomial. Yeah, I used rational root theorem then synthetic division and got lucky when I hit x=1 right away.

Answer 21

Yeah, I ended up getting x=1 as well.

Answer 22

Might want to try that perpendicular slope idea too. It might be easier.

Answer 23

Yeah, I dont know but I feel like the number should be much smaller lol.

Answer 24

hi guys, if y=u^1/2
the derivative of y = y'=(1/2)u(1/2 -1)=(1/2)u^-1/2 du/dx

du/dx
y'=--------- yes or no? ...:D
2sqrt u

so it will be u=2x^2 -6x+9
4x-6
y'= ------------------- =0
2sqrt(2x^2 -6x+9)
4x-6=0
x=1.5 ?
y=....?

Answer 25

^ " u=2x^2 -6x+9" ?

Answer 26

u=(2x^2 -6x+9)^1/2
du/dx=(1/2)[(2x^2 -6x+9)^-1/2](4x -6)

4x - 6
du/dx= -------------------- =0
2(2x^2 -6x+9)^1/2
4x-6=0
x=6/4= 1.5
sub x on u=(2x^2 -6x+9)^1/2

Answer 27

Yeah, I get all that. It looks like you're taking the derivative correctly, but you are taking the derivative of the wrong function. @mark_o. The distance function is
\[\large \sqrt{x^4+x^2-6x+9}\]
Getting the function and differentiating it was never an issue. I helped @tara_magallon find the critical point, and the correct solution of (1,1) was found. (See above).

The distance is minimized at (1,1); the distance is \(\sqrt{5}\) ≈2.2
If you use x=1.5, then the distance is \(\large \frac{\sqrt{117}}{4}\) ≈2.7

Answer 28

ah ok, im sorry i thought you guys want to do derivative on u?.... :D