Question

The point on the curve x^2+2y=0 that is nearest to the point (0,-1/2) occurs where y is
A) 1/2
B) 0
C) -1/2
D) -1
E) none

I don't know how to start this problem.

Answer 1

given a point in the curve (x,y), what is the distance from that point to (0,-1/2) ?

Answer 2

sorry, i don't understand what to do. :(

Answer 3

given any two points \((x_0,y_0),~(x_1,y_1)\), the distance between them is given by\[d=\sqrt{(x_1-x_0)^2-(y_1-y_0)^2}\]

Answer 4

typo, meant\[d=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}\]

Answer 5

so we are given one point, which we can plug into this formula, giving\[d=\sqrt{(x-0)^2+(y-(-\frac12))^2}=\sqrt{x^2+(y+\frac12)^2}\]with me so far?

Answer 6

Yes

Answer 7

ok, so x and y are somewhere on the curve, and d is what we want to minimize
notice that
1: we need to get all this in terms of x to minimize it
2:if we minimize the distance, we minimize the *square* of the distance as well, so we can make our distance formula much cleaner by minimizing its square

Answer 8

as per observation #2 we can write\[d=x^2+(y+\frac12)^2\]as the function we want to minimize. Now we just need to substitute x in terms of y, or vice versa, and take the derivative

Answer 9

What do you mean by substitute x in terms of y?

Answer 10

well to minimize you have to take the derivative, but if I ask you to take the derivative of this right now, you'd be stuck, right? because we need everything in terms of x or y
go back to your equation for the curve, solve for either x or y (or maybe something slightly different, hint hint), then substitute one for the other

Answer 11

so y=-1/2(x^2) ?
Then I plug that into d=x^2+(y+1/2)^2 and then take the derivative?

Answer 12

yes, you could do that
let me point out a major shortcut: try solving the curve for x^2

Answer 13

I plugged in y=-1/2(x^2), did the derivative, and got x=0. Is that right? And how do I find y?

Answer 14

you would find y by using the equation for the curve
let me see if I get the same thing

Answer 15

no that's not right
try my little tip of solving for x^2; it will make your life so much easier

Answer 16

so x^2=-2y?

Answer 17

yes, now sub that into the distance formula, and take the derivative wrt y

Answer 18

so is that -2y'+2y'(y+1/2) ?

Answer 19

you took the derivative wrt x out of habit, but here you want to take the derivative wrt y

Answer 20

:( please explain

Answer 21

if you had instead put the whole thing in terms of x, which is equally valid, and I told you to take the derivative, you wouldn't be writing x', would you?
that's because you are too used to thinking that y is a function of x. In this case, we can flip the script and put everything in terms of y, then take the derivative wrt y

i.e f(y)=y^2+y
f'(y)=2y+1

see, I just treated it like d/dx of f(x), but instead it's d/dy of f(y)
make any sense?

Answer 22

I get it. So why isn't -2y'+2y'(y+1/2) right?

Answer 23

because what is y' ?

Answer 24

y'=dy/dx
but you didn't take the derivative wrt x at all! you took it wrt y
see the problem here?

Answer 25

so y'=1?

Answer 26

right!
dy/dy=1

Answer 27

Okay. So if I take the derivative, I get -2+2(x+1/2)?

Answer 28

that x should be a y, but otherwise, yes
all set =?

Answer 29

so y=1/2?

Answer 30

that's what I got

Answer 31

I know this problem was probably pretty tough for you, but rest assured that the next time you have to minimize a distance, it won't be so many new tricks to learn. Just noticed what we did to make our lives easy:

1. minimized the *square* of the distance to avoid that nasty radical
2. made the simplest substitution possible: x^2=-2y
3. took the derivative wrt the variable we put everything in terms of

the first two tricks are especially handy. Try to let them sink in ;)

Answer 32

Alright. Thank you so much for your help!

Answer 33

welcome!