Question

(calculus) The point on the curve x^2+2y = 0 that is nearest to the point (0,-1/2) occurs where y is?

Answer 1

So you know we are talking about distance?
And we want to minimize the distance?

Answer 2

ohh. ok. and how would i do that?

Answer 3

do you know the distance formula

Answer 4

yes

Answer 5

ok use that to find the distance between (x,y) and (0,-1/2)
also if x^2+2y=0 then how can we rewrite y?

Answer 6

y= -x^2/2

Answer 7

right
now use the distance formula one (x,-x^2/2) and (0,-1/2)

Answer 8

\[\sqrt{-x^{2} + (-\frac{ 1 }{ 2 } + \frac{ x ^{2} }{ 2 }})^2\]

Answer 9

is that right?

Answer 10

how did that - end up on the outside of that square?

Answer 11

i did (0 - x)^2

Answer 12

which is x^2

Answer 13

not -x^2

Answer 14

ohh right.

Answer 15

you could say (-x)^2 but not -x^2

Answer 16

but the square of (-x) is also the same as saying the square of x

Answer 17

since the square of -1 is 1

Answer 18

alright now you can find derivative of d^2 to find your critical numbers

Answer 19

you could find it of d but d^2 is easier

Answer 20

so the derivative is x^3+x

Answer 21

now set that equal to 0 to find critical numbers

Answer 22

x=0

Answer 23

of we have the relationship between x and y is y=-x^2/2
so if x=0 then y=?

Answer 24

y=0

Answer 25

so 0 is the answer?

Answer 26

yep that's what i got