Question

Determine the point on the curve y=x^2 which is nearest point (8, 0)

Answer 1

well we know the shortest distance between a point and a line is a straight line whose slope is the opposite reciprocal of the given function's slope....Which we can find by taking the derivative of the function.
y'=2x
So the slope of the line is we seek is -1/2... and our given point is (8,0)
Hope that just about gives you what you need to know!

Answer 2

I need to find a point, not a line though

Answer 3

the nearest point is perp to the curve

Answer 4

you need a point with a slope of -1/2x that hits the point (8,0)

Answer 5

or better yet, prolly define a generic point that slopes out correctly with (8,0)

Answer 6

given points (x,x^2) and (8,0)

slope is defined as:

x^2
-----
x-8

and this needs to equal -1/2x

Answer 7

put it all together and solve for x :)

Answer 8

okay that makes sense.. thanks

Answer 9

you might get an extra value so chk both results

Answer 10

I got 2x^3+x-8=0

Answer 11

how do I solve that?

Answer 12

carefully :)

Answer 13

thats gonna have min max at 6x^2 + 1 = 0; so at x = +- sqrt(1/6) which can narrow down the domain

Answer 14

how am I supposed to find the y coordinate from there?

Answer 15

a cubic looks like this:so knowing where the bends are can narrow down the interval we are looking for is my idea for that

Answer 16

when x=0 we have a change of concavity, so its a good assumption that x is going to rott out between sqrt(1/6) and 8

Answer 17

id start with 1 and try to work out a few integers to see if you can hit a good integer

Answer 18

2+1-8 is negative

2(2)^3+2-8=0

16 - 6 is postitive, so it crosses someplace between 1 and 2

Answer 19

I think this is a lot more difficult than it needs to be

Answer 20

prolly, but i have to rely upon my wits since I aint got your material to guide me :)

Answer 21

if you start from the beginning, the point I'm looking for is (x, x^2) so couldn't I use the distance formula and set it equal to zero then find the derivative to solve for x?

Answer 22

you dont want a distance of zero; 8,0 aint on the curve

Answer 23

but when you're finding max and mins with derivatives, don't you want to set the derivative equal to zero?

Answer 24

hmmm, so your idea is to find the min distance with a derivative of the distance formula? might work ...

Answer 25

Yeah, I vaguely remember someone mentioning that in class last week

Answer 26

id have to remember back many years to recall someone mentioning it in class lol

Answer 27

to find the derivative of a sqrt you would have to use the chain rule, right?

Answer 28

if you derive the distance formula you only gotta worry about the derivative of the innards that make up the numerator

Answer 29

chain rule would work; yes

Answer 30

sqrt(x) derives to x'/2sqrt(x)

Answer 31

and a fraction is only zero when the numerator is zero; so focus on the derivative of the innards

Answer 32

x^2^2 +(x-8)^2

4x^3 +2(x-8) = 2(2x^3+x-8) lol same thing

Answer 33

yeah so would my answer just be (-sqrt(1/6), 1/6)

Answer 34

no, that is just a bending point; the solution is in the interval between 1 and 2

Answer 35

we can do a newton rhapsody method to narrow it closer and closer if need be

Answer 36

I don't understand...