Question

Calculus: Find the shortest distance from the positive x-axis to the positive y-axis passing through the point (8, 1)...my work attached, but doing something wrong.

Answer 1

posting work now

Answer 2

HI!!

Answer 3

i suggest working with the square of the distance, not the distance
not sure if you did that or not

Answer 4

whoa, made a mistake in the setup I see...bad equation to begin with

Answer 5

y/x = 1/(x-1) ???

Answer 6

I think misty1212 is right you should have worked with H^2.

Answer 7

it is one equation that I use to get to one variable

Answer 8

y/1

Answer 9

x/x-8 = y/1

Answer 10

I think your equation is identical to that which I used to sub in for x^2 + y^2 = H^2

Answer 11

yes, sounds right

Answer 12

so x-8/x = 1/y

Answer 13

I end up with\[H=\sqrt{x^2+1/(x-8)^2}\]

Answer 14

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

Answer 15

yes

Answer 16

I think you missed x right ?

Answer 17

I guess I would have square rooted the H before taking the derivative

Answer 18

H^2 = x^2 + (x/(x-8))^2

Answer 19

Bro, take the derivative to this

Answer 20

0=(1/2)(x^2 + (x-8)^(-2))(2x-2((x-8)^(-3))

Answer 21

2(x/x-8) * (-6/(x-8)^2) + 2x = 0 * (x-8)^3
(x-8)^3 - 8 = 0
x-8 = 2
x = 10
y = 10/10-8 = 5
the distance = rootof(25 + 100) = 5 rootof(5)