Calculus

[Cogman]

[edit]Forgot the Cream Filling[/edit]
Ok, Im in AP Calculus. We are at dirivitives and tangent lines (not too far). Just so you know this is one of the last problems and missing it really will not hurt me much, Plus Im going to get help from the teacher tommorrow anyways . The questions goes something like this (btw Our second math teacher who is pretty smart could not figure it out ether.)

Find Constant K such that the line is tangent to a point on the function

Function Line
k/x -3/4*x + 3

we are dealing with dirivitives and So we know that it has to deal with the slope, Right? so

f'(x)= -kx^-2 and the slope of the line -3/4 Therefore we should be able to conclude that -kx^-2 = -3/4 Thats about how far we got . Im going to ramble here, but I dont know if im on the right track or not.

k/x = -3/4*x+3 so k= -3/4*x^2 + 3x so 4k = -3x^2 + 12x 4k = x(-3x + 12).

Any help?

 

[Cogman]

nm Figured it out. in a bizzare sort of way. Ill see if I can explane it.

f'(x) = -k/x^2
-k/x^2 = -3/4
k = 3x^2/4

now, we know that the x has to be common on the tangent line right? so. we take the origional forumula of k/x and say

3x^2/4/x = -3x/4 + 3
3x/4 = -3x/4 + 3
6x = 12
x = 2
(skiped some basic steps, but you should be able to find them) So you take the dirivitive of k/x again and say
-k/x^2 = -3/4
k/2^2 = 3/4
k/4 = 3/4
k = 3

and it works. That was fun. Tune in next time I get confused. Same time, same place.

 

[Chaotic42]

Man, I wish we could get some kind of mathematical formatting for these forums.

 

[Stealth1024]

an equation toolbar would be awesome... what would be cooler is if we could insert pictures into our posts

 

[Mday]

THIS IS NOT A HOMEWORK FORUM.