Determining slope of the tangent line.

1. Sep 20, 2009

fghtffyrdmns

1. The problem statement, all variables and given/known data
I've never seen such a problem before: Find the two straight lines that are perpendicular to $$y=0.25x$$ and tanget to the curve $$f(x)=1/x$$

2. Relevant equations

$$y=0.25x$$ ; $$f(x)=1/x$$

3. The attempt at a solution

Using the power rule, I found the derivatives of: $$f(x)=-x^{-2}$$ and $$y=0.25$$

But I am not sure what I do now? Do I find the slope of the equations? Do I equate y=f(x)?

Last edited: Sep 20, 2009
2. Sep 20, 2009

Gregg

If they are perpendicular to y=x/4 then the slope of the line is what?

Now, to find a tangent to y=1/x.

you need the equation for the straight line:

$$y-y_0 = (x-x_0)f'(x_0)$$

what x0 is required to for the desired slope?

3. Sep 20, 2009

fghtffyrdmns

Is it -4?

4. Sep 20, 2009

fghtffyrdmns

Any help anyone? :(

5. Sep 20, 2009

Gregg

All you have to do is not get mixed up, we need a line that is tangent to 1/x. It also needs to be perpendicular to that line you described.So, to get a line that is normal to x/4, you do need a line that has a gradient or slope of -4.

$$y-y_0 = (x-x_0)m$$

First of all in order for the line to be tangent to 1/x it needs to parallel to it at a given point x0. The slope of f(x) is, as you say -1/x^2. The line is going to look like this:

$$y-y_0 = (x-x_0)f'(x_0)$$

and

$$f'(x_0)=-4$$

Get the x0s then find y0 and plug them into the line equation.

6. Sep 20, 2009

fghtffyrdmns

"the x0s then find y0"

What are these? What are the equations that you come up with?

7. Sep 20, 2009

Gregg

Well the general equation for the line is going to take the form y=mx+c, in order to get that equation you need the slope (m) and a point on the line, which will be (x0), y0. For example if I knew my slope was 3 and that a point on my curve was (1,2) then I can work out that the equation for line is:

$$y-2=3(x-1), y=3x-1$$

The same is with this problem with have here, we need to work out at what point the slope is -4 don't we. The derivative of the curve is -1/x^2.

$$\frac{1}{x_0^2} = 4 \Rightarrow x_0 = \pm \frac{1}{2}$$

So you have two. Now all you do is work out the corresponding y values and use the equation for a straight line to work it out.

8. Sep 20, 2009

fghtffyrdmns

I figred out that the functions cross at 2 and -2. I can't use these points?

9. Sep 20, 2009

Gregg

Just find out what value -1/x^2 = -4, that's x1. (there's two of them)

Then plug it into y=1/x to get the y1. (there's two of them)

Then plug it into

y-y1=-4(x-x1)

and you will get 2 lines.

10. Sep 20, 2009

fghtffyrdmns

Okay, I got +/- 1/2 which gets me +/- 2

Now I'm confused which is x and which is y :/

11. Sep 20, 2009

Bohrok

You have two x values; plug them into f(x) = 1/x and you'll get the y values of the coordinates of the points tangent to 1/x. The plug them into the line equation y = mx + b, solve for b, then get the two equations of the tangent lines.

12. Sep 20, 2009

Gregg

x=-1/2
so what is y at this point, 1/(-1/2) = -2
So the corresponding y for -1/2 is -2. you can do the rest?

13. Sep 20, 2009

fghtffyrdmns

Alright. I got y=-4x-4 and y=-4x+4

Are these correct?

14. Sep 20, 2009

Bohrok

They are correct. You should try graphing on a calculator or by hand, or look for an online graphing calculator. 1/x should be easy to graph by hand.

15. Sep 20, 2009

Gregg

Red is -4x-4
Yellow is -4x+4
Green is x/4
Blue is 1/x

Nice!

16. Sep 20, 2009

fghtffyrdmns

Thank you guys :)!