Determining slope of the tangent line.

[fghtffyrdmns]

Homework Statement

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

Homework Equations

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

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)?

 

[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?

 

[fghtffyrdmns]

Is it -4?

I'm confused already :(.

 

[fghtffyrdmns]

Any help anyone? :(

 

[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.

 

[fghtffyrdmns]

"the x0s then find y0"

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

 

[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.

 

[fghtffyrdmns]

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

 

[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.

 

[fghtffyrdmns]

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

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

 

[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.

 

[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?

 

[fghtffyrdmns]

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

Are these correct?

 

[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.

 

[Gregg]

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

Nice!

 

[fghtffyrdmns]

Thank you guys :)!