Finding the slope of the tangent

[meeklobraca]

Homework Statement

Find the slope(s) of the tangent at x = 1 for 2y^2 - xy -x^2 = 0

Homework Equations

The Attempt at a Solution

Ive figured the derivative to be 2x+4/4y+x

But I don't know how to make this equation work to find a y value so I can find the slope. I tried singling out a y value and got y = 1 but I am not sure if that's the correct way to go about doing this question.

Thank You for your help!

 

[Dick]

You have a sign error in the derivative, I'd check it. But as to your original question the original curve is quadratic in y. You can have two different values of y that are both on the curve at x=1. They will give you two different slopes. If the problem doesn't give you any clue as to which to use, you'd better work out both.

 

[meeklobraca]

Can you point out the sign error for me? I tried working it out again and I got the following

4y dy/dx - y + x dy/dx - 2x = 0

Which led me to 2x + y / 4y + x = dy/dx

As for the slopes, I am not sure how to go about finding y for the quadratic. Its the xy that's throwing me off.

Thank you for your help though!

 

[Dick]

I get 4y dy/dx - y - x dy/dx - 2x = 0. How did you get a + sign on the x dy/dx term? Don't worry about the xy term. Substitute x=1 into the equation first, then solve for y.

 

[meeklobraca]

I used the product rule for derivatives to differentiate the xy.

 

[Dick]

Sure, I'm with you so far. But then shouldn't both of the terms you get from the product rule have a minus sign in front of them, since xy is SUBTRACTED?

 

[meeklobraca]

okay i just thought of this and I get it now, my equation should have been

4y dy/dx -(y + x dy/dx) - 2x = 0


which leads me to the derivative that you got. Thank you. Now onto the slope. If I plug 1 into the equation to get a y value I get

2y^2 - 1y - 1^2 = 0
= 2y^2 -y - 1 = 0

Do I combine the y in the equation before solving for y?

 

[Dick]

You must have solved a quadratic equation before, right? You can either try to factor, complete the square or use the quadratic equation. Your choice.

 

[meeklobraca]

Thank you, i remember that now.

For y I got 1 and -1/2

What do you think?

 

[Dick]

That's what I got.

 

[meeklobraca]

Thanks Dick! I apprecaite your help very much.

Would you be able to look at the other question I posted up in a couple of threads below?

 

[meeklobraca]

Oh one more thing, for the slope for this question, I got 1 and 2/3. Is the 2/3 right or is it - 1/6?

 

[Dick]

If I put y=(-1/2) into (2+y)/(4y-1) I don't get either 2/3 or -1/6. What are you doing?

 

[meeklobraca]

I got the 2/3 by putting x and y into the deriviative giving me, 2(1) - 1/2 / 4(-1/2) - 1. I canceled out the -1/2

I got the - 1/6 by mistake, I actually meant -2/3 but I got that by working out the equation. I got 3/2 / -3 which I turned into - 2/3

 

[Dick]

(3/2)/(-3)=(-1/2). I think you might want to review arithmetic with fractions, it looks like you have similar problems in the other thread.

 

[meeklobraca]

arrgg. I am not even thinking straight. I added the numerator while multiplying the denomenator. Its such a stupid mistake. Thanks for the help though!