Homework Help: Finding the slope of the tangent

1. Dec 28, 2008

meeklobraca

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

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

But I dont 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 im not sure if thats the correct way to go about doing this question.

2. Dec 28, 2008

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.

3. Dec 28, 2008

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, Im not sure how to go about finding y for the quadratic. Its the xy thats throwing me off.

Thank you for your help though!

4. Dec 28, 2008

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.

5. Dec 28, 2008

meeklobraca

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

6. Dec 28, 2008

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?

7. Dec 28, 2008

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?

8. Dec 28, 2008

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.

9. Dec 28, 2008

meeklobraca

Thank you, i remember that now.

For y I got 1 and -1/2

What do you think?

10. Dec 28, 2008

Dick

That's what I got.

11. Dec 28, 2008

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?

12. Dec 28, 2008

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?

13. Dec 28, 2008

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?

14. Dec 28, 2008

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 cancelled 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

15. Dec 28, 2008

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.

16. Dec 28, 2008

meeklobraca

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