Finding the slope of the tangent

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Homework Help Overview

The discussion revolves around finding the slope of the tangent line to the curve defined by the equation 2y² - xy - x² = 0 at the point where x = 1. Participants are exploring the implications of the quadratic nature of the equation in y and how it affects the number of possible slopes at that point.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the process of differentiating the equation and question the correctness of the derivative. There is an exploration of how to isolate y from the quadratic equation and the implications of having multiple y values for a single x value. Some participants express uncertainty about the arithmetic involved in calculating slopes.

Discussion Status

The discussion is active, with participants providing feedback on each other's attempts at differentiation and solving for y. There is recognition of the possibility of multiple slopes due to the quadratic nature of the equation, and some guidance has been offered regarding the methods to solve the quadratic equation for y.

Contextual Notes

Participants are navigating the complexities of differentiating a product and dealing with a quadratic equation, which introduces potential for multiple solutions. There are indications of arithmetic errors that may affect the final slope calculations.

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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!
 
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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.
 
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!
 
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.
 
I used the product rule for derivatives to differentiate the xy.
 
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?
 
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?
 
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.
 
Thank you, i remember that now.

For y I got 1 and -1/2

What do you think?
 
  • #10
That's what I got.
 
  • #11
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
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
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
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
 
  • #15
(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
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!
 

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