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Help finding equation of the tangent line to the graph

  • Thread starter hauk-gwai
  • Start date
  • #1
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Homework Statement


Find the equation of the tangent line to the graph of x^2 - xy + y^2 = 19 (where y=y(x)) at (3,-2).


Homework Equations





The Attempt at a Solution



So, this is what I did:
d/dx(x^2-xy+y^2) = (19)d/dx

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

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

And I am stuck, the answer to this question is:
y+2=8/7(x-3)

and I don't know how to get the slope.
 

Answers and Replies

  • #2
Gib Z
Homework Helper
3,346
4
Well you have your values of x and y, (3,-2). Sub that into the 2x+2y(dy/dx)=0 you got to solve for dy/dx, then you have the gradient of the tangent. Now use the gradient-point form of a straight line to find the equation of the tangent.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,772
911

Homework Statement


Find the equation of the tangent line to the graph of x^2 - xy + y^2 = 19 (where y=y(x)) at (3,-2).


Homework Equations





The Attempt at a Solution



So, this is what I did:
d/dx(x^2-xy+y^2) = (19)d/dx

2x*dx/dx-dy/dx+dy/dx+2y(dy/dx)=0
d(x^2)/dx= 2x and d(y^2)/dx= 2y(dy/dx) but d(-xy)/dx is NOT -dy/dx+ dy/dx!

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

And I am stuck, the answer to this question is:
y+2=8/7(x-3)

and I don't know how to get the slope.
 
  • #4
3
0
I am pretty sure I got the x^2 and the y^2 part right.

I am just having trouble taking the deriv of the bolded part:
x^2-xy+y^2=19

Do I apply the product rule for the d/dx(-xy)?
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,772
911
Yes! It's a product, isn't it?
 
  • #6
3
0
Yeah, but the thing is, I tried applying the product rule, and my results were different, I answer did not match with the answer key, so somewhere within the product rule, I screwed up somewhere, so how do you take d/dx(-xy)?
 
  • #7
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,098
129
The product rule:
d/dx(u v)
= (du/dx) v + u (dv/dx)

In this case, uv = xy, so ... ?
 

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