Find the equation of the tangent line to the curve x^2 + xy + y^2 = 3 at the point (1,1).

Question

 the might be the most confusing question for me, in my opinion.

Answer 1

Naturally, you have to find the derivative of y with respect to x.

 

But you can't easily solve for y, so how do you do that?

 

Remember the chain rule and the product rule and forge ahead:

 

d/dx (x² + xy + y²) = d/dx 3 = 0

 

Break the left side into separate terms

 

d/dx (x²) = 2x

d/dx (xy) = x dy/dx + y dx/dx = x dy/dx + y

d/dx (y²) = 2y dy/dx

 

So the whole left side is:

 

2x + x dy/dx + y + 2y dy/dx, which equals 0

 

Gather terms and solve for dy/dx:

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

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

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

 

Plug the point (1,1) into the expression for dy/dx:

dy/dx= - (2 + 1) / (1 + 2) = -3/3 = -1

 

Now you know the slope of the tangent line at (1,1).  From that you can find the equation for the line.