Posted by erika on .
The ellipse x^2+3y^3=13, has two points when x+1
. Find the slope of the ellipse at both these points.
2. Thoroughly explain the method you used to find the slope in part 1. What are the strengths
and weaknesses of the method you used?
3. Explain whether your method will work to find the slope of x^2+xy+y^2=3 at points 1, 2 and
(1, 1 . If it succeeds, please show your work as you find these two slopes. If your method fails,
that’s fine, just explain thoroughly why it failed.

calculus 
Reiny,
I will assume you meant
x = 1 , (the + and + are on the same key)
x^2 + 3y^2 = 13
2x + 6y dy/dx = 0
dy/dx = 2x/6y = x/3y
when x = 1 ......
1 + 3y^2 = 13
3y^2 = 12
y^2 = 4
y = ± 2
the two points are (1,2) and (1, 2)
at (1,2) , slope = 1/6
at (1,2) , slope = 1/6 = 1/6
2. your call
3. x^2 + xy + y^2 = 3
2x + xdy/dx + y + 2ydy/dx = 0
dy/dx(x + 2y) = 2x  y
dy/dx = (2x  y)/(x+2y)
at (1,2) , slope= (22)/(1+4) = 0
at (1,1) , slope = (2+1)/(12) = 1
I don't get the instructions.
" If your method fails, that’s fine, just explain thoroughly why it failed."  > very odd! 
calculus 
erika,
Thank you! helped a lot and yes I meant '=' ,
basically I think he just wants us to explain on why it did or didn't fail, yes I know very strange! :)