posted by jasmineT on .
Find the slope of the tangent line to the curve (a lemniscate)
2(x^2+y^2)^2=25(x^2–y^2) at the point
I know that we are supposed to use chain rule for this problem but i think i am messing up on the algebra. Can someone please help me, i keep on getting m= (42x^3-42y^2x)/(58x^2+58y^2)
Differentiate (implicitly) with respect to x, denote dy/dx as y'
Collect terms and solve for y' in terms of x and y:
At (-3,1), x=-3, y=1
Check for arithmetic errors.
4(x^2 + y^2)(2x + 2yy') = 50x - 50y y'
y' = ((50x - 8x^3 - 8xy^2)/[ (8x^2)y + 8y^3 - 50y)
plug in your point (-3,1) to get the slope
Take it from there.
Oh my goodness, thank you soooo much for writing it out i totally wasn't getting it! THank you so much :)
Do you think you coul check my arithmatic for another problem? It's
Use implicit differentiation to find the slope of the tangent line to the curve
y/x+6y=x^2–6 at the point (1–5/31)
I got [(x+6y)(y')-(y)(1+6y')]/(x+6y)^2=2x then y'=(2x+x^2+12yx+6y^2-y)/(x+6y+6). Is that right?
y/(x+6y) or (y/x)+6y?
As it is, the expression means the latter.
Since I see (x+6y)y' somewhere, I think you mean the first interpretation.
Cross multiply to get
Check my arithmetic and differentiation and take it from here.