Question

All > Homework Help

829ffff: 1 year ago

2x^2 - 3xy - y^3 = 1.
a) Verify the slope is dy/dx=(4x-3y)/(3x+3y^2)
b) Find the point on C where the curve has a horizontal tangent line.

need help with part bSuppose curve C is defined by the relation 2x^2 - 3xy - y^3 = 1.
a) Verify the slope at any point on C is dy/dx=(4x-3y)/(3x+3y^2)
b) Find the point on C where the curve has a horizontal tangent line.Justify your answers.

I did part a.For part b I set the numerator equal to zero and got x=3/4y. I plugged it in to the original equation and got -y^3-18/16y^2-1=0. If i plug this into the cubic equation i get the square root of a negative. Where is my mistake?

Answer This Question

Report as

Answer 1

katandmah: 1 year ago via iPhone

If I remember correctly, and I don't remember what it is, but I think there is a standard equation for tangent of line using dx/dy. Sorry but I hope it helps.

Helpful Fun Thanks for voting Comments (0)

Report as

Add a comment...

Answer 2

NilesAnderson: 1 year ago

Asking for help on math home work :P

Helpful Fun Thanks for voting Comments (0)

Report as

Add a comment...

Answer 3