The slope of the tangent line to a curve at any point (x, y) on the curve is x divided by y. What is the equation of the curve if (2, 1) is a point on the curve?

Question

anonymous · Answer

I know using separation of variables you would eventually get to y^2=x^2+c and have to solve for c using (2,1), but I'm not sure whether I need to simplify the equation more or not like making it y=x+c

mathmate · Answer

y^2=x^2+c does not equal y=x+c, so you just have to keep it as it is.  Put (2,1) in and solve for C to get your equation.

anonymous · Answer

Well, couldn't I do the sqrt of c and solve for that? I don't know how to tell which form the equation should be in

irishboy123 · Answer

dy/dx = x/y y dy = x dx y^2/2 = x^2/2 + K y^2 = x^2 + C 1 = 4 + C; C = -3 y^2 = x^2 -3 check slope 2ydy/dx = 2x dy/dx = x/y: TICK check (2,1) 1^2 = 2^2 -3: TICK VOILA: http://www.wolframalpha.com/input/?i=y%5E2+%3D+x%5E2-3

mathmate · Answer

y^2=x^2+c is the right form of the equation.
If in doubt, differentiate to check if you still get y'=x/y

anonymous · Answer

Okay, I wasn't sure because the condition for c was y(2)=1 and I thought I would have to have the sqrt of y and x to be able to use that condition