Posted by Catherine on Wednesday, March 16, 2011 at 7:06pm.
A) If x^2+y^3−xy^2=5, find dy/dx in terms of x and y.
dy/dx=________
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2 x dx + 3 y^2 dy - x (2 y dy) - y^2dx = 0
(3 y^2 -2 x y )dy = (y^2 - 2x)dx
dy/dx = (y^2-2x)/(3y^2-2xy) agree
for the second part I tried to plug 1 for x and all the other values for y and then in the third part I plugged 2 for y and 0.96 for x and I got them all wrong:
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No !
That first column is various values of x.
You compute the column of y s for those different values of x.
You do it by using the point (1,2) for initial (xo,yo)
then compute dy/dx there.
dy/dx = (y^2-2x)/(3y^2-2xy)
= (4-2)/(12-4) = 2/8 = 1/4
so for a point (x,y) close to (1,2)
y = yo + (dy/dx) (x-xo)
for example for x = 1.02:
y = 2 + (1/4)(1.02-1) = 2+.02/4 = 2.005
That might get you headed in the right direction
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