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=________
B) Using your answer for dy/dx, fill in the following table of approximate yvalues of points on the curve near x=1 y=2.
0.96 ______
0.98 ______
1.02 ______
1.04 ______
C) Finally, find the yvalue for x=0.96 by substituting x=0.96 in the original equation and solving for y using a computer or calculator.
y(0.96)= ________
D) How large (in magnitude) is the difference between your estimate for y(0.96) using dy/dx and your solution with a computer or calculator?
___________
For the first part I get (2x+y^2)/(3y^2x2y) and 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:
my values are:
0.96 = 1.2765
0.98= 1.1285
1.02= 0.8875
1.04= 0.7885
and for part C) 0.255
and For part D) 1.5315
What am I doing wrong in the last 3 parts?

Calculus (pleas help!!!)  Damon, Wednesday, March 16, 2011 at 7:19pm
A) If x^2+y^3−xy^2=5, find dy/dx in terms of x and y.
dy/dx=________

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^22x)/(3y^22xy) agree

Calculus (pleas help!!!)  Damon, Wednesday, March 16, 2011 at 7:29pm
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:
=============================
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^22x)/(3y^22xy)
= (42)/(124) = 2/8 = 1/4
so for a point (x,y) close to (1,2)
y = yo + (dy/dx) (xxo)
for example for x = 1.02:
y = 2 + (1/4)(1.021) = 2+.02/4 = 2.005
That might get you headed in the right direction
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