# Calculus (pleas help!!!)

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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 y-values of points on the curve near x=1 y=2.

0.96 ______
0.98 ______
1.02 ______
1.04 ______

C) Finally, find the y-value 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^2-x2y) 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!!!) - ,

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^2-2x)/(3y^2-2xy) agree

• Calculus (pleas help!!!) - ,

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^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