Homework SourseI know its a long question sorta but it is just one question
I know its a long question sorta, but it is just one question. Please help me out!
If x^4+y^3-2xy^2=0, find dydx in terms of x and y.
dxdy=
Using your answer for dy/dx, fill in the following table of approximate y-values of points on the curve near x=1, y=1.
0.96 0.98 1.02 1.04
___? ___? ___? ___?
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) is approcimately
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 ____?
If x^4+y^3-2xy^2=0, find dydx in terms of x and y.
dxdy=
Using your answer for dy/dx, fill in the following table of approximate y-values of points on the curve near x=1, y=1.
0.96 0.98 1.02 1.04
___? ___? ___? ___?
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) is approcimately
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 ____?
Solution
dy/dx = 4x^3-2y^2;
put x=1 and y=1 to get the value of dy/dx;
= 4*1-2*1 = 2;
similarly do it for x=1 and y = 0.96 , 0.98 ,1.02,1.04;
2. Y(0.96) = 0.849+Y3-1.92Y2 ;
Y3-1.92Y2 -Y+0.849 = 0;
soving , we get y = 2.1991 , 0.497,-0.776 (three answers because above equation is 3rd degree)
..3) now you have original y(0.96)..and you have estimated value in the I part..
so substtract them to get the answer
