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?
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
Oh boy, looks like you're having some trouble with these calculations! But fear not, Clow Bot is here to help add some laughter to the equation!
For part B), instead of plugging in random values for y, let's use the provided y-value of 2 and calculate the corresponding x-values using the dy/dx expression. So, plug y = 2 into dy/dx and we get:
dy/dx = (-2x + 4)/(12 - 2x)
Now, let's find the x-values near x = 1:
For x = 0.96:
dy/dx = (-2(0.96) + 4)/(12 - 2(0.96))
= 2.08/10.08
≈ 0.2063
For x = 0.98:
dy/dx = (-2(0.98) + 4)/(12 - 2(0.98))
= 2.04/10.04
≈ 0.2032
For x = 1.02:
dy/dx = (-2(1.02) + 4)/(12 - 2(1.02))
= 1.96/9.96
≈ 0.1968
For x = 1.04:
dy/dx = (-2(1.04) + 4)/(12 - 2(1.04))
= 1.92/9.92
≈ 0.1935
Now, let's move on to part C). Substituting x = 0.96 into the original equation, we get:
(0.96)^2 + y^3 - (0.96)(y^2) = 5
Solving this equation for y using a calculator or computer, we find that y ≈ 2.0803.
Lastly, in part D), we want to find the magnitude of the difference between your estimate for y(0.96) using dy/dx and the actual solution using a calculator or computer. In this case, it would be the absolute value of your estimate (-1.2765) minus the actual solution (2.0803). So the magnitude is |(-1.2765) - 2.0803|.
I hope this helps clarify things for you! Keep on clowning around!
Let's go step by step and address each part of the question.
A) To find dy/dx, we need to use implicit differentiation. Start by differentiating the equation x^2 + y^3 - xy^2 = 5 with respect to x on both sides:
d/dx(x^2) + d/dx(y^3) - d/dx(xy^2) = d/dx(5)
Using the chain rule, we get 2x + 3y^2 * dy/dx - y^2 - 2xy * dy/dx = 0
Rearranging the terms, we get: dy/dx = (2x - y^2) / (3y^2 - 2xy)
So the correct derivative is dy/dx = (2x - y^2) / (3y^2 - 2xy).
B) Now, using the equation for dy/dx, you can approximate the y-values for points on the curve near x=1 and y=2. By plugging in the x-values into the equation and solving for y, you can complete the table:
For x = 0.96:
Substituting x = 0.96 into dy/dx: dy/dx = (2*0.96 - y^2) / (3y^2 - 2*0.96y)
Since y = 2, plug in y = 2 into dy/dx: dy/dx = (2*0.96 - 2^2) / (3*2^2 - 2*0.96*2)
Simplifying this expression will give you the approximate value of dy/dx at x = 0.96.
For x = 0.98:
Repeat the same process as above.
Continue this process for x = 1.02 and x = 1.04.
C) To find the y-value for x = 0.96, you need to substitute x = 0.96 into the original equation x^2 + y^3 - xy^2 = 5. Then solve the equation for y using a computer or calculator. The resulting value of y will give you the y-coordinate at x = 0.96.
D) The difference between your estimate for y(0.96) using dy/dx and the solution obtained with a computer or calculator is the absolute value of the difference between the two values. Simply subtract the estimate from dy/dx from the solution obtained using a computer or calculator and take the absolute value.
It seems that you made mistakes in your calculations for parts B, C, and D. Make sure to double-check your calculations and use parentheses correctly when substituting values into equations.
If you provide your work for these parts, I can help review your calculations and point out any errors.
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