[SOLVED] *THIRD prime? What? If x^2 + xy + y^3 = 1, find the value of y''' at the point where x = 1.?

Question

What? If x^2 + xy + y^3 = 1, find the value of y''' at the point where x = 1.?

If

x2 + xy + y3 = 1,

find the value of

y'''

at the point where

x = 1.

:(

I think second prime is 2 when x=1

...

stuck after that

Answer 1

The Best Answer for What? If x^2 + xy + y^3 = 1, find the value of y''' at the point where x = 1.?

When x = 1, what is y? We have:

1 + y + y^3 = 1

y + y^3 = 0

y(y^2 + 1) = 0

y = 0 ... (since y^2 + 1 > 0)

Thus we are interested in the point (x, y) = (1, 0). Implicitly differentiating with respect to x:

2x + y + xy' + 3y^2y' = 0

When (x, y) = (1, 0), we have:

2(1) + 0 + 1y' + 3(0)^2y' = 0

y' = -2

Implicitly differentiating again:

2 + y' + xy'' + y' + 6y(y')^2 + 3y^2y'' = 0

2 + 2y' + xy'' + 6y(y')^2 + 3y^2y'' = 0

When (x, y, y') = (1, 0, -2), we have:

2 + 2(-2) + 1y'' + 6(0)(-2)^2 + 3(0)^2y'' = 0

y'' = 2 ... (good!)

Implicitly differentiating again:

2y'' + xy''' + y'' + 6(y')^3 + 12yy'y'' + 6yy'y'' + 3y^2y''' = 0

3y'' + xy''' + 6(y')^3 + 18yy'y'' + 3y^2y''' = 0

When (x, y, y', y'') = (1, 0, -2, 2), we have:

3(2) + 1y''' + 6(-2)^3 + 18(0)(-2)(2) + 3(0)^2y''' = 0

y''' = 42

Hope that helps!

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.

SHOW ANSWER