HELP! Calculus question?

Question

Find where the slope is defined:

x^3 + y^3 = xy

Wouldn't it just be the following:

3x^2 + 3y^2*(dy/dx) = x*(dy/dx)+y

3y^2*(dy/dx) -x*(dy/dx) = -3x^2+y
(dy/dx)(3y^2-x) = -3x^2

so: (dy/dx) = (-3x^2 + y)/(3y^2 -x) ??

Thanks a lot, 10 pts to first correct answer

Answer 1

Your differentiation is good. Now the slope is not defined when the denominator is zero, ie when x = 3y^2.
So the slope is defined everywhere except (t, 3t^2)

Answer 2

Yes, that's the start, although there's a + y missing from the end of your second last line. But you have to find where this is undefined, i.e. where 3y^2 - x = 0.
(Actually, to know the slope is undefined we need 3y^2 - x = 0 and -3x^2 + y ≠ 0. If they are both zero the equation (dy/dx)(3y^2-x) = -3x^2 + y becomes 0 dy/dx = 0, so we have no information about dy/dx.)

Note that if x = 3y^2, then -3x^2 + y = y - 27y^4 = y (1 - 27y^3).

So we have x = 3y^2 and x^3 + y^3 = xy. Substitution gives
27y^6 + y^3 = 3y^3 <=> 27y^6 - 2y^3 = 0
<=> y^3 (27y^3 - 2) = 0
<=> y = 0 or 27y^3 = 2
<=> y = 0 or (cube root of 2) / 3.

When y = 0 we get x = 0 also, so -3x^2 + y = 0 and we have no information about the slope of the graph. For the other value we have -3x^2 + y = y - 27y^4 = y (1 - 27y^3) = -y ≠0, so the slope is definitely undefined at the point ((cube root 4)/3, (cube root 2)/3).

To determine the slope at (0, 0) we need to do a bit more work. Let x be very small and consider the equation y^3 - xy + x^3 = 0 (just rewriting the main equation). Write the first few terms of a power series for y in terms of x, noting that the constant term is of course 0: y = ax + bx^2 + O(x^3).

Putting this into the equation gives us
a^3x^3 - ax^2 - bx^3 + x^3 + O(x^4) = 0
<=> (-a)x^2 + (a^3-b+1)x^3 + O(x^4) = 0
and so we get a=0, b=1. So y = x^2 + O(x^3), which means that near (0, 0) the slope will be 2x + O(x^2), and so the slope at (0, 0) is well defined and equals 0.

So there is exactly one point on this curve where the slope is undefined, and it is ((cube root 4)/3, (cube root 2)/3).

Answer 3

(dy/dx)(3y^2-x) = -3x^2
you forgot about the y on the right hand side