Find the slope of the tangent line to the graph at the given point.
x^3 + y^3 – 6xy = 0, (4/3, 8/3)
x^3 + y^3 - 6xy = 0
3x^2 + 3y^2 y' - 6y - 6xy' = 0
(3y^2-6x)y' = 6y-3x^2
y' = (2y-x^2)/(y^2-2x)
So, at (4/3,8/3) the slope is (2(8/3)-(4/3)^2)/((8/3)^2-2(4/3)) = 4/5
Using the point-slope form of the line, its equation is
y - 8/3 = 4/5 (x - 4/3)
see the graphs at
www.wolframalpha.com/input/?i=plot+x%5E3+%2B+y%5E3+-+6xy+%3D+0,+y+%3D+4%2F5+(x-4%2F3)%2B8%2F3
To find the slope of the tangent line to the graph at the given point, we need to take the derivative of the equation with respect to x and evaluate it at the given point.
Taking the derivative of the equation implicitly with respect to x, we get:
3x^2 + 3y^2(dy/dx) - 6(y + x(dy/dx)) = 0
Simplifying the equation, we have:
3x^2 + 3y^2(dy/dx) - 6y - 6x(dy/dx) = 0
Rearranging the equation and isolating dy/dx, we get:
dy/dx = (6y - 3x^2) / (3y^2 - 6x)
Now we can substitute the given point (4/3, 8/3) into the derivative equation:
dy/dx = (6(8/3) - 3(4/3)^2) / (3(8/3)^2 - 6(4/3))
Simplifying further, we have:
dy/dx = (16 - 16/3) / (8 - 8/3)
dy/dx = (48/3 - 16/3) / (24/3 - 8/3)
dy/dx = (32/3) / (16/3)
dy/dx = 32/3 * 3/16
dy/dx = 32/48
dy/dx = 2/3
Therefore, the slope of the tangent line to the graph at the point (4/3, 8/3) is 2/3.
To find the slope of the tangent line to the graph at a given point, you need to find the derivative of the function and then evaluate it at that point. Here's how you can do it step by step:
1. Start with the equation of the curve: x^3 + y^3 - 6xy = 0.
2. Take the derivative of both sides of the equation with respect to x: (d/dx)(x^3 + y^3 - 6xy) = 0
3. Use the chain rule to find the derivative of y^3 with respect to x. Since y is a function of x, we have to use the chain rule to differentiate it. The chain rule states that (d/dx)(f(g(x))) = f'(g(x)) * g'(x). In this case, f(u) = u^3 and g(x) = y, so the derivative of y^3 with respect to x is 3y^2 * dy/dx.
4. Apply the chain rule to the other terms as well. The derivative of x^3 with respect to x is simply 3x^2, and the derivative of -6xy with respect to x is -6x * dy/dx.
5. Simplify the equation to isolate dy/dx on one side. The equation becomes 3x^2 + 3y^2 * dy/dx - 6x * dy/dx = 0.
6. Rearrange the equation to solve for dy/dx: 3y^2 * dy/dx - 6x * dy/dx = -3x^2.
7. Factor out dy/dx: (3y^2 - 6x) * dy/dx = -3x^2.
8. Divide both sides by (3y^2 - 6x) to isolate dy/dx: dy/dx = -3x^2 / (3y^2 - 6x).
9. Now that you have the derivative expression, you can find the slope of the tangent line at the given point, (4/3, 8/3), by substituting the x and y values into the equation. Plug in x = 4/3 and y = 8/3 into dy/dx = -3x^2 / (3y^2 - 6x):
dy/dx = -3 * (4/3)^2 / [3 * (8/3)^2 - 6 * (4/3)].
Simplify the expression:
dy/dx = -16/9 / (8 - 8) = -16/9 /0.
Since the denominator is zero, the slope of the tangent line is undefined at the point (4/3, 8/3). This means that the tangent line is vertical at that point.