find the slope of the curve x^3 y^3 + y^2 = x + y at the point (1,1) and (1,-1)??
x3y3 + y2 = x + 1
3x2y3 + 3x3y2y' = 1 + y'
At (1,1)
3 + 3y' = 1 + y'
y' = -1
At (1,-1)
-3 + 3y' = 1 + y'
y' = 2
WHAT IS THAT x3y3 + y2 = x + 1
??
To find the slope of a curve at a given point, we can use the concept of derivatives.
Step 1: Differentiate both sides of the equation.
The given equation is x^3 * y^3 + y^2 = x + y.
Differentiating both sides with respect to x will give us:
(d/dx) (x^3 * y^3 + y^2) = (d/dx) (x + y)
Using the power rule, the derivative of x^3 * y^3 with respect to x is 3x^2 * y^3. The derivative of y^2 with respect to x is 0 because y is not a function of x. The derivative of x with respect to x is 1, and the derivative of y with respect to x is dy/dx. Simplifying the right side gives us dy/dx + 1.
So, we have: 3x^2 * y^3 + 0 = 1 + dy/dx
Step 2: Substitute the given points into the equation.
For the point (1,1):
Substituting x = 1 and y = 1 into the equation, we get:
3(1)^2 * (1)^3 + 0 = 1 + dy/dx
3 + 0 = 1 + dy/dx
3 = 1 + dy/dx
dy/dx = 3 - 1
dy/dx = 2
For the point (1,-1):
Substituting x = 1 and y = -1 into the equation, we get:
3(1)^2 * (-1)^3 + 0 = 1 + dy/dx
-3 + 0 = 1 + dy/dx
-3 = 1 + dy/dx
dy/dx = -3 - 1
dy/dx = -4
So, the slope of the curve at the point (1,1) is 2 and at the point (1,-1) is -4.