Find the slope of the tangent line for the curve xy^3-yx^3=6 at the point(1,1)
To find the slope of the tangent line at a given point on a curve, we need to find the derivative of the curve with respect to x and evaluate it at the given point.
First, rearrange the equation to isolate y:
xy^3 - yx^3 = 6
y(x - x^3) = 6
y = 6 / (x - x^3)
Now, take the derivative of y with respect to x:
dy/dx = d(6 / (x - x^3))/dx
Using the quotient rule,
dy/dx = (0 - 6 * (1 - 3x^2)) / (x - x^3)^2
= -6(1 - 3x^2) / (x - x^3)^2
To find the slope at the point (1,1), substitute x = 1 into the expression for the derivative:
dy/dx = -6(1 - 3(1)^2) / (1 - 1^3)^2
= -6(1 - 3)/ (1 - 1)^2
= -6(-2)/0
The expression is undefined when the denominator is zero, indicating that the slope at the point (1,1) is undefined. This means that the tangent line at that point is vertical.
To find the slope of the tangent line at a specific point on a curve, we need to find the derivative of the curve and substitute the coordinates of the point into the derivative equation. Let's start solving it step-by-step.
Step 1: We have the equation of the curve: xy^3 - yx^3 = 6.
Step 2: To find the derivative, we need to differentiate both sides of the equation with respect to x. Let's differentiate each term separately:
Differentiating the left side: d/dx(xy^3) - d/dx(yx^3)
Using the product rule, we have:
y^3 + 3xy^2(dy/dx) - 3x^2y(dx/dx) - x^3(dy/dx)
Step 3: Simplify the derivative equation:
3xy^2(dy/dx) - 3x^2y - x^3(dy/dx) = 0
Step 4: We need to find dy/dx, which represents the slope of the tangent line. Let's simplify the equation further by combining like terms:
3xy^2(dy/dx) - x^3(dy/dx) = 3x^2y
Factor out dy/dx:
dy/dx(3xy^2 - x^3) = 3x^2y
dy/dx = (3x^2y) / (3xy^2 - x^3)
Step 5: Now, we substitute the coordinates of the point (1, 1) into dy/dx to find the slope at that point:
slope at (1,1) = (3(1)^2(1)) / (3(1)(1)^2 - (1)^3)
slope at (1,1) = 3/2
Therefore, the slope of the tangent line for the curve xy^3 - yx^3 = 6 at the point (1, 1) is 3/2.