Find the slope of the tangent line to the curve 0x^2−2xy+2y^3=10 at point (-4,1)
Help please? Thanks in advance
To find the slope of the tangent line to the curve at a specific point, we can use the concept of calculus.
Step 1: Take the derivative of both sides of the equation with respect to x.
The given equation is 0x^2 - 2xy + 2y^3 = 10.
Taking the derivative of each term with respect to x, we get:
d/dx(0x^2) - d/dx(2xy) + d/dx(2y^3) = d/dx(10)
0 - 2y - 6y^2(dy/dx) = 0
Step 2: Solve the equation for dy/dx.
The dy/dx represents the slope of the tangent line.
Rearrange the equation to isolate dy/dx:
-2y - 6y^2(dy/dx) = 0
-2y = 6y^2(dy/dx)
dy/dx = -2y / (6y^2)
Step 3: Substitute the x and y values of the given point into the equation to find the slope at that point.
The given point is (-4,1). Substitute these values into the equation:
dy/dx = -2(1) / (6(1)^2)
dy/dx = -2 / 6
dy/dx = -1/3
Therefore, the slope of the tangent line to the curve at the point (-4,1) is -1/3.