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
Don't know why you bother with the zero:
−2xy+2y^3=10
(yes, (-4,1) is on that graph.
just use the product rule, remembering that x' = 1:
-2yx' - 2xy' + 6yy' = 0
-2y + (6y-2x)y' = 0
y' = 2y/(6y-2x)
at (-4,1),
y' = 2(1)/(6(1)-2(-4)) = 2/(6+8) = 1/7
To find the slope of the tangent line to the given curve at the point (-4, 1), we need to find the derivative of the curve equation with respect to x and evaluate it at (-4, 1).
Step 1: Find the derivative of the curve equation with respect to x.
To do this, we differentiate each term of the equation separately using the rules of differentiation.
0x^2: The derivative of a constant (0 in this case) is zero.
-2xy: We differentiate each term separately:
The derivative of -2x with respect to x is -2.
The derivative of y with respect to x is dy/dx.
2y^3: We differentiate each term separately:
The derivative of 2y^3 with respect to x is 6y^2 * dy/dx.
10: The derivative of a constant (10 in this case) is zero.
Putting it all together, the derivative of the curve equation is:
0 + (-2) * (x * dy/dx) + 6y^2 * dy/dx + 0 = -2xy + 6y^2 * dy/dx = 0
Step 2: Evaluate the derivative at the given point (-4, 1).
We substitute x = -4 and y = 1 into the equation -2xy + 6y^2 * dy/dx = 0.
-2(-4)(1) + 6(1^2) * dy/dx = 0
8 + 6 * dy/dx = 0
6 * dy/dx = -8
dy/dx = -8/6
dy/dx = -4/3
So, the slope of the tangent line to the curve at the point (-4, 1) is -4/3.