use implicit differentiation to find the slope of the tangent line to the curve of x^2/3+y^2/3=4 at the point (-1,3sqrt3)
Rodriguez is your school subject? Please follow directions.
x^2/3+y^2/3=4
2/3 x^-1/3 + 2/3 y^-1/3 y' = 0
y' = -(y/x)^1/3
So, at (-1,3√3), y' = -(3√3/-1)^(1/3) = √3
You can see the graphs at
http://www.wolframalpha.com/input/?i=plot++x^%282%2F3%29%2By^%282%2F3%29%3D4+%2C+y+%3D+-%E2%88%9A3%28x-1%29+%2B+3%E2%88%9A3
The fractional power makes wolframalpha blink, so using symmetry I showed the line in the first quadrant, rather than the second.
To find the slope of the tangent line to a curve using implicit differentiation, follow these steps:
Step 1: Differentiate both sides of the equation with respect to x.
In this case, we differentiate the equation x^(2/3) + y^(2/3) = 4 with respect to x. To differentiate y with respect to x, we treat it as an implicit function and use the chain rule.
Differentiating x^(2/3) gives us:
(d/dx)(x^(2/3)) = (2/3)x^(-1/3) = 2/3x^(-1/3)
Differentiating y^(2/3) with respect to x involves the chain rule:
(d/dx)(y^(2/3)) = (2/3)y^(-1/3) * (dy/dx)
Step 2: Solve for dy/dx.
Now we need to isolate dy/dx. By rearranging the equation, we have:
(2/3)x^(-1/3) + (2/3)y^(-1/3) * (dy/dx) = 0
Multiply through by 3:
2x^(-1/3) + 2y^(-1/3) * (dy/dx) = 0
Subtract 2x^(-1/3) from both sides:
2y^(-1/3) * (dy/dx) = -2x^(-1/3)
Divide both sides by 2y^(-1/3):
(dy/dx) = -x^(-1/3) / y^(-1/3)
Simplify the expression:
(dy/dx) = -y^(1/3) / x^(1/3)
Step 3: Plug in the coordinates of the given point.
Now plug in the x-coordinate, -1, and y-coordinate, 3√3, of the point (-1, 3√3) into the equation we derived in step 2 to find the slope at that point.
(dy/dx) = -y^(1/3) / x^(1/3) = -(3√3)^(1/3) / (-1)^(1/3)
Simplifying the expression:
(dy/dx) = -(3√3)^(1/3) / (-1)^(1/3) = -(3)^(1/6) / (-1)^(1/3) = -√3 / (-1) = √3
Therefore, the slope of the tangent line to the curve x^(2/3) + y^(2/3) = 4 at the point (-1, 3√3) is √3.