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)
2 answers
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.
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.