Find the slope of the tangent to the curve: x^3 +y^2 =9 @ (0,3).

Find the slope of the tangent to the curve: x^3 +y^2 =9 @ (0,3).

Slope=

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y = sqrt (9 - x^3)

y ' = -3x^2 / sqrt(9 - x^3)

for x = 0 ===> y ' = 0

so, if m = 0

tangent :

y = mx + b

(y - 3) / x = 0 ===> the tangent is the line y = 3

the slope of the tan line at (0,3) is 0
basically you do implicit differentiation and end up with the expression 3x^2+2y(dy/dx)=0 and then you solve for dy/dx and find (-3x^2/2y) if you plug in your pt you will find the answer to be 0

x^3 + y^2 = 9

Differentiating implicitly:

3x^2 + 2yy' = 0

(-3x^2)/2y = y'

Slope of tangent line at (0, 3) = 0

Eqn of tangent:

y - 3 = 0

y = 3