Question

What is the equation of the line tangent to the curve x^2 + y^3 = 9 at the point (-1, 2)? I am thrown off because it's no longer a circle. Does it matter?

Answer 1

`y=x/6+13/6`

Explanation:

The function `x^2+y^3=9`...........`[1]` can be differentiated to give us the gradient [ slope] of the line required.

This can be done either implicitly or explicitly, it's easier done explicitly since we are given the `x and y` values.

Differentiating both sides of `[1]` implicitly,

`d/dx[x^2+y^3]`=`[d/dx 9]` and thus,

`2x+3y^2dy/dx=0`, Rearranging, `dy/dx=[-2x]/[3y^2]` and plugging in `x-1, y=2`

`dy/dx=1/6`.

From the equation for a straight line...`[y-y1]=1/6[[x-x1]` and so `[y-2]=1/6[x-[-1]]` which gives us `y=1/6x+13/6`.

You can also use the chain rule to differentiate the function explicitly, ie, `dy/dx=dy/dz*dz/dx` where `z` represents the contents of a bracket or similar function.

From`[1]` ...`y=[9-x^2]^[1/3`, so `dy/dx=1/3[9-x^2]^[-2/3` `d//dx[9-x^2]`

=`[-2x]/3 [[sqrt[9-x^2]]^2]^[1/3]*`, and substituting `x=-1` , will give`dy/dx=1/6`.