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?

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Mar 9, 2018

Answer:

#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#.

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