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February 6th, 2012, 06:06 AM   #1
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Equation of the tangent line

Hi!

I need to find dy/dx explicitly (i.e. in the form dy/dx=g(x,y) if

x^3+y^2+2y=2.

Then I need to determine the equation of the tangent to the curve represented by the above equation at the point (-1,1).

Can anyone help?

Thanks
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February 6th, 2012, 06:30 AM   #2
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Re: Equation of the tangent line

This is called "implicit differentiation", since you haven't/won't solve for y explicitly before differentiating.

x^3+y^2+2y=2

We are going to use the chain rule on EVERY TERM. I will write dx/dx, just so you can see that the chain rule is always in play. However, dx/dx is just equal to 1, so we will make the appropriate simplification...


(x^3) ' + (y^2) ' + (2y) ' = (2) ' ......here I have just put a ' after every term to indicate that we are differentiating.

(3x^2 dx/dx) + (2y dy/dx) + (2 dy/dx) = 0 ... chain rule, power rule, and/or constant rule

Factor out the dy/dx and simplify...

(3x^2) + (dy/dx)*(2y + 2) = 0

Solve for dy/dx...

(dy/dx)*(2y + 2) = -3x^2


after dividing.

Plug in x = -1, y = 1 to THIS formula to get the slope at the point (-1, 1). Then use point slope formula to get an equation of the tangent line.
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February 6th, 2012, 06:36 AM   #3
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Re: Equation of the tangent line

Thanks, I thought that.

Can you please help me with the point slope formula.
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February 6th, 2012, 07:05 AM   #4
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Re: Equation of the tangent line

Actually Is the following correct,

plug in x=-1 and y=1, i get the slope m = -3/4

using the point slope fomula i get

y-y1=m(x-x1)

y-1=-3/4(x--1)

y=-3/4x+1/4
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February 6th, 2012, 07:24 AM   #5
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Re: Equation of the tangent line

Yes, that is correct.
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February 9th, 2012, 01:29 AM   #6
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Re: Equation of the tangent line

Thank you for your help
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