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Parametric Tangent Problem driving me insane

  • Thread starter Andrusko
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  • #1
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Homework Statement



[tex] x = e^{t} , y = (t-1)^{2} , (1,1) [/tex]
Find an equation of the tangent to the curve at a given point by two methods. Without eliminating the parameter and by first eliminating the parameter.

The answer in the book says [tex] y = -2x + 3 [/tex] and I cannot see how you get it.

So I can't do it either way.


Homework Equations



[tex]\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]


The Attempt at a Solution



[tex] \frac{dy}{dt} = 2(t-1)[/tex]

[tex]\frac{dx}{dt} = e^{t}[/tex]

[tex]\frac{dy}{dx} = \frac{2(t-1)}{e^{t}}[/tex]



okay so now I need to substitute in (1,1)

Rearranging x and y in terms of t does no good:

[tex]x = e^{t}[/tex]

so[tex] t = lnx ...(1)[/tex]

[tex]\sqrt{y} = t - 1[/tex]

so [tex]t = \sqrt{y} + 1 ... (2)[/tex]

At point (1,1):

t = 0 by equation (1)

t = 2 by equation (2)

So which t to use?

Lets try a different approach:

[tex] \frac{dy}{dx} = \frac{2(\sqrt{y} + 1 - 1)}{e^{lnx}}[/tex]

so [tex]\frac{dy}{dx} = \frac{2\sqrt{y}}{x}[/tex]

Substituting (1,1)

[tex]\frac{dy}{dx} = 2 = m[/tex]

so [tex]y - y_{1} = m(x - x_{1})[/tex]

[tex]y = 2x - 1[/tex]

I really can't see what I'm doing wrong. Any help appreciated.
 

Answers and Replies

  • #2
Ben Niehoff
Science Advisor
Gold Member
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[tex] t = lnx ...(1)[/tex]

[tex]\sqrt{y} = t - 1[/tex]

so [tex]t = \sqrt{y} + 1 ... (2)[/tex]

At point (1,1):

t = 0 by equation (1)

t = 2 by equation (2)

So which t to use?
Remember every positive number has two real square roots. Try using the other one.
 
  • #3
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Ah of course. That sorts it out:

[tex]\frac{dy}{dx} = \frac{2(0-1)}{e^{0}}[/tex]

[tex]\frac{dy}{dx} = -2[/tex]

[tex] y - 1 = -2x + 2[/tex]
[tex] y = -2x + 3[/tex]

Thanks!
 

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