Parametric Tangent Problem driving me insane

[Andrusko]

Homework Statement

x = e^{t} , y = (t-1)^{2} , (1,1)
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 y = -2x + 3 and I cannot see how you get it.

So I can't do it either way.

Homework Equations

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

The Attempt at a Solution

\frac{dy}{dt} = 2(t-1)

\frac{dx}{dt} = e^{t}

\frac{dy}{dx} = \frac{2(t-1)}{e^{t}}



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

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

x = e^{t}

sot = lnx ...(1)

\sqrt{y} = t - 1

so t = \sqrt{y} + 1 ... (2)

At point (1,1):

t = 0 by equation (1)

t = 2 by equation (2)

So which t to use?

Lets try a different approach:

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

so \frac{dy}{dx} = \frac{2\sqrt{y}}{x}

Substituting (1,1)

\frac{dy}{dx} = 2 = m

so y - y_{1} = m(x - x_{1})

y = 2x - 1

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

 

[Ben Niehoff]

t = lnx ...(1)

\sqrt{y} = t - 1

so t = \sqrt{y} + 1 ... (2)

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.

 

[Andrusko]

Ah of course. That sorts it out:

\frac{dy}{dx} = \frac{2(0-1)}{e^{0}}

\frac{dy}{dx} = -2

y - 1 = -2x + 2
y = -2x + 3

Thanks!