What Am I doing wrong in trying to find the tangent to this parametric equ?(W/ANSWER)

**skinsdomination09** · September 27th 2012, 03:47 PM

find the tangent to the curve at the point corresponding to the given value of the parameter

x=e^(√t)
y= t - ln t²
t = 1

I found...

dx/dt = (e√x)/(2x)
dy/dt = 1 - (2t/t²)

then you do

dy/dt / dx/dt

I got -e from that. I plugged in my coordinates and got

y-1 = -e(x-e)

the book says -(2/e)x+3. How am I so off?

Thanks guys. and no offense but it'd be cool if you could do an entire "debug" in one post instead of asking questions that I probly couldn't answer. Not trying to sound like a jerk but i'm just not the sharpest calc student on the planet so theres no point in wasting our time.

**skeeter** · September 27th 2012, 05:06 PM

$x = e^{\sqrt{t}}$

$\frac{dx}{dt} = \frac{e^{\sqrt{t}}}{2\sqrt{t}}$

$y = t - 2\ln{t}$

$\frac{dy}{dt} = 1 - \frac{2}{t}$

at $t = 1$ , $x = e$ , $y = 1$ , $\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = -\frac{2}{e}$

$y - 1 = -\frac{2}{e} (x - e)$

finish it ...

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