at , , ,
finish it ...
find the tangent to the curve at the point corresponding to the given value of the parameter
x=e^(√t)
y= t - ln tē
t = 1
I found...
dx/dt = (e√x)/(2x)
dy/dt = 1 - (2t/tē)
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.