math
posted by sara .
Find
dy/dx and d2y/dx2.
x = et, y = te−t

x = e^t
y = te^t
dy/dx = y'/x'
= (1t)e^t/e^t
= (1t)e^2t
d^2/dx^2 = (x'y"x"y')/x'^3
= ((e^t)(t2)e^t  (e^t)(1t)e^t)/e^3t
= (2t3)e^3t
or, do it directly
x = e^t, so
y = lnx/x
y' = (1lnx)/x^2 = (1t)e^2t
y" = (2lnx3)/x^3 = (2t3)e^3t 
thanks. so how woudl i find For which values of t is the curve concave upward?

just as usual. Where is y" positive?
e^3t is always positive, so all you need is
2t3 > 0
t > 3/2
e^3/2 = 4.48, and you can see from the graph that it is concave upward for x > 4.48
http://www.wolframalpha.com/input/?i=plot+x+%3D+e^t%2C+y+%3D+te^t+for+1%3C%3Dt%3C%3D2