Monday

March 2, 2015

March 2, 2015

Posted by **sara** on Monday, July 28, 2014 at 2:05pm.

dy/dx and d2y/dx2.

x = et, y = te−t

- math -
**Steve**, Monday, July 28, 2014 at 3:09pmx = e^t

y = te^-t

dy/dx = y'/x'

= (1-t)e^-t/e^t

= (1-t)e^-2t

d^2/dx^2 = (x'y"-x"y')/x'^3

= ((e^t)(t-2)e^-t - (e^t)(1-t)e^-t)/e^3t

= (2t-3)e^-3t

or, do it directly

x = e^t, so

y = lnx/x

y' = (1-lnx)/x^2 = (1-t)e^-2t

y" = (2lnx-3)/x^3 = (2t-3)e^-3t

- math -
**sara**, Monday, July 28, 2014 at 4:55pmthanks. so how woudl i find For which values of t is the curve concave upward?

- math -
**Steve**, Monday, July 28, 2014 at 6:37pmjust as usual. Where is y" positive?

e^-3t is always positive, so all you need is

2t-3 > 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

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