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 3:09pm
x = e^t
y = te^-t
dy/dx = y'/x'
d^2/dx^2 = (x'y"-x"y')/x'^3
= ((e^t)(t-2)e^-t - (e^t)(1-t)e^-t)/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 4:55pm
thanks. so how woudl i find For which values of t is the curve concave upward?
math - Steve, Monday, July 28, 2014 at 6:37pm
just 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
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