Question

If x=e^t and y=t^2-3, then for what values of t, t>=0, is the function y=f(x) concave up?

Answer 1

let's eliminate the parameter t
from x = e^t
lnx = t
then y = (lnx)^2 - 3

to be concave up, the second derivative must be positive
y' = 2(lnx)(1/x)
y'' = 2lnx (-1/x^2) + (1/x)(1/x)
= (-2lnx + 1)/x^2

x^2 is always positive
so -2lnx + 1 > 0
lnx < -1/2 ----> x < .6065...

so e^t < .6065
t < ln .6065
t < -.5

Answer 2

x=e^t and y=t^2-3
dx/dt = e^t
dy/dt = 2t

dy/dx = (dy/dt)/(dx/dt) = 2t/e^t

d^y/dx^2 = d/dt (dy/dx) * dt/dx

d/dt(dy/dx) = -2(t-1)/e^t
dt/dx = 1/(dx/dt) = 1/e^t

d^2y/dx^2 = -2(t-1)/e^2t

e^2t is always positive, so y">0 when t<1

Hmmm. From above,
y' = 2lnx/x
y" = 2lnx (-1/x^2) + <u><b>2</b></u>(1/x)(1/x)
= 2(1-lnx)/x^2

so, y" > 0 when lnx < 1
or, when t < 1