`x=t^2 , y=lnt` Determine the open t-intervals on which the curve is concave downward or concave upward.

Expert Answers

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`dy/dx = (dy/dt)/(dx/dt) = (1/t)/(2t) = 1/(2t^2) `

`(d^2y)/(dx^2) = d/dx(dy/dx) = d/dt(dy/dx) dt/dx `
`= -1/2 t^{-3} (1/(2t)) = -1/(4t^4)`

This is always positive, so the function is concave upward for all t where its second derivative is defined, which is to say all `t ne 0` .

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