`x=3t^2 , y=t^3-t` Determine the open t-intervals on which the curve is concave downward or concave upward.

Expert Answers

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Given parametric equations are:

`x=3t^2,y=t^3-t`

We need to find the second derivative, to determine the concavity of the curve.

`dy/dx=(dy/dt)/(dx/dt)`

Let's take the derivative of x and y with respect to t,

`dx/dt=3*2t=6t`

`dy/dt=3t^2-1`

`dy/dx=(3t^2-1)/(6t)`

`dy/dx=(3t^2)/(6t)-1/(6t)`

`dy/dx=t/2-1/(6t)`

`(d^2y)/dx^2=d/dx[dy/dx]`

`=(d/dt[dy/dx])/(dx/dt)`

`=(d/dt(t/2-1/(6t)))/(6t)`

`=(1/2-1/6(-1)t^(-2))/(6t)`

`=(1/2+1/(6t^2))/(6t)`

`=((3t^2+1)/(6t^2))/(6t)`

`=(3t^2+1)/(6t^2(6t))`

`=(3t^2+1)/(36t^3)`

Curve is concave upwards if second derivative is positive and concave downwards if it is negative,

So, the curve is concave upward for `t>0`

Curve is concave downward for `t<0`

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