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

`x=2+t^2`


`y=t^2+t^3`


First, take the derivative of x and y with respect to t.


`dx/dt =2t`


`dy/dt =2t + 3t^2`


Then, determine the first derivative,`dy/dx` . Take note that in parametric equation, the formula of `dy/dx` is:


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


Applying this formula, the first derivative is:


`dy/dx = (2t+3t^2)/(2t)`


`dy/dx= (2t)/(2t) + (3t^2)/(2t)`


`dy/dx=1+3/2t`


Then, determine the second derivative of the parametric equation. To do so, apply the formula:


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


So the second derivative is:


`(d^2y)/(dx^2) = (d/dt (1+3/2t))/(2t)`


`(d^2y)/(dx^2) = (3/2)/(2t)`


`(d^2y)/(dx^2) = 3/2 * 1/(2t)`


`(d^2y)/(dx^2) = 3/(4t)`


Take note that the concavity of the curve changes when the second derivative is zero or does not exist.


`(d^2y)/(dx^2)= 0`     or     `(d^2y)/(dx^2)= DNE`


Since the second derivative of the given parametric equation is


`(d^2y)/(dx^2) = 3/(4t)`


there are no values of t in which it results to zero.


However, the second derivative does not exist when its denominator is zero. Setting 4t equal to zero yields


`4t=0`


`t=0/4`


`t=0`


So the concavity of the parametric curve changes at t=0.


Now that the inflection is known, apply the second derivative test.


Take note that when the value of the second derivative on an interval is positive, the curve on that interval is concave up.


`(d^2y)/(dx^2)gt0 `      `:.`  concave up


And when the value of the second derivative on an interval is negative, the curve on that interval is concave down.


`(d^2y)/(dx^2)lt0`     `:.` concave down


So assign a test value for each interval t<0 and t>0. And plug-in the test values to the second derivative.


For the interval t<0, let the test value be t=-1.


`(d^2y)/(dx^2) = 3/(4t)`


`(d^2y)/(dx^2) = 3/(4(-1))=3/-4=-3/4`


So the parametric curve is concave down on the interval t<0.


For the interval t>0, let the test value be t=1.


`(d^2y)/(dx^2) = 3/(4(1))=3/4`


So the parametric curve is concave up on the interval t>0.



Therefore, the graph of the given parametric equation is concave down on the interval t<0 and it is concave up on the interval t>0.

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