`x=4cost , y= 2sint , 0<t<2pi` Determine the open t-intervals on which the curve is concave downward or concave upward.

Expert Answers
lemjay eNotes educator| Certified Educator

`x=4cost`

`y=2sint`

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

`dx/dt=-4sint`

`dy/dt=2cost`

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= (2cost)/(-4sint)`

`dy/dx=-1/2cott`

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(-1/2cott))/(-4sint)`

`(d^2y)/(dx^2) = (1/2csc^2t)/(-4sint)`

`(d^2y)/(dx^2)=-1/8csc^3t`

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`

Setting the second derivative equal to zero result to no solution.

`-1/8csc^3t = 0`

`t={O/}`

Since there are no angles in which cosecant will be zero.

However, on the interval 0<t<2pi, cosecant does not exist at angle pi.

`-1/8csc^3t= DNE`

`t=pi`

So the concavity of the parametric curve changes at `t=pi` .

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 divide the given interval `0lttlt2pi` into two regions. The regions are `0lttltpi` and `pilttlt2pi` . Then, assign a test value for each region. And, plug-in the test values to the second derivative.

For the first region `0lttltpi` , let the test value be `t=pi/2` .

`(d^2y)/(dx^2)=-1/8csc^3(pi/2) = -1/8(1)^3=-1/8`

So the parametric curve is concave down on the interval `0lttltpi` .

For the second region `pilttlt2pi` , let the test value be `t=(3pi)/2` .

`(d^2y)/(dx^2)=-1/8csc^3((3pi)/2) = -1/8(-1)^3=-1/8*(-1)=1/8`

So it is concave up on the interval `pilttlt2pi` .

 

Therefore, the graph of the given parametric equation is concave down on the interval `0lttltpi` and it is concave up on the interval `pilttlt2pi` .