2nd Derivative for Parametric Equations: x = t^2 - 5, y = sin(2t)

[koreamaniac101]

a curve is defined by the parametric equations below:

x = t² - 5
y = sin(2t)

Find the left endpoint and the right endpoint of the interval on which the curve is concave up.

So I managed to find the 2nd derivative for this:

(-2tsin2t - cos2t) / 2t³

But when i try to graph this in my calculator, it says all points are undefined.

Is my 2nd derivative wrong? Is there another way to find the range where the graph is concave up?

 

[ksdhart2]

Okay, so the rules for derivatives of parametric functions tell us that, if x = x(t) and y = y(t) and \(\displaystyle \dfrac{dx}{dt} \ne 0\), then:

\(\displaystyle \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\)

We know that \(\displaystyle \dfrac{dx}{dt}=\dfrac{d}{dt}(t^2-5)=2t\) and \(\displaystyle \dfrac{dy}{dt}=\dfrac{d}{dt}(sin(2t))=2cos(2t)\). That means that \(\displaystyle \dfrac{dy}{dx}=\dfrac{2cos(2t)}{2t}=\dfrac{cos(2t)}{t}\).

Finding the second derivative is then just a matter of taking the derivative of that with respect to t and dividing by dx/dt. Let's do that now.

\(\displaystyle \dfrac{d^2y}{dx^2}=\dfrac{\frac{d}{dt}\left( \dfrac{cos\left(2t\right)}{t}\right)}{2t}= \dfrac{\dfrac{cos\left(2t\right)}{t^2}-\dfrac{2sin\left(2t\right)}{t}}{2t}= \dfrac{cos\left(2t\right)-2tsin\left(2t\right)}{2t^3}\)

That checks with what you got, so you're good on that front. The only thing I can think of is that you messed up on inputting it into your calculator. I used Desmos online graphing calculator and it worked just fine. The only place its undefined is at t=0. The real question is, what did you get when you tried to find the intervals where f''(t) is positive/negative?

 

[koreamaniac101]

Solved!

I figured out the problem! I got a new calculator, and I imputed xsinx instead of x * sinx. It thought that I was calling a new variable with the name xsinx. Thanks for your help!!