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cheng's question at Yahoo! Answers involving parametric equations and concavity

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MarkFL

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Feb 24, 2012
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Here is the question:

MATH 141 For which values of t is the curve concave upward? (Enter your answer using interval notation.)?

Consider the following.
x = t^3 − 12t, y = t^2 − 6

(a) Find
dy/dx and d^2y/dx^2

(b) For which values of t is the curve concave upward? (Enter your answer using interval notation.)
t = _____________
Here is a link to the question:

MATH 141 For which values of t is the curve concave upward? (Enter your answer using interval notation.)? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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MarkFL

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Feb 24, 2012
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Hello cheng,

We are given the parametric equations:

\(\displaystyle x(t)=t^3-12t\)

\(\displaystyle y(t)=t^2-6\)

Now, using the chain rule, we may state:

\(\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}\)

And so we have:

\(\displaystyle \frac{dy}{dx}=\frac{2t}{3t^2-12}\)

Now, using the quotient and chain rules, we find:

\(\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx} \right)=\frac{(3t^2-12)(2)-(2t)(6t)}{(3t^2-12)^2}\cdot\frac{1}{3t^2-12}=\frac{6\left((t^2-4)-2t^2 \right)}{(3t^2-12)^3}=-\frac{2(t^2+4)}{9(t^2-4)^3}\)

We see our critical values are \(\displaystyle t=\pm2\) and since these roots are of odd multiplicity, concavity will alternate across the resulting intervals, and at \(\displaystyle t=0\) we find \(\displaystyle \frac{d^2y}{dx^2}>0\), hence concavity is up on the interval for $t$:

$(-2,2,)$.

To cheng and any other guests viewing this topic, I invite and encourage you to post other calculus questions here in our Calculus forum.

Best Regards,

Mark.
 
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