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

#### MarkFL

Staff member
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.

#### MarkFL

Staff member
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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