Concavity

shamieh

Active member
Find dy/dx and d^2y/dx^2. For which value of t is the curve concave upward?

$$\displaystyle x = t^3 + 1$$
$$\displaystyle y = t^2 - t$$

Here is what I have so far and where I got stuck.

$x' = 3t^2$
$y' = 2t - 1$

$$\displaystyle \frac{2t - 1}{3t^2} /9t^4 = \frac{2t - 1}{3t^2} * \frac{1}{9t^4}$$

I'm confused. Am I doing this correctly? I took the derivative of x and y and then did y' / x' / (x')^2

MarkFL

Staff member
I would use clear notation to indicate what you are doing. We may use the chain rule as follows:

$$\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}$$

Now, as you found:

$$\displaystyle \frac{dy}{dt}=2t-1$$

$$\displaystyle \frac{dx}{dt}=3t^2$$

Hence:

$$\displaystyle \frac{dy}{dx}=\frac{2t-1}{3t^2}$$

Now differentiate again with respect to $x$, carefully applying the quotient, power and chain rules. What do you find?

Prove It

Well-known member
MHB Math Helper
Why did you divide by (x')^2?