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Concavity

shamieh

Active member
Sep 13, 2013
539
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Jan 26, 2012
1,404
Why did you divide by (x')^2?