Question

For what t-values is the curve concave upward? x = 2sint, y = 3cost, 0 < t < 2pi

Answer 1

\[\frac{ dy }{ dx }=\frac{ -3 }{ 2 }\tan(t)\]\[\frac{ d^2y }{ dx^2 }=\frac{ -3 }{ 4 }\sec^3(t)\]\[\frac{ -3 }{ 4 }\sec^3(t) > 0\]\[\sec^3(t)<0\]
I am stuck

Answer 2

All of the steps are correct, so you don't have to verify the work.

Answer 3

I would find t values for which dy/dx=0
then I would inspect it below and above those values, to see if it is concave up

this is only a suggestion since there may be a more efficient way

Answer 4

but peresonally

x = 2sint, y = 3cost

x/2=sint
y/3=cost

square both

(x/2)^2=sin^2t
(y/3)^2=cos^2t

add together

sin^2t+cos^2t=1

(x/2)^2 + (y/3)^2=1


maybe this can help somehow, I gotta go to sleep good night!

Answer 5

Forgot which quadrants your secant is negative in?
Can convert to cosines if that helps.
\(\large\rm \frac{1}{\cos^3t}<0\)

Cosine is our x-coordinate as we're traveling along the unit circle, ya?
So I guess we're negative in quadrants 2 and 3.

Answer 6

so the interval would be ( pi/2, 3pi/2 )

Answer 7

Mmmmmmmmm ya that sounds right.

Answer 8

would I state the interval as concave up or down?

Answer 9

Since the equality sign got changed, and the question is asking for concave up.

Answer 10

or do I just leave it as (pi/2, 3pi/2) and not state the concavity

Answer 11

\[\large\rm y''=-\frac34\sec^3t\]And you made the statement that this expression needs to be larger than 0 for `concave up`.\[\large\rm 0<-\frac34\sec^3t\]

The only way we can get concave up is if sec^3t is negative,
(That makes the entire thing positive, right?)

Answer 12

yes

Answer 13

So you found the interval for `concave up`.
Don't worry about the inequality flip and all that fancy stuff that happened in between. It didn't change what we were looking for.

Answer 14

Ah I see, thanks for the rescue :)

Answer 15

yay team!