Question

Find where the points on the given curve on concave up in interval notation.

I've managed to get the derivatives of the curve, and I think I've got the critical points, but I'm having trouble figuring out if the graph is concave up at all, and if it's not, I forgot how to state that in interval notation. I'll put the equations in the next post.

Answer 1

Have you taken the second derivative?

Answer 2

Yeah I've got the first and second. I just wanted to use the equation thing to post them because they're kinda messy.

Answer 3

\[\frac{ dy }{ dx } = \frac{ (\sin(t) }{ 2\sin(2t) }\] \[\frac{ d^2dy }{ dx^2 } = \frac{ 2\sin(2t)*\cos(t)-\sin(t)*4\cos(2t) }{ 2\sin(2t)^2*-\sin(2t) }\]

Answer 4

Forgot to add, the boundary is from 0 < t < pi.

So going off the first derivative the critical points should be 0 and pi right?

Answer 5

For the the second derivative, you used f'(x)g(x)-g'(x)f(x) / (g(x))^2?

Answer 6

Yeah and I verified it's correct.

Answer 7

And yes, those critical points seem right.

Answer 8

You just have to find the points of inflection and plug them in then.

Answer 9

You want to find concavity, right?

Answer 10

Yes. I'm looking for where the curve is concave up.

Answer 11

Are you having trouble finding the inflection points then?

Answer 12

I don't think so. I think I've managed to solve the second derivative down so that the inflection points are also 0 and pi.

My problem is, I'm supposed to run a value between the critical points to determine concavity right? Well most of the values I tried (pi/2, pi/4 etc) just caused the denominator to zero out. I managed to find a couple of values that actually gave me a result, but they were all negative. So I think the curve is concave down over the entire interval. If that's correct, I don't know how to state that in interval notation.

Answer 13

Are you sure your second derivative is right? I'm not entirely certain why you multiplied (2sin(2t))^2 by -sin(2t)

Answer 14

Yes, I'm using an online homework submission program, and it's saying that second derivative is correct.

Answer 15

I'll see what I get for inflection points then first.

Answer 16

I'm getting something different, but it doesn't seem to really work either.

Answer 17

Oh, never mind, I messed up a negative.

Answer 18

You said you got 0 and pi?

Answer 19

Do you know if they're right?

Answer 20

Yes.

Answer 21

No, I'm not sure if those are right. On second thought, it seems odd that the critical points would also be inflection points. Is that possible?

Answer 22

No, that's possible. I think you're right though, I got those numbers too once I plugged in the identity right.

Answer 23

You were right from the start, unless you plugged something in wrong.

Answer 24

Ok. So if the curve isn't concave up along the interval, is there a way to say that in interval notation?

Answer 25

Isn't it positive from 0 to pi though?

Answer 26

I put pi/4 into my graphing calculator and the value came back as -1 in the equation for the second derivative. Unless I entered it wrong, I'll check.

Answer 27

I plugged in 1.5 and it came out positive.

Answer 28

and then pi/4 comes out negative.

Answer 29

I double checked the equation I put in and I'm getting a negative value for 1.5 and pi/4. Even if that's wrong though, if 1.5 is positive and pi/4 is negative... That doesn't track with the critical points.

Answer 30

Yeah. I'm probably doing something stupid. @sourwing can probably help you better.

Answer 31

I doubt it's you. This question seems pretty ridiculous. I've run it by a tutor it seems to have stumped them as well.

Answer 32

what are the y(t) and x(t). There is something not quite right in your second derivative. But I need y(t) and x(t)

Answer 33

x = cos(2t) y = cos(t)

Answer 34

The second derivative should be right unless there's some kind of error with the program though.

Answer 35

ok, did you do
d/dt (-sint / (-2sin(2t))
-------------------------
d/dt (cos(2t))

??

Answer 36

Yes and I ended up with the second derivative I posted.

Answer 37

ok, did you get
(1/4) (sect - 3cost)
after simplifying?

Answer 38

No, I didn't simplify. I just put the answer in like that, and was using my graphing calculator with the unsimplified equation.

Answer 39

well, you should simply so that it is less work when you solve

Answer 40

Ok, I'll get to work on simplifying it. Might take me a bit though.

Answer 41

I would suggest you do the second derivative again. Yours didn't match mime

https://www.wolframalpha.com/input/?i=%5Cfrac%7B+2%5Csin%282t%29*%5Ccos%28t%29-%5Csin%28t%29*4%5Ccos%282t%29+%7D%7B+2%5Csin%282t%29%5E2*-%5Csin%282t%29+%7D

Answer 42

Oh jeez, I missed a 2 in there and didn't notice this whole time. It should be

\[\frac{ 2\sin(2t)*\cos(t)-\sin(t)*4\cos(2t) }{ 2\sin(t)^2-2\sin(2t) }\]

Sorry for wasting all of yours and greenglasses' time trying to solve the wrong equation.

Answer 43

lol, ok you found the mistake. But notice that sin(2t) = 2sint cost

dy/dx = sint / (4sint cost) = (1/4)sect

the second derivative is much easier to do if you simplify first derivative first

Answer 44

So \[\frac{ 1 }{ 4\cos(t) } = 0\] =
\[\frac{1}{\cos(t)} = 0\]

Are there real solutions to this?

Answer 45

also, forget about my simplified version earlier, it is also wrong lol

Answer 46

(1/4)sect is only the first derivative. Second derivative is
(1/4)sect / (-2sin(2t))
(1/4)sect / (-4sint cost)
(1/4)csct sec^2(t)

Answer 47

(1/4)csct sec^2(t) = 0 has no solution

Answer 48

I forgot what that means for the concavity of the original curve.

Answer 49

if the second derivative is positive, then it's concave up

Answer 50

Ok, I see what you're saying. So the final answer in interval notation should be (0, pi)?

Answer 51

it's (-pi,0)

https://www.wolframalpha.com/input/?i=parametric%7Bcos%282t%29%2C+sin%28t%29%7D%2C+-pi%3C%3D+t+%3C%3D+0

Answer 52

But if the boundaries in the initial problem are 0 < t < pi ?