Critical numbers related

[farealol]

I was trying to solve number (ii) but I found the second derivative of s (derivative of velocity) is unreasonably long and complicated to solve for critical numbers. Is it the only way or are there other ways to do it?

Thanks for any help!

 

[Steven G]

You do not need to find the zeros of the 2nd derivative of s. You just need to see if the sign is positive or negative for s'' at the critical values (which you find from s' not s"

If you really want help then you should show us your work so that we can see where you went wrong. So please follow the forum guidelines and show your work.

 

[farealol]

You do not need to find the zeros of the 2nd derivative of s. You just need to see if the sign is positive or negative for s'' at the critical values (which you find from s' not s"

If you really want help then you should show us your work so that we can see where you went wrong. So please follow the forum guidelines and show your work.

I only did a very rough calculation before, it’s a bit messy, that’s why I didn’t post it. I did it again. But it is really long.

But don’t you need to find the roots of s’’ in order to find the critical numbers? And the critical numbers may be the local extrema?

 

[HallsofIvy]

No, that's the "first derivative test"- if f'(x)< 0 for x< a (so f is decreasing going up to x= a), f'(x)> 0 for x> a (so f is increasing going away from x=a) then (a, f(a)) is a minimum. If f'(x)> 0 for x< a (so f is increasing going up to a), f'(x)< 0 for x> a (so f is decreasing for x going away from a) then (a, f(a)) is a maximum.

 

[Steven G]

View attachment 19333

I only did a very rough calculation before, it’s a bit messy, that’s why I didn’t post it. I did it again. But it is really long.

But don’t you need to find the roots of s’’ in order to find the critical numbers? And the critical numbers may be the local extrema?

You really need to do as suggested. However if you do not understand then you should ask! But you chose to ignore! I'll say it one more time (and combined with HallofIvy) that will be a total of three times now that you were told that you do not need to find the 2nd derivative and certainly not its roots. Find the 1st derivative, set it equal to 0 and then you will have your critical values. Post back with the critical values and we will go from there.

 

[Dr.Peterson]

View attachment 19322
I was trying to solve number (ii) but I found the second derivative of s (derivative of velocity) is unreasonably long and complicated to solve for critical numbers. Is it the only way or are there other ways to do it?

Perhaps the others are missing the fact that you are asked for the maximum velocity, not the maximum displacement, so you do indeed need to solve v' = 0, not s' = 0, and v' is s". (I missed that myself, at first.)

On the other hand, you are complicating v', rather than simplifying it. I'm going to look for a way to simplify it as early as possible; it isn't immediately obvious. While I do that, can you tell us where this problem comes from? It does seem intentionally (or maybe unintentionally) overcomplicated.

Actually, you should try simplifying the original s(t) using the double-angle formula. That will make things a lot easier.

 

[farealol]

You really need to do as suggested. However if you do not understand then you should ask! But you chose to ignore! I'll say it one more time (and combined with HallofIvy) that will be a total of three times now that you were told that you do not need to find the 2nd derivative and certainly not its roots. Find the 1st derivative, set it equal to 0 and then you will have your critical values. Post back with the critical values and we will go from there.

I am asked the maximum velocity, not the maximum displacement. The function in the question is position, so the velocity is already the first derivative of s and I’m looking for maximum velocity that I should set v’=0, that is s’’=0, no?

 

[farealol]

Perhaps the others are missing the fact that you are asked for the maximum velocity, not the maximum displacement, so you do indeed need to solve v' = 0, not s' = 0, and v' is s". (I missed that myself, at first.)

On the other hand, you are complicating v', rather than simplifying it. I'm going to look for a way to simplify it as early as possible; it isn't immediately obvious. While I do that, can you tell us where this problem comes from? It does seem intentionally (or maybe unintentionally) overcomplicated.

Actually, you should try simplifying the original s(t) using the double-angle formula. That will make things a lot easier.

Thank you. I was really doubting myself a second. I will try it myself first.
This question was from my teacher and I don’t know if she made it herself. And it should be high school level.