Spring question

[/0]

Possible Spoiler! This is taken from the 2006 Kilbaha exam 2.



A mass is suspended from the ceiling on the end of a spring. It is pulled down and then released. The mass oscillates up and down. The length of the spring x cm at a time t seconds, where t ≥ 0 is given by the equation:



1. Find the times when the length of the spring is a maximum and a minimum, and find the maximum and minimum lengths of the spring, giving all answers correct to four decimal places.

2. Is the motion periodic? Explain your answer.

Thanks

Also another,

The area of the region bounded by the graphs of and is closest to
A. 0.576 (I chose)
B. 0.612 (Answer)
C. 1.188
D. 6.601
E. 11.11

[shinny]

1. I assume its a tech-active, so just graph the function and set the window to x:[0,20] and y:[0,100]. Maximum should be easy to spot at t=0.70, with x=96.34. Similar for the minimum, t=2.20 and x=0.0056.

Not sure what you're really required for this one without knowing the marking scheme, hopefully just the answer was enough. Product rule to differentiate and then solve seems a bit too far fetched for methods, especially since doing something like that wouldn't help much knowing that theres an infinite number of stationary points anyhow.

2. No. As t approaches infinity, e^-0.2t approaches 0, and the whole function approaches 41, as shown by the asymptotish thing (not sure if its truly an asymptote as it still oscillates a bit, but then again, I guess thats negligible in terms of t->infinity anyhow) on the graph if u widen the domain.

[/0]

Thanks shinjitsuzx

With the solutions they found what t must equal in terms of a parameter and then just popped out the answer. Don't you need justification?



But I don't know how to prove when there is a maximum as opposed to a minimum... it gets tricky

[shinny]

Sign table it to prove min/max. And well yeh, I guess with what they did, you'll have to diff it and get up to that.....pretty hideous equation and then yeh, just jump straight to the answer and say (from calc) or something like that.