VERY hard maths question - please help me lol?

Question

One end of a piece of elastic is attached to a point at the top of a door frame and the other end hangs freely. A small ball is attached to the free end of elastic. When the ball is hanging freely it is pulled down a small distance and then released, so that the ball oscillates up and down on the elastic. The depth d centimetres of the ball from the top of the door frame after t seconds is given by:
d = 100 + 10cos500t

Find
a) the greatest and least depths of the ball,

b) the time at which the ball first reaches its highest position,

c) the time taken for a complete oscillation,

d) the proportion of the time during a complete oscillation for which the depth of the ball is less then 99 centimeters.

Answer 1

a) d max = 110 cm ,dmin 90cm
b) cos 500 t = -1 so 500t = pi ( first height)
t=pi/500 s
c) 500= 2pi/T so T (period)=2pi/500 = pi/250
d) 100+10cos500t <99
10cos500t<-1 cos 500t <-1/10 1,67/500 total time= (2pi-3.34)/500 and proportion = (2pi-3.34)/2pi=
46.84%

Answer 2

d = 100 + 10cos500t

For a) the greatest and least depths of the ball,

We need to maximize and minimize the function d. SInce
-1 <= cos500t <= 1 We have,
90 = 100 + 10 ( -1 ) <= d <= 100 + 10 (1) = 110,
That is,
90 <= d <= 110.

That is the 'greatest and least depths of the ball are 90 and 110 centimetres respectively'.

b) the time at which the ball first reaches its highest position,

At the 'hightest position', the 'distance is 90' by the inequality above, so... 90 = 100 + 10cos500t, yields,
cos500t = -1 or 500t = 180 degrees = 'pi' radian. Hence,
t = pi' / 500 = 0.0063 sec approximately. Note that we have to get the answer in 'radian' and not 'degrees' because thats the 'convention' on how the cos functions are defined. Lets not get too complicated. hehehe. Also, its 180 degrees and not 360 degrees because

at 0 degrees, the ball is at the 'lowest position' for the first time while

at 90 degrees, its 'at the middle position' at first time

and at 180 degrees, its at the 'highest position' for the first time

c) the time taken for a complete oscillation,

One complete oscilation is how long it takes for '360' and by symmetry its the time taken 2 times the time taken for the 'highest point'. That is about
2 * 0.0063 = 0.013 seconds

d) the proportion of the time during a complete oscillation for which the depth of the ball is less then 99 centimeters.
For d < 99.
100 + 10cos500t < 99
10cos500t < -1
cos500t < -0.1

Since arccos ( -0.1 ) = 1.67096375 radian and Cos is a 'decreasing function' for our consideration , this means that
500t > 1.67096375 or rouhgly,
t > 0.0033.

Our '1 total oscillation time' is about 0.013 seconds as we calculated earlier.

Thus depth of the ball is less than 99 centimetres for the times t > 0.0033 as shown above. Hence the 'amount of time for which this happens is roughly' = 0.013 - 0.0033 = 0.0097 seconds.

Thus the proiportion we seek is therefore

= ( amount of time for which d < 99 centimetres ) /
( Total amount of time for which one oscillation takes )

= 0.0097 / 0.013

= 74.62% ( roughly )


Hope this helped.

:)

Answer 3

The following is a guide to the weary student of mathematics who is often confronted with terms which are commonly used but rarely defined. In the search for proper definitions for these terms we found no authoritative, nor even recognized, source. Thus, we followed the advice of mathematicians handed down from time immortal: "Wing It."
CLEARLY:
I don't want to write down all the "in-between" steps.
TRIVIAL:
If I have to show you how to do this, you're in the wrong class.
OBVIOUSLY:
I hope you weren't sleeping when we discussed this earlier, because I refuse to repeat it.
RECALL:
I shouldn't have to tell you this, but for those of you who erase your memory tapes after every test...
WLOG (Without Loss Of Generality):
I'm not about to do all the possible cases, so I'll do one and let you figure out the rest.
IT CAN EASILY BE SHOWN:
Even you, in your finite wisdom, should be able to prove this without me holding your hand.
CHECK or CHECK FOR YOURSELF:
This is the boring part of the proof, so you can do it on your own time.
SKETCH OF A PROOF:
I couldn't verify all the details, so I'll break it down into the parts I couldn't prove.
HINT:
The hardest of several possible ways to do a proof.
BRUTE FORCE (AND IGNORANCE):
Four special cases, three counting arguments, two long inductions, "and a partridge in a pair tree."
SOFT PROOF:
One third less filling (of the page) than your regular proof, but it requires two extra years of course work just to understand the terms.
ELEGANT PROOF:
Requires no previous knowledge of the subject matter and is less than ten lines long.
SIMILARLY:
At least one line of the proof of this case is the same as before.
CANONICAL FORM:
4 out of 5 mathematicians surveyed recommended this as the final form for their students who choose to finish.
TFAE (The Following Are Equivalent):
If I say this it means that, and if I say that it means the other thing, and if I say the other thing...
BY A PREVIOUS THEOREM:
I don't remember how it goes (come to think of it I'm not really sure we did this at all), but if I stated it right (or at all), then the rest of this follows.
TWO LINE PROOF:
I'll leave out everything but the conclusion, you can't question 'em if you can't see 'em.
BRIEFLY:
I'm running out of time, so I'll just write and talk faster.
LET'S TALK THROUGH IT:
I don't want to write it on the board lest I make a mistake.
PROCEED FORMALLY:
Manipulate symbols by the rules without any hint of their true meaning (popular in pure math courses).
QUANTIFY:
I can't find anything wrong with your proof except that it won't work if x is a moon of Jupiter (Popular in applied math courses).
PROOF OMITTED:
Trust me, It's true.

Answer 4

Easy

Answer 5

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