Application of Calculus In the real world Q's (1 Viewer)

[SouL Shinobi]

Yo. im having problems with these questions, the teacher gave some examples but i still dont get it (im still new to this log and exp stuff) so could you please show me how to do these 3 q's.

The distance X metres at time t seconds of a particale moving in a straight line from a fixed point O is given by "x=2-2cos2t"

a) sketch the graph of the equation for 0<t<2N (N=Pie)

b) Find the times when the particale is at rest and where the particale is at these times

c) what is the maximum distance of the particale from O

the answers was some wierd a** stuff which i would never get

answers: t=21N/36, t=5N/6; x=3.37, x=4 (yeah answers a bit confusing as they didnt number which answer is for which question... )

Peace out Jack

 

[who_loves_maths]

Yo. im having problems with these questions, the teacher gave some examples but i still dont get it (im still new to this log and exp stuff) so could you please show me how to do these 3 q's.

The distance X metres at time t seconds of a particale moving in a straight line from a fixed point O is given by "x=2-2cos2t"

a) sketch the graph of the equation for 0<t<2N (N=Pie)

b) Find the times when the particale is at rest and where the particale is at these times

c) what is the maximum distance of the particale from O

the answers was some wierd a** stuff which i would never get

answers: t=21N/36, t=5N/6; x=3.37, x=4 (yeah answers a bit confusing as they didnt number which answer is for which question... )

Peace out Jack

Part a)

{apologies for my poor illustration skills...}


Part b)

Differentiate the expression with respect to t to obtain velocity v = dx/dt:

v = dx/dt = 4Sin(2t)

Rest means v = 0:

0 = 4Sin(2t) = Sin(2t) -----> t = (1/2)k.pi ;
where k is a nonnegative integer and "pi" is your N which you denote to be "pie".

To find positions of your particle, substitute this time back to your expression for x:

x = 2 - 2Cos(2t) = 2 - 2Cos(k.pi)

Cos(k.pi) = {1, -1} [whether it's 1 or -1 depends on actual value of k]

i.e. x = {0, 4}


Part c)

Two ways of doing this:

1) Looking at the expression for x, it's easy to see the particle undergoes oscillatory motion about the centre x = 2 with amplitude = 2.
Thus, max. dist. relative to O = 2 + 2 = 4 units.

2) Consider:
-1 <= Cos(2t) <= 1 , for all t in R.
---> -2 <= -2Cos(2t) <= 2
---> 0 <= 2 - 2Cos(2t) <= 2 + 2 = 4

is the required inequality.

Hence, max. x = max{2 - 2Cos(2t)} = 4 units.


Hope this helps.

P.S. The answers given are "weird" because they are wrong.