help (1 Viewer)

[Arithela]

The displacement of a particle P moving along a straight line, in metres, from the origin after t seconds is given by x = 6 - 2sin2t. Find the maximum displacement of P.

thanks

 

[Arithela]

thanks - the book used the 2nd method but subbed in -1 instead of 1 --> 6m. also, how do you know that the maximum value of sin 2t is 1 and not the other?

 

[dwarven]

isnt it because you need the acute angle?

 

[Arithela]

i dont know..

 

[Affinity]

x = 6 - 2sin 2t
Maximum occurs when dx/dt = 0
dx/dt = -4cos 2t
t = π/4 is the first instance this occurs
dx²/dt² = -8sin2t <--- error here
At t = π/4, dx²/dt² = -8 < 0, so maximum occurs, here
so max displacement is: x = 6 - 2 = 4

Alternatively, you could reason that the maximum value of sin 2t is 1, so to maximise x, you have:
x = 6 - 2(1)
= 4

wrong.. obviously max = 8 at x = 3pi/4
you can do this one qithout much calculus.. you know that sine ranges from -1 to 1.. so max x is when that sine function takes a value of -1, so max x = 8

 

[lyounamu]

The displacement of a particle P moving along a straight line, in metres, from the origin after t seconds is given by x = 6 - 2sin2t. Find the maximum displacement of P.

thanks

x = 6 - 2sin2t
dx/dt = -2cos2t
The maximum displacement occurs when dy/dx = 0
i.e. when 2cos2t = 0
cos2t = 0
t = pi/4, 3pi/4 ....
since cos pi/2 = 0 (so 2t = pi/2, i.e. t = pi/4) and cos 3pi/2 = 0 (so 2t = 3pi/2, i.e. t= 3pi/4)
The maximum is at t = 3pi/4

Now, x = 6-2sin2t
= 6 - 2sin(3pi/2)
= 6 + 2
= 8

 

[Arithela]

sorry how did u get t = pi/4, 3pi/4 in the first place?

 

[lyounamu]

sorry how did u get t = pi/4, 3pi/4 in the first place?

cos x = 0
So what must x be? x must be pi/2 or 3pi/2, yes?

Since cos 2t =0, 2t must be equal to either pi/2 or 3pi/2 so t = pi/4 or 3pi/4

So the minimum or maximum point occurs at pi/4 or 3pi/4

If you do your own test, you will see that the MAXIMUM turning point occurs at 3pi/4. Therefore, t = 3pi/4 since 2t = 3pi/2

 

[Arithela]

thanks i get it now! can u tell me how u would test to see if its a max turning point or do u just sub in the 2 t values and see?

 

[lyounamu]

thanks i get it now! can u tell me how u would test to see if its a max turning point or do u just sub in the 2 t values and see?

Well, to test whether it is a maximum or minimum. You first find the t-value (which are pi/4 and 3pi/4) and then you sub them back into the 2nd derivative. If it is negative, it is a maximum turning point, it if is positive, it is a minimum turning point.

 

[Arithela]

thank you