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Math Help - trig function - proof

  1. #16
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    Quote Originally Posted by thermalwarrior View Post
    okay so now i understand how you got to the final part apart from one thing - where does the -1 come from???

    thanks
    Factorise and see it ......
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  2. #17
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    think it is sorted now thanks for the help. there are more questions to come though!!!
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  3. #18
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    hello again, 2 more questions:

    1) I know that the solving equation mention above (2sin(2t)cos(2t)-1) will give values of 1/8pi and -3/8pi with the negative value excluded due to the nature of the question. However, this was done with a computer programme - how do i find the answers by hand? is it a rearrangement?

    2) the final part of the question asks what is the maximum distance the object will travel? is this as simple as it sounds - i.e. for this interval t=2pi and put this into the displacement equation? or is it more involved??

    cheers again
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  4. #19
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    Quote Originally Posted by thermalwarrior View Post
    hello again, 2 more questions:

    1) I know that the solving equation mention above (2sin(2t)cos(2t)-1) will give values of 1/8pi and -3/8pi with the negative value excluded due to the nature of the question. However, this was done with a computer programme - how do i find the answers by hand? is it a rearrangement?

    2) the final part of the question asks what is the maximum distance the object will travel? is this as simple as it sounds - i.e. for this interval t=2pi and put this into the displacement equation? or is it more involved??

    cheers again
    If for the first one you are asking how to by hand solve

    2\sin(2\theta)\cos(2\theta)=1

    Using a trig identity we have

    \sin(4\theta)=1\Rightarrow{\arcsin(1)=4\theta}

    and I think you can go form there
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  5. #20
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    Quote Originally Posted by thermalwarrior View Post
    hello again, 2 more questions:

    1) I know that the solving equation mention above (2sin(2t)cos(2t)-1) will give values of 1/8pi and -3/8pi with the negative value excluded due to the nature of the question. However, this was done with a computer programme - how do i find the answers by hand? is it a rearrangement?
    [snip]
    Those solutions are wrong. You have obviously entered the wrong equation into your CAS.

    The equation being solved is 2sin(2t)(cos(2t)-1) = 0.

    It follows that either:

    1. sin(2t) = 0 => 2t = n pi => t = npi/2 where n is an integer, or

    2. cos(2t) - 1 = 0 => cos(2t) = 1 => etc.
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  6. #21
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    Quote Originally Posted by thermalwarrior View Post
    [snip]
    2) the final part of the question asks what is the maximum distance the object will travel? is this as simple as it sounds - i.e. for this interval t=2pi and put this into the displacement equation? or is it more involved??

    cheers again
    No.

    Distance is not the same as displacement.

    And I assume you mean maximum distance from the origin? Or from the starting point?

    Assuming you want maximum distance from the origin, note that D^2 = \sin^2 (2t) + 4 \cos^2 (t). You need to find the maximum value of D ......


    You need to go back to your class notes and/or textbook and thoroughly review all of the mathematical knowledge this question has demanded you know.
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  7. #22
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    okay so now i have the fact that pi/2 and -pi/2 gives the answers I want excluding the negative value in this case. Sound better??
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  8. #23
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    Quote Originally Posted by thermalwarrior View Post
    okay so now i have the fact that pi/2 and -pi/2 gives the answers I want excluding the negative value in this case. Sound better??
    If you want the maximum distance from the origin (and you haven't yet made it clear where the distance is being measured from), then after differentiating D^2 with respect to t and equating the result to zero you have:

    2 \sin (2t) \cos (2t) - 8 \cos (t) \sin (t) = 0

    \Rightarrow \sin (2t) [2 \cos (2t) - 2] = 0.

    Solutions in the domain 0 \leq t \leq 2 \pi are: t = \frac{\pi}{2}, ~  \frac{3 \pi}{2}, ~ 0, ~ \pi, ~ 2 \pi.

    Now you need to test their nature. The easiest way to do this is to substitute into D^2 and see which ones give the largest value of D^2.

    The one you found, t = pi/2, gives the minimum value (of zero) of D^2 and therefore the minimum value (of zero) of D. I think you'll find the maximum distance from the origin is 2 .......
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  9. #24
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    sorry thats my fault is was refer to the fact pi/2 is the other time the velocity and acceleration are perpendicular to each other

    and yes it is meant to be from the origin
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  10. #25
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    Quote Originally Posted by thermalwarrior View Post
    okay so now i have the fact that pi/2 and -pi/2 gives the answers I want excluding the negative value in this case. Sound better??
    2 \sin (2t) (\cos (2t) - 1) = 0 has more solutions than just t = pi/2.

    You have to solve two equations: sin(2t) = 0 and cos(2t) = 1.

    Clearly solutions are of the form t = n pi/2 where n is an integer.
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