Results 1 to 7 of 7

Math Help - Trig

  1. #1
    Banned
    Joined
    Feb 2008
    Posts
    24

    Trig

    The displacement of a certain object is given as a function of the time t (in seconds) by s = sin \;t + cos\; 2t. Find the smallest positive value of t for which the displacement is zero.

    Is the answer t = 90
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by TI-84 View Post
    The displacement of a certain object is given as a function of the time t (in seconds) by s = sin \;t + cos\; 2t. Find the smallest positive value of t for which the displacement is zero.

    Is the answer t = 90
    Use the equation:

    \cos(2t) = 1-2\sin^2(t)

    With s = 0 your equation becomes:

    0 = \sin(t) + 1 - 2\sin^2(t) ..... Use the substitution:

    y = \sin(t) ..... Then your equation becomes:

    0= y + 1 - 2y^2..... Solve for y. Afterwards re-substitute and solve for t. Your result is OK if 1 s correspond to 1.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Feb 2008
    Posts
    24
    I'm a little confused after working your method out. Here's the way I did it.

    S = Sin\;t + cos\;2t

    0 = Sin\;t + cos\;2t

    sin\;t + (1 - 2\;sin^2\;t) = 0

    -2\;sin^2\;t + sin\;t + 1 = 0

    (-sin\;t + 1) (2\;sin\;t + 1) = 0

    -sin\;t = -1\Longrightarrow sin\;t = 1\Longrightarrow t = sin^{-1}(1)

    2\;sin\;t = -1\Longrightarrow sin\;t = \frac{-1}{2}\Longrightarrow t = sin^{-1}\left(\frac{-1}{2}\right)

    t = 90\;or\; t = -30

    The t = -30 gets crossed out because its negative which leaves t = 90 as the final answer. Or did I mess up somewhere?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by TI-84 View Post
    I'm a little confused after working your method out. Here's the way I did it.

    S = Sin\;t + cos\;2t

    0 = Sin\;t + cos\;2t

    sin\;t + (1 - 2\;sin^2\;t) = 0

    -2\;sin^2\;t + sin\;t + 1 = 0

    (-sin\;t + 1) (2\;sin\;t + 1) = 0

    -sin\;t = -1\Longrightarrow sin\;t = 1\Longrightarrow t = sin^{-1}(1)

    2\;sin\;t = -1\Longrightarrow sin\;t = \frac{-1}{2}\Longrightarrow t = sin^{-1}\left(\frac{-1}{2}\right)

    t = 90\;or\; t = -30

    The t = -30 gets crossed out because its negative which leaves t = 90 as the final answer. Or did I mess up somewhere?
    Believe it or not, your method (which by the way is correct) is more or less the same as what earboth proposed ....!

    However ...... your answer is wrong.

    The correct answer is  t = {\color{red}\frac{\pi}{2}} seconds ...... Capisce?

    Radians, NOT degrees is the implied unit .....
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by TI-84 View Post
    I'm a little confused after working your method out. Here's the way I did it.

    S = Sin\;t + cos\;2t

    0 = Sin\;t + cos\;2t

    sin\;t + (1 - 2\;sin^2\;t) = 0

    -2\;sin^2\;t + sin\;t + 1 = 0

    (-sin\;t + 1) (2\;sin\;t + 1) = 0

    -sin\;t = -1\Longrightarrow sin\;t = 1\Longrightarrow t = sin^{-1}(1)

    2\;sin\;t = -1\Longrightarrow sin\;t = \frac{-1}{2}\Longrightarrow t = sin^{-1}\left(\frac{-1}{2}\right)

    t = 90\;or\; t = -30

    The t = -30 gets crossed out because its negative which leaves t = 90 as the final answer. Or did I mess up somewhere?
    By the way .....

    Even though they are not used here, you do understand how to get positive solutions for t from \sin t = -\frac{1}{2}, right?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Feb 2008
    Posts
    24
    Does it have something to do with reference angles. One more thing. How did you know it was radians as the units I should use? Unless its just common sense. Then my bad.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by TI-84 View Post
    Does it have something to do with reference angles. One more thing. [snip]
    Yes. The third and fourth quadrant angles are  t = \frac{7\pi}{6} and t = \frac{11 \pi}{6} ~ ( = -\frac{\pi}{6} \, , the 'calculator answer') respectively.

    Then  t = \frac{7\pi}{6} + 2n\pi and t = \frac{11 \pi}{6} + 2n\pi where n is an integer.

    Quote Originally Posted by TI-84 View Post
    [snip]How did you know it was radians as the units I should use? [snip]

    When workng with models involving trigonometric functions, the implied unit for the angle will always be radians. Note that the formulae for the derivative of sin x, cos x, tan x etc are for x measured in radians.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compute Trig Function Values, Solve Trig Equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: September 8th 2011, 08:00 PM
  2. Replies: 7
    Last Post: April 15th 2010, 09:12 PM
  3. Replies: 6
    Last Post: November 20th 2009, 05:27 PM
  4. Replies: 1
    Last Post: July 24th 2009, 03:29 AM
  5. Trig Equations with Multiple Trig Functions cont.
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 7th 2008, 06:50 PM

Search Tags


/mathhelpforum @mathhelpforum