Results 1 to 8 of 8
Like Tree1Thanks
  • 1 Post By skeeter

Thread: Solve for t

  1. #1
    Super Member sakonpure6's Avatar
    Joined
    Sep 2012
    From
    Canada
    Posts
    833
    Thanks
    83

    Solve for t

    I have the following equation $\displaystyle y=-20cos225t+50$ If i set y= 60 and i solve i get root 2 over 2, howevef if i sub that back in the equation, i don't get 60 but I get about 68! Any help is appreciated. Here is what i did:

    $\displaystyle 60=-20cos225t+50$
    $\displaystyle t=10/-20cos225$
    $\displaystyle t= root 2/2$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702

    Re: Solve for t

    is the equation ...

    $\displaystyle y = -20\cos(225t) + 50$

    or

    $\displaystyle y = -20\cos(225) \cdot t + 50$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member sakonpure6's Avatar
    Joined
    Sep 2012
    From
    Canada
    Posts
    833
    Thanks
    83

    Re: Solve for t

    Quote Originally Posted by skeeter View Post
    is the equation ...

    $\displaystyle y = -20\cos(225t) + 50$

    or

    $\displaystyle y = -20\cos(225) \cdot t + 50$

    This one $\displaystyle y = -20\cos(225t) + 50$

    if it's the other equation then root2/2 works... so what did i do wrong?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702

    Re: Solve for t

    Quote Originally Posted by sakonpure6 View Post
    This one $\displaystyle y = -20\cos(225t) + 50$

    if it's the other equation then root2/2 works... so what did i do wrong?
    the whole thing ...

    $\displaystyle 60 = -20\cos(225t) + 50$

    $\displaystyle -10 = -20\cos(225t)$

    $\displaystyle \frac{1}{2} = \cos(225t)$

    $\displaystyle \cos^{-1}\left(\frac{1}{2}\right) = 225t$

    if $\displaystyle t$ is in radians ...

    $\displaystyle \frac{\pi}{3} = 225t$

    $\displaystyle t = \frac{\pi}{675}$

    if $\displaystyle t$ is in degrees ...

    $\displaystyle 60 = 225t$

    $\displaystyle t = \frac{60}{225} = \frac{4}{15}$

    note that these are the principle solutions ... there are more.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849

    Re: Solve for t

    Hello, sakonpure6!

    I don't understand your difficulty.


    $\displaystyle y\:=\:-20(\cos225)t+50$
    If I set $\displaystyle y= 60$ and I solve, I get $\displaystyle \tfrac{\sqrt{2}}{2}$
    However if I sub that back in the equation, I don't get 60 but I get about 68! . How?

    If $\displaystyle t = \tfrac{\sqrt{2}}{2}$, we have:

    . . $\displaystyle \text{-}20(\cos225)\left(\tfrac{\sqrt{2}}{2}\right) + 50 \;=\;\text{-}20\left(\text{-}\tfrac{\sqrt{2}}{2}\right)\left(\tfrac{\sqrt{2}}{ 2}\right) + 50 $

    . . . . $\displaystyle =\;\text{-}20\left(\text{-}\tfrac{1}{2}\right) + 50 \;=\;10 + 50 \;=\;60$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member sakonpure6's Avatar
    Joined
    Sep 2012
    From
    Canada
    Posts
    833
    Thanks
    83

    Re: Solve for t

    Quote Originally Posted by Soroban View Post
    Hello, sakonpure6!

    I don't understand your difficulty.



    If $\displaystyle t = \tfrac{\sqrt{2}}{2}$, we have:

    . . $\displaystyle \text{-}20(\cos225)\left(\tfrac{\sqrt{2}}{2}\right) + 50 \;=\;\text{-}20\left(\text{-}\tfrac{\sqrt{2}}{2}\right)\left(\tfrac{\sqrt{2}}{ 2}\right) + 50 $

    . . . . $\displaystyle =\;\text{-}20\left(\text{-}\tfrac{1}{2}\right) + 50 \;=\;10 + 50 \;=\;60$
    I multiplied root2/2 by 225 before taking it's cosine, that was my mistake there and for skeeter how did you get -10? shouldnt it be 10
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702

    Re: Solve for t

    my mistake ...

    $\displaystyle 60 = -20\cos(225t) + 50$

    $\displaystyle 10 = -20\cos(225t)$

    $\displaystyle -\frac{1}{2} = \cos(225t)$

    $\displaystyle \cos^{-1}\left(-\frac{1}{2}\right) = 225t$

    if $\displaystyle t$ is in radians ...

    $\displaystyle \frac{2\pi}{3} = 225t$

    $\displaystyle t = \frac{2\pi}{675}$

    if $\displaystyle t$ is in degrees ...

    $\displaystyle 120 = 225t$

    $\displaystyle t = \frac{120}{225} = \frac{8}{15}$
    Thanks from sakonpure6
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member sakonpure6's Avatar
    Joined
    Sep 2012
    From
    Canada
    Posts
    833
    Thanks
    83

    Re: Solve for t

    alright thank you very much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: Jul 16th 2010, 10:29 PM
  2. Solve dy/dx + xy/4-x^2 =1
    Posted in the Differential Equations Forum
    Replies: 7
    Last Post: May 19th 2010, 07:39 PM
  3. Solve 2/[(x^2)-x] = (1/x) + 4/(x^2-1)
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: May 11th 2010, 06:01 PM
  4. Replies: 1
    Last Post: Jun 9th 2009, 10:37 PM
  5. solve
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Sep 6th 2008, 07:59 AM

Search Tags


/mathhelpforum @mathhelpforum