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Thread: How to simplify velocity magnitude equations?

  1. #1
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    How to simplify velocity magnitude equations?

    In my calculus class a few days ago, my professor was going over how to find three dimensional velocity and its magnitude. In the problem, the magnitude of the velocity came out to be $\displaystyle \sqrt{9+36t^2+36t^4}=3+6t^2$.

    My question is, how did he simplify the equation under the square root so that it became $\displaystyle 3+6t^2$?

    For the homework, I need to simplify these two square root equations to get something without a square root sign (if possible). Can you guys help me out with this? Thanks.
    1) $\displaystyle \sqrt{9t^4+36+36t^2}$

    2) $\displaystyle \sqrt{(-3cost)^2+(-4cost)^2+(-5sint)^2)}$
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  2. #2
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    Hello Infernorage!

    Quote Originally Posted by Infernorage View Post
    In my calculus class a few days ago, my professor was going over how to find three dimensional velocity and its magnitude. In the problem, the magnitude of the velocity came out to be $\displaystyle \sqrt{9+36t^2+36t^4}=3+6t^2$.
    Do you know this rule:

    $\displaystyle (a+b)^2 = a^2+2ab+b^2$

    It is not difficult to see that $\displaystyle 9+36t^2+36t^4 = (6t^2+3)^2$

    So

    $\displaystyle \sqrt{9+36t^2+36t^4}=3+6t^2 = \sqrt{(6t^2+3)^2} = 3+6t^2$


    Quote Originally Posted by Infernorage View Post
    My question is, how did he simplify the equation under the square root so that it became $\displaystyle 3+6t^2$?

    For the homework, I need to simplify these two square root equations to get something without a square root sign (if possible). Can you guys help me out with this? Thanks.
    1) $\displaystyle \sqrt{9t^4+36+36t^2}$
    This one is tricky.
    I told you $\displaystyle (a+b)^2 = a^2+2ab+b^2$. You can't use it here unless :

    $\displaystyle \sqrt{9t^4+36+36t^2} = \sqrt{9*(t^4+4+4t^2)}$

    Hence

    $\displaystyle \sqrt{9*(t^4+4+4t^2)} = \sqrt{9(t^2+2)^2}$


    Quote Originally Posted by Infernorage View Post
    2) $\displaystyle \sqrt{(-3cost)^2+(-4cost)^2+(-5sint)^2)}$
    It is $\displaystyle sin^2+cos^2=1$

    and furthermore

    $\displaystyle \sqrt{(-3cost)^2+(-4cost)^2+(-5sint)^2)}$

    $\displaystyle =\sqrt{9cos^2t+16cos^2t+25sin^2t}$

    $\displaystyle =\sqrt{25cos^2t+25sin^2t}$

    $\displaystyle =\sqrt{25(cos^2t+sin^2t)}$

    $\displaystyle =\sqrt{25*(1)} = \sqrt{25}$

    Do you understand?

    Yours
    Rapha
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