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Differential equations with rectilinear motion watch

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    I'm currently studying Mechanical Engineering 1st year, however, I feel this problem falls into Physics/Maths also. The following question is in "Engineering Mechanics, Dynamics, 6th Edition", by "J. L. Mriam and L. G. Kraige", page 29 but here are pictures:

    The question:
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    The soltuion:
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    The question is with regards to the integration of the equation which is next to the number 2. This is ds/(sqrt(v^2-(k^2*s^2))). I do not understand how this then equals 1/k(*(sin^-1(ks/v))).

    Appreciate any help.
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    (Original post by No-Idea.)
    ..
    It's integrated using a trigonometric substitution. It's in the form \sqrt{a^2 - b^2x^2}, so we look for a substitution that would cancel this down to a single term. I.e, if you subbed in some form of sin\theta, and use the identity 1 - sin^2\theta = cos^2\theta.

    Click the spoiler for the full working. I wouldn't usually do this, but it's a common integral that you'll learn to know by heart, and the maths is slightly tricky.

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    \displaystyle \int \dfrac{ds}{\sqrt{v_0^2 - k^2s^2}}

    Let s = \dfrac{v_0}{k}sin\theta

    \dfrac{ds}{d\theta} = \dfrac{v_0}{k}cos\theta \, \, \, \, \Rightarrow \, \, \, ds = \dfrac{v_0}{k}cos\theta d\theta

    \displaystyle \int \dfrac{ds}{\sqrt{v_0^2 - k^2s^2}}

    = \displaystyle \dfrac{v_0}{k} \int \dfrac{cos\theta}{\sqrt{v_0^2 - v_0^2sin^2\theta}}\, d\theta

    = \displaystyle \dfrac{1}{k} \int \dfrac{cos\theta}{cos\theta}\, d\theta ---> Taken a factor of v_0 out the root, and cancelled it with the \dfrac{v_0}{k}. Used the identity 1 - sin^2\theta = cos^2\theta

    = \displaystyle \dfrac{1}{k} \int 1 \, d\theta

    = \displaystyle \dfrac{1}{k}(\theta) ---> Using the initial substitution. \theta = sin^{-1}\left(\dfrac{ks}{v_0}\right)

    = \dfrac{1}{k}sin^{-1}\left(\dfrac{ks}{v_0}\right)

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    (Original post by Phichi)
    It's integrated using a trigonometric substitution. It's in the form \sqrt{a^2 - b^2x^2}, so we look for a substitution that would cancel this down to a single term. I.e, if you subbed in some form of sin\theta, and use the identity 1 - sin^2\theta = cos^2\theta.

    Click the spoiler for the full working. I wouldn't usually do this, but it's a common integral that you'll learn to know by heart, and the maths is slightly tricky.

    Spoiler:
    Show


    \displaystyle \int \dfrac{ds}{\sqrt{v_0^2 - k^2s^2}}

    Let s = \dfrac{v_0}{k}sin\theta

    \dfrac{ds}{d\theta} = \dfrac{v_0}{k}cos\theta \, \, \, \, \Rightarrow \, \, \, ds = \dfrac{v_0}{k}cos\theta d\theta

    \displaystyle \int \dfrac{ds}{\sqrt{v_0^2 - k^2s^2}}

    = \displaystyle \dfrac{v_0}{k} \int \dfrac{cos\theta}{\sqrt{v_0^2 - v_0^2sin^2\theta}}\, d\theta

    = \displaystyle \dfrac{1}{k} \int \dfrac{cos\theta}{cos\theta}\, d\theta ---> Taken a factor of v_0 out the root, and cancelled it with the \dfrac{v_0}{k}. Used the identity 1 - sin^2\theta = cos^2\theta

    = \displaystyle \dfrac{1}{k} \int 1 \, d\theta

    = \displaystyle \dfrac{1}{k}(\theta) ---> Using the initial substitution. \theta = sin^{-1}\left(\dfrac{ks}{v_0}\right)

    = \dfrac{1}{k}sin^{-1}\left(\dfrac{ks}{v_0}\right)

    I see, thanks a lot
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    (Original post by Phichi)
    It's integrated using a trigonometric substitution. It's in the form \sqrt{a^2 - b^2x^2}, so we look for a substitution that would cancel this down to a single term. I.e, if you subbed in some form of sin\theta, and use the identity 1 - sin^2\theta = cos^2\theta.

    Click the spoiler for the full working. I wouldn't usually do this, but it's a common integral that you'll learn to know by heart, and the maths is slightly tricky.

    Spoiler:
    Show


    \displaystyle \int \dfrac{ds}{\sqrt{v_0^2 - k^2s^2}}

    Let s = \dfrac{v_0}{k}sin\theta

    \dfrac{ds}{d\theta} = \dfrac{v_0}{k}cos\theta \, \, \, \, \Rightarrow \, \, \, ds = \dfrac{v_0}{k}cos\theta d\theta

    \displaystyle \int \dfrac{ds}{\sqrt{v_0^2 - k^2s^2}}

    = \displaystyle \dfrac{v_0}{k} \int \dfrac{cos\theta}{\sqrt{v_0^2 - v_0^2sin^2\theta}}\, d\theta

    = \displaystyle \dfrac{1}{k} \int \dfrac{cos\theta}{cos\theta}\, d\theta ---> Taken a factor of v_0 out the root, and cancelled it with the \dfrac{v_0}{k}. Used the identity 1 - sin^2\theta = cos^2\theta

    = \displaystyle \dfrac{1}{k} \int 1 \, d\theta

    = \displaystyle \dfrac{1}{k}(\theta) ---> Using the initial substitution. \theta = sin^{-1}\left(\dfrac{ks}{v_0}\right)

    = \dfrac{1}{k}sin^{-1}\left(\dfrac{ks}{v_0}\right)

    Sorry, just one question going over your reply.

    How did you go about getting s=(v/k)*sin(theta)?

    Thanks
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    (Original post by No-Idea.)
    Sorry, just one question going over your reply.

    How did you go about getting s=(v/k)*sin(theta)?

    Thanks
    Bump..anyone..?
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    (Original post by No-Idea.)
    Sorry, just one question going over your reply.

    How did you go about getting s=(v/k)*sin(theta)?

    Thanks
    Ever so sorry, for some reason I've only just seen this!

    Like I mentioned in the post, I was looking for a substitution to narrow the expression down to a single term.

    We have this:

    \sqrt{v_0^2 - k^2s^2}

    And to apply sin^2x + cos^2x = 1

    I wanted the root to be in the form:

    \sqrt{v_0^2 - v_0^2p^2} \, \, \, \Rightarrow v_0\sqrt{1 - p^2}

    Where p would be either sin\theta or cos\theta

    I picked to make it sin\theta, so I could avoid having the negative in the differentiation when switching the variables.

    So, we have, \sqrt{v_0^2 - k^2s^2} and we want \sqrt{v_0^2 - v_0^2sin^2\theta}.

    So we need a substitution for s, where k^2s^2 becomes v_0^2sin^2\theta

    Thus:

    k^2s^2 = v_0^2sin^2\theta \, \, \, \, \Rightarrow \, \, \, s = \dfrac{v_0}{k}sin\theta

    Does that help? Sorry for the long winded answer, you'll learn to do this sort of thing via inspection however, someday.
 
 
 
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