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Easy Implicit Differentiation Question

  1. Jul 31, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]y^2 = 3x^3 + 2x[/tex] and y must be positive.
    Find the normal component of acceleration when:
    [tex]x=3m[/tex]
    [tex]\dot x = 5ms^{ - 1}[/tex]
    [tex]\ddot x = 5ms^{ - 2} [/tex]

    2. The attempt at a solution
    Well my approach would be to differentiate it implicitly twice and solve for [tex]{\ddot y}[/tex].
    Then I have the x and y components of acceleration, not sure how I would find the normal acceleration from here. But here goes anyways:

    [tex]2y.\frac{{dy}}{{dt}} = 9x^2 .\frac{{dx}}{{dt}} + 2.\frac{{dx}}{{dt}}[/tex]

    Now I looked at this and thought ... "how the hell am I gonna differentiate that", then I though I could perhaps use the product rule:

    [tex]2.\frac{{dy}}{{dt}} + 2y.\frac{{d^2 y}}{{dt^2 }} = 18x.\frac{{dx}}{{dt}} + 9x^2 .\frac{{d^2 x}}{{dt^2 }} + 2.\frac{{d^2 x}}{{dt^2 }}[/tex]

    Now I can chuck in my known values and solve for [tex]{\ddot y}[/tex].

    [tex]
    \begin{array}{l}
    2y.\frac{{d^2 y}}{{dt^2 }} = \left( {9x^2 + 2} \right)\frac{{d^2 x}}{{dt^2 }} + 18x.\frac{{dx}}{{dt}} - 2.\frac{{dy}}{{dt}} \\
    \frac{{d^2 y}}{{dt^2 }} = \frac{{\left( {9x^2 + 2} \right)\frac{{d^2 x}}{{dt^2 }} + 18x.\frac{{dx}}{{dt}} - 2.\frac{{dy}}{{dt}}}}{{2y}} \\
    \end{array}
    [/tex]

    The answer I get for [tex]{\ddot y}[/tex] is ~32.9

    Any ideas where to go from here?
    Cheers
    Steven
     
  2. jcsd
  3. Jul 31, 2008 #2

    Defennder

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    It's odd that you are given how y is a function of x, but not how either x,y are functions of t. Maybe I'm not interpreting the question properly but what I'll do is to first find a position vector r(t) and then apply the formula to find the normal component of acceleration.
     
  4. Jul 31, 2008 #3
    Thanks for your reply.

    Yeah the question does seem a little odd.
    So you would find a position vector from the given equation in terms of y and x?
    not sure what you mean here...
     
  5. Jul 31, 2008 #4

    Defennder

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    Did you post the question exactly as where you read it? Some important information might have been omitted. I would like to assume that we could take x=t as the free variable, but in this case t is given a physical interpretation and it wouldn't be justified if this assumption is made without more info.
     
  6. Jul 31, 2008 #5
    OK, here is the exact question:

    Given [tex]y^2 = 3x^3 + 2x[/tex] where x and y are in metres and y is positive,
    what is the normal component of the acceleration when:

    [tex]x=3m[/tex]

    [tex]\dot x=5m[/tex]

    [tex]\ddot x=5m[/tex]
     
  7. Jul 31, 2008 #6

    Dick

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    If you look at that equation and it's first and second derivative with respect to t, that's three equations in the six unknowns x,x',x'',y,y',y''. The problem statement gives you values for three of them. You should be able to find the other three. Then you know everything.
     
  8. Jul 31, 2008 #7
    Thanks for your reply.

    OK, so I also have:
    [tex]y = + \sqrt {3x^3 + 2x} = \sqrt {3\left( 3 \right)^3 + 2\left( 3 \right)} = \sqrt {87}
    [/tex]
    thats 4/6 unknowns found.

    I need to find the derivate with respect to time, how should I approach this?
    Implicit Differentiation - I tried this but got a bit stuck.

    [tex]\begin{array}{l}
    2y.\frac{{dy}}{{dt}} = 9x^2 .\frac{{dx}}{{dt}} + 2.\frac{{dx}}{{dt}} \\
    \frac{{dy}}{{dt}} = \frac{{\frac{{dx}}{{dt}}\left[ {9x^2 + 2} \right]}}{{2y}} = \frac{{5\left[ {9\left( 3 \right)^2 + 2} \right]}}{{2\sqrt {87} }} = \frac{{415}}{{2\sqrt {87} }} \approx 22.246\;ms^{ - 1} \\
    \end{array}[/tex]

    Correct?

    Now i'm stuck with the second derivative.
     
  9. Jul 31, 2008 #8

    Dick

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    That looks fine. For the second derivative you've got the right idea to use the product rule, but you haven't quite got everything right. Take y^2. The first derivative is 2*y*y'. The second derivative is 2*y'*y'+2*y*y''. Compare that with your left side.
     
  10. Jul 31, 2008 #9
    OK thanks, how about:

    [tex]\begin{array}{l}
    2.\frac{{dy}}{{dt}}.\frac{{dy}}{{dt}} + 2y.\frac{{d^2 y}}{{dt^2 }} = 18x.\frac{{dx}}{{dt}}.\frac{{dx}}{{dt}} + 9x^2 .\frac{{d^2 x}}{{dt^2 }} + 2.\frac{{d^2 x}}{{dt^2 }} \\
    2y.\frac{{d^2 y}}{{dt^2 }} = \frac{{d^2 x}}{{dt^2 }}\left[ {9x^2 + 2} \right] + 18x.\left( {\frac{{dx}}{{dt}}} \right)^2 - 2.\left( {\frac{{dy}}{{dt}}} \right)^2 \\
    \frac{{d^2 y}}{{dt^2 }} = \frac{{\frac{{d^2 x}}{{dt^2 }}\left[ {9x^2 + 2} \right] + 18x.\left[ {\frac{{dx}}{{dt}}} \right]^2 - 2.\left[ {\frac{{dy}}{{dt}}} \right]^2 }}{{2y}} = \frac{{5\left[ {9\left( 3 \right)^2 + 2} \right] + 18\left( 3 \right) \times \left[ 5 \right]^2 - 2\left[ {\frac{{415}}{{2\sqrt {87} }}} \right]^2 }}{{2\sqrt {87} }} \approx 41.55\;ms^{ - 2} \\
    \end{array}[/tex]

    So I know have all of my unknowns and need to find the normal component of acceleration.
    I have my 2 components of acceleration in x/y form, how can I go about converting this to a normal/tangential form?
    I guess I can find the magnitude and direction of the total acceleration using simple trig, but where to from here?
     
  11. Jul 31, 2008 #10

    Dick

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    I'll trust you on the arithmetic for that one. The normal component of acceleration (a) is the part that's perpendicular to the velocity (v). So find a unit vector parallel to v and split a into normal and parallel parts relative to that direction.
     
  12. Jul 31, 2008 #11
    lol :P

    OK, so the velocity is:
    [tex]\begin{array}{l}
    \left| v \right| = \sqrt {\left( {\frac{{dy}}{{dt}}} \right)^2 + \left( {\frac{{dx}}{{dt}}} \right)^2 } = \sqrt {\left( {\frac{{415}}{{2\sqrt {87} }}} \right)^2 + \left( 5 \right)^2 } \approx 22.80\;ms^{ - 1} \\
    \tan \theta = \frac{{dy/dt}}{{dx/dt}} = \frac{{\left( {\frac{{415}}{{2\sqrt {87} }}} \right)}}{5} = 4.45 \Rightarrow \theta = 77.33^ \circ \\
    \end{array}[/tex]

    The normal acceleration is perpendicular to the velocity, so we can make a right angled triangle as such:
    [tex]
    \begin{array}{l}
    \tan \theta = \frac{{a_n }}{{\left| v \right|}} \Rightarrow a_n = \left| v \right|.\tan \theta \\
    a_n = 22.80 \times 4.45 = 101.45\;ms^{ - 2} \\
    \end{array}[/tex]

    Is this what you mean here?
    - The answer seems wayy to high...
     
  13. Jul 31, 2008 #12

    Dick

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    No. That's not it. It's easier if you use vectors. As I said, find a unit vector along the same direction as the velocity. Rotate it 90 degrees and find the dot product with the acceleration vector. Have you done that sort of thing before?
     
  14. Jul 31, 2008 #13
    OK ta

    No I haven't done dot products in this sense before.
    I now realise the second part of my above working is a bit stupid...

    OK so converting the velocity into vector form:
    [tex]\left\langle {\begin{array}{*{20}c}
    {22.8\cos 77.33} \\
    {22.8\sin 77.33} \\
    \end{array}} \right\rangle = \left\langle {\begin{array}{*{20}c}
    5 \\
    {22.246} \\
    \end{array}} \right\rangle[/tex]

    I must admit I am a little lost now :(
     
  15. Jul 31, 2008 #14

    Dick

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    You don't really have to convert to vector form. v=(5,22.246), a=(5,41.55). They were already in vector form. What I would do is write down the unit vector u=v/|v|. Then a.u (dot product) is the component of acceleration parallel to v. That would mean the perpendicular component is sqrt(|a|^2-(u.v)^2). Or I would rotate u 90 degrees and get the component straight from the dot product. If you aren't comfortable with this, you could just keep going with your previous approach. Find |a| and an angle for the acceleration as well and do some trig.
     
  16. Jul 31, 2008 #15
    Of course, yes they were already .. argh i'm complicating it even more.

    I would prefer to use the latter method as it makes more sense to me:
    [tex]
    \begin{array}{l}
    \tan \alpha = \frac{{d^2 y/dt^2 }}{{d^2 x/dt^2 }} = \frac{{41.55}}{5} = 8.31 \Rightarrow \alpha = 83.14^ \circ \\
    \left| a \right| = \sqrt {\left[ {\frac{{d^2 y}}{{dt^2 }}} \right]^2 + \left[ {\frac{{d^2 x}}{{dt^2 }}} \right]^2 } = \sqrt {41.55^2 + 5^2 } = 41.85\;ms^{ - 2} \\
    \end{array}
    [/tex]

    I think I am getting on the right track now; here is a diagram I made:
    http://img105.imageshack.us/img105/9264/normalaccelerationln8.jpg

    So I am trying to find the component of acceleration, parallel to the velocity vector.
    Should be:
    [tex]\begin{array}{l}
    \alpha - \theta = 5.806^ \circ \Rightarrow \cos \left( {\alpha - \theta } \right) = \frac{{component\;of\;acceleration\;parallel\;to\;velocity}}{{\left| a \right|}} \\
    \Rightarrow component\;of\;acceleration\;parallel\;to\;velocity = \left| a \right|\cos \left( {\alpha - \theta } \right) = 41.64\;ms^{ - 2} \\
    \end{array}
    [/tex]

    Look ok?
    Now I am looking for the perpendicular vector to this component I just found?

    [tex]\tan \theta = \frac{{a_n }}{{component\;of\;acceleration\;parallel\;to\;velocity}} \Rightarrow a_n = 185.27\;ms^{ - 2}
    [/tex]

    Still seems too high though :(

    Thanks once again for your help, this has turned into a bit of a slog.
     
    Last edited: Jul 31, 2008
  17. Jul 31, 2008 #16

    Dick

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    You've got the parallel part right. p=41.64m/s^2. You want to find the normal part n. |a|^2=|p|^2+|n|^2, right?
     
  18. Jul 31, 2008 #17
    [tex]\begin{array}{l}
    \left| a \right| = \sqrt {a_n ^2 + a_t ^2 } \\
    \Rightarrow \left| {a_n } \right| = \sqrt {\left| a \right|^2 - \left| {a_t } \right|^2 } = \sqrt {41.85^2 - 41.64^2 } = 4.23\;ms^{ - 2} \\
    \end{array}[/tex]

    Ahhhhhhh thanks so much mate, makes sense now :)
     
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