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Math Help - Related Rates Problem

  1. #1
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    Related Rates Problem

    Thank you <33
    Last edited by onhcetum; December 2nd 2009 at 12:36 PM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    There are some missed details the knowledge of which can help very much...

    a) what is the wheight of the missile at the time t=0?...

    b) what is the 'rate of combustion' of the fuel in kilograms?... is this rate constant with time?...

    c) what is the 'propulsive boost' of the missile in kilograms?... is that constant with time?...

    d) is air resistance negligible?...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by onhcetum View Post
    What does this even look like?
    Yes, start with a drawing, in this case a right triangle with base 5 and angle of elevation theta. Call the height s.

    Quote Originally Posted by onhcetum View Post
    I don't know what my original function should even be.
    The relation between s and theta probably very useful...

    s = 5 \tan \theta

    Just in case a picture also helps with the logic of related rates... You might want to try filling up this pattern...



    ... where straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case time), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).



    So differentiate with respect to the inner function, and the inner function with respect to t. Then sub in the given value of \frac{d\theta}{dt} and \theta.

    Spoiler:

    __________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus: Standard Integrals, Derivatives and Methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; November 26th 2009 at 03:44 AM. Reason: simplified
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  4. #4
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    Hey, thank you very much.

    I got 13.333 miles/sec as my answer.

    Is this correct?
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  5. #5
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    Not what I'm getting... radians/degrees?
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  6. #6
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    I was using degrees.

    (5) x sec(30)^2 x 2 = 13.333 miles/sec.
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  7. #7
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    When I used radians, I got 0.233 miles/sec.
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  8. #8
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    Yeh, me too.

    I can see the 30 is OK if the sec function is set to degrees, but I think you need to express the angular rate as pi/90. Can't put my finger on why... (except radians are usually more correct!)
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  9. #9
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    Can you tell me why we used radians instead of degrees? considering how everything was given to us in degrees.

    I understand both ways.

    Thank you
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  10. #10
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    D'oh! Either, of course, but as a fraction of a whole revolution!
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  11. #11
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    Double d'oh! The differentiation formula for tan does assume radians, so if we want to sub in an angle in degrees it'll affect the chain rule too...



    So that's why.

    Anyway... mods, very grateful if you can restore the original post, as Mr F was able to do on a previous occasion - for which, thanks.
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