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Math Help - Partial Differentiation

  1. #1
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    Partial Differentiation

    hi all,
    can anyone help me with this?how do you get ∂A/∂q ? i amv.v. confused
    i dont need all details or workings,just shorthand way of seeing how to get ∂A/∂q. how do ppl do this quickly?



    A = q [(a – qsinѲ)secѲ + (b – qcosѲ)cosecѲ]
    ∂A/∂q = (a – qsinѲ)secѲ + (b – qcosѲ) – q(sinѲsecѲ + cosѲcosecѲ)





    thanks!!
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  2. #2
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    Quote Originally Posted by matlabnoob View Post
    hi all,
    can anyone help me with this?how do you get ∂A/∂q ? i amv.v. confused
    i dont need all details or workings,just shorthand way of seeing how to get ∂A/∂q. how do ppl do this quickly?



    A = q [(a – qsinѲ)secѲ + (b – qcosѲ)cosecѲ]
    ∂A/∂q = (a – qsinѲ)secѲ + (b – qcosѲ) – q(sinѲsecѲ + cosѲcosecѲ)





    thanks!!
    One way to do this would be to use the product rule. The derivative of "q" is 1, of course, and the derivative of (a- qsin\theta)sec(\theta)+ (b- qcos(\theta))cosec(\theta) is -sin(\theta)sec(\theta)- cos(\theta)cosec(\theta) because the derivative of "Cq" where C is not a function of q, is just C itself.

    Putting those together, using the product rule:
    (q)'(a- qsin\theta)sec(\theta)+ (b- qcos(\theta))cosec(\theta))+ (q)(a- qsin\theta)sec(\theta)+ (b- qcos(\theta))cosec(\theta))'
    = a- qsin\theta)sec(\theta)+ (b- qcos(\theta))cosec(\theta)+ q(sin(\theta)sec(\theta)- cos(\theta)cosec(\theta))

    Another way to do this is to go ahead and multiply that out:
    aq sec(\theta)- q^2 sin(\theta)sec(\theta)+ bq cosec(\theta)- q^2 cos(\theta)cosec(\theta).

    Now, the derivative of Cq is just C and the derivative of Cq^2 is 2Cq so the derivative of that is
    a sec(\theta)- 2q sin(\theta)sec(\theta)+ b cosec(\theta)- 2q cos(\theta)cosec(\theta).

    Those two methods should give you the same thing.
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  3. #3
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    Suppose A = q [(5 – 3q)*2 + (11 – 7q)*13]. I assume you can find \tfrac{dA}{dq} rather “quickly” as you say (just several applications of the product rule).

    Partial differentiation is just like the above. Any term which involves a variable which is not the variable you’re considering can be treated as a constant (e.g., when finding \tfrac{\partial A}{\partial q} you can treat anything involving a or \theta as numbers). I hope this has clarified and not exacerbated your confusion.

    P.S.
    I think you are missing a factor of \csc\theta in the second term of your answer.
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  4. #4
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    aha..thank you both of you!
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