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Math Help - Calculus

  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Prophet View Post
    So, then you take the derivative of this function and then set it equal to 0, I took the derivative and got that function... but I was just wondering if my differentiation was correct.
    it isn't correct. you would need the chain rule, which you didn't use
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  2. #17
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    Quote Originally Posted by Jhevon View Post
    it isn't correct. you would need the chain rule, which you didn't use
    I did it again and got 2~3^(1/2) - 4x - (16-x^2)^(-1/2) = D'

    ~ = times
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  3. #18
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    Quote Originally Posted by Prophet View Post
    I did it again and got 2~3^(1/2) - 4x - (16-x^2)^(-1/2) = D'

    ~ = times

    Correct that, -2~3^1/2 - (16-x^2)^(-1/2)
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  4. #19
    is up to his old tricks again! Jhevon's Avatar
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    D = (\sqrt{3} - x)^2 + \left(1 - \sqrt{16 - x^2} \right)^2

    \Rightarrow D' = -2(\sqrt{3} - x) -\frac {2x \left( 1 - \sqrt{16 - x^2} \right)}{\sqrt{16 - x^2}}

    now simplify
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  5. #20
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    Does the F^2 have anything to do with the final result at all?
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  6. #21
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    I set that equation equal to 0 and got x = 3^(1/2) and :P 2~3^(1/2)= (-)[1+(16-x^2)^(1/2)]/(16-x^2)^(1/2) do you know how to simplify that :P?
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  7. #22
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    And for number 3.
    S=k(144w-w^{3})<br />

    When we differentiate that, do we have to use implicit differentiation... since there is a k and a w involved, or do we leave out the k because the type of wood isn't required?
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  8. #23
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Prophet View Post
    And for number 3.
    S=k(144w-w^{3})<br />

    When we differentiate that, do we have to use implicit differentiation... since there is a k and a w involved, or do we leave out the k because the type of wood isn't required?
    k is a constant, it is the constant of proportionality
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  9. #24
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    Oh okay, I understand, so we can just move it out of the differential, I see thanks.
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  10. #25
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Prophet View Post
    Does the F^2 have anything to do with the final result at all?
    no, we are just maximizing the square of the distance, it is the same as maximizing the distance itself
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