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Math Help - Having Lots Of Calculus III Problems

  1. #1
    Newbie amanda1410's Avatar
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    Having Lots Of Calculus III Problems

    My brain has gone dead, got a few days to do the following questions:






    would appreciate any help with any of the above question and methods

    thank you
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by amanda1410 View Post
    My brain has gone dead, got a few days to do the following questions:



    would appreciate any help with any of the above question and methods

    thank you
    Q4a. Let:

    <br />
f(x)=x^3-2x^2+x-3<br />

    then f(2)=-1 and f(3)=9. Also as f(x) is continuous by the intermediate value theorem it takes all values between f(2) and f(3) between x=2 and x=3, but as f(x) changes sign in this interval one of these values is 0. Hence we conclude that f(x) has a root in (2,3).

    RonL
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  3. #3
    Newbie amanda1410's Avatar
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    cheers for that, appreciate all the help possbile with the rest, at uni pullin my hair out trying to do this and my electronics at the same time
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    First I would like to mention that f''_{xy} and f''_{yx} are the same! Because the function is continous on some open disk.
    (But since you are in mechanical engineering you probably do not care about that).

    f(x,y)=x\ln y+e^{xy}+\cos (2x-3y)
    Thus,
    f'_x(x,y)=\ln y+ye^{xy}-2\sin (2x-3y)
    f'_y(x,y)=\frac{x}{y}+xe^{xy}+3\sin (2x-3y)

    Now take the derivative along y on the second one,
    f''_{yy}(x,y)=-\frac{x}{y^2}+x^2e^{xy}-9\cos (2x-3y)

    And finally,
    f''_{xy}=\frac{1}{y}+e^{xy}+xye^{xy}+6\cos (2x-3y)
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  5. #5
    Forum Admin topsquark's Avatar
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    Question 1 a) I presume.

    i) y = (2x - 1)3^{5x+1} - tan(2x)

    First note that \frac{d}{dx}a^x = ln(a) \cdot a^x. The rest is an exercise in the chain rule:

    So
    y' = 2 \cdot 3^{5x+1} + (2x - 1) \cdot ln(3) \cdot 3^{5x+1} \cdot 5 - sec^2(2x) \cdot 2

    y' = [ 2 + 5(2x - 1)ln(3) ] 3^{5x+1} - 2 sec^2(2x)

    ================================================== ======
    ii) y = ln \left ( 2x + e^{2x} \right ) + sin( \pi x)

    y' = \frac{1}{2x + e^{2x}} \cdot (2 + e^{2x} \cdot 2 ) + cos( \pi x) \cdot \pi

    y' = \frac{2(1 + e^{2x})}{2x + e^{2x}} + \pi cos( \pi x)

    ================================================== ======
    iii) y = x + cos^{-1}(y)

    First, this needs to be done by implicit differentiation.
    Second: \frac{d}{dx}cos^{-1}(x) = - \frac{1}{\sqrt{1-x^2}}.

    So:
    y' = 1 + -\frac{1}{\sqrt{1-y^2}} \cdot y'

    y' + \frac{1}{\sqrt{1-y^2}} \cdot y' = 1

    y' \left ( 1 + \frac{1}{\sqrt{1-y^2}} \right ) = 1

    y' = \frac{1}{1 + \frac{1}{\sqrt{1-y^2}}}

    This needs to be neated up a bit:
    y' = \frac{1}{1 + \frac{1}{\sqrt{1-y^2}}} \cdot \frac{\sqrt{1-y^2}}{\sqrt{1-y^2}}

    y' = \frac{\sqrt{1-y^2}}{\sqrt{1-y^2} + 1}

    y' = \frac{\sqrt{1-y^2}}{\sqrt{1-y^2} + 1} \cdot \frac{\sqrt{1-y^2} - 1}{\sqrt{1-y^2} - 1}

    y' = \frac{(\sqrt{1-y^2})(\sqrt{1-y^2} - 1)}{1 - y^2 - 1}

    y' = -\frac{1 - y^2 - \sqrt{1-y^2}}{y^2}

    I'll leave it like this.

    ================================================== ======
    iv) Ah, a practically straightforward problem!
    y = \frac{\sqrt{2x+1}}{x}

    Using the quotient rule:
    y' = \frac{\left ( \frac{1}{2} \frac{1}{\sqrt{2x+1}} \cdot 2 \right ) x - \sqrt{2x+1} \cdot 1}{x^2}

    y' = \frac{\left ( \frac{1}{\sqrt{2x+1}} \right ) x - \sqrt{2x+1}}{x^2}

    y' = \frac{\left ( \frac{1}{\sqrt{2x+1}} \right ) x - \sqrt{2x+1}}{x^2} \cdot \frac{\sqrt{2x+1}}{\sqrt{2x+1}}

    y' = \frac{x - (2x+1)}{x^2 \sqrt{2x+1}}

    y' = \frac{-x - 1}{x^2 \sqrt{2x+1}} \cdot \frac{\sqrt{2x+1}}{\sqrt{2x+1}}

    y' = -\frac{(x+1)\sqrt{2x+1}}{x^2 (2x+1)}

    -Dan
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