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Math Help - Help w/ Problems for Test Tomorrow

  1. #1
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    Help w/ Problems for Test Tomorrow

    There are several I need help with.

    Find the derivative

    1. y=ln(x^2+y^2)
    2. f(u)=(2^u+2^{-u})^{10}
    3. y=x^x

    Use the substitution rule
    4. \int \frac{cos\sqrt{t}}{\sqrt{t}}
    5. \int \cos\theta\sin^6\theta d\theta

    thanks
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  2. #2
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    Quote Originally Posted by c_323_h
    4. \int \frac{cos\sqrt{t}}{\sqrt{t}}
    Express as,
    2\int \frac{\cos \sqrt{t}}{1} \cdot \frac{1}{2\sqrt{t}} dt
    Let,
    u=\sqrt{t}, thus \frac{du}{dt}=\frac{1}{2\sqtr{t}}
    Thus,
    2\int \cos u \frac{du}{dt} dt=2\int \cos u du
    Thus,
    2\sin u+C
    Substitute back,
    2\sin \sqrt{t}+C
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  3. #3
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    Quote Originally Posted by c_323_h
    5. \int \cos\theta\sin^6\theta d\theta
    Let,
    u=\sin \theta then, \frac{du}{d\theta }=\cos \theta
    Thus,
    \int u^6 \frac{du}{d\theta} d\theta=\int u^6 du
    Thus,
    \frac{1}{7}u^7+C=\frac{1}{7}\sin^7 \theta +C
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  4. #4
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    Quote Originally Posted by c_323_h
    1. y=ln(x^2+y^2)
    IMPLICITLY.
    (Remember to use chain rule now)
    y'=\frac{(x^2+y^2)'}{x^2+y^2}
    Thus,
    y'=\frac{2x+2yy'}{x^2+y^2}
    Thus,
    x^2y'+y^2y'=2x+2yy'
    Thus,
    x^2y'+y^2y'-2yy'=2x
    Thus,
    y'(x^2-2y+y^2)=2x
    Thus,
    y'=\frac{x^2-2y+y^2}{2x}
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  5. #5
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    Quote Originally Posted by c_323_h
    3. y=x^x
    Note,
    x^x=(e^{\ln x})^x=e^{x\ln x}
    Now just use chain rule,
    y'=(x\ln x)'e^{x\ln x}
    Product rule,
    y'=(1+\ln x)e^{x\ln x}
    Express back,
    y'=(1+\ln x)x^x
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  6. #6
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    Quote Originally Posted by c_323_h
    2. f(u)=(2^u+2^{-u})^{10}
    Use chain rule,
    10(2^u+2^{-u})'(2^u+2^{-u})^9
    Thus,
    10(\ln 2\cdot 2^u-\ln 2\cdot  2^{-u})(2^u+2^{-u})^9
    Thus,
    10\ln 2(2^u-2^{-u})(2^u+2^{-u})^9
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  7. #7
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    Thank you very much for the quick response...and for solving all the problems.
    Last edited by c_323_h; July 4th 2006 at 03:01 PM.
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  8. #8
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    Hello, c_323_h!

    Find the derivative:

    1)\; y \:=\:\ln(x^2+y^2)

    Differentiate implicitly: . y' \:=\:\frac{1}{x^2+y^2}(2x + 2xy\cdot y')

    Then we have: . (x^2 + y^2)y' \:=\:2x + 2y\cdot y'

    Rearrange terms: . x^2\cdot y' + y^2\cdot y' - 2y\cdot y' \:= \:2x

    Factor: . (x^2 + y^2 - 2y)y' \:= \:2x

    Therefore: . y' \;= \;\frac{2x}{x^2 + y^2 - 2y}



    2^u + 2^{-u})^{10}" alt="2)\;f(u)\:=\2^u + 2^{-u})^{10}" />

    f'(u)\;=\;10\left(2^u + 2^{-u}\right)^9\cdot\left(2^u\cdot\ln 2 + 2^{-u}[-1]\cdot\ln 2\right)

    f\(u) \;= \;10\cdot\ln 2\left(2^u + 2^{-u}\right)^9\left(2^u - 2^{-u}\right)


    3)\;y\;=\;x^x

    Take logs: . \ln(y) \;=\;\ln\left(x^x\right) \;= \;x\cdot\ln x

    Differentiate implicitly: . \frac{1}{y}\,y' \;= \;x\cdot\frac{1}{x} + 1\cdot\ln x

    We have: . \frac{y'}{y}\;=\;1 + \ln x\quad\Rightarrow\quad y' \;= \;y(1 + \ln x)

    Since y = x^x, we have: . y' \;= \;x^x(1 + \ln x)



    Use the substitution rule:

    4)\;\int \frac{\cos\sqrt{t}}{\sqrt{t}}\,dt

    We have: . \int t^{-\frac{1}{2}}\cdot\cos\left(t^{\frac{1}{2}}\right) \,dt

    Let: u = t^{\frac{1}{2}}\quad\Rightarrow\quad du = \frac{1}{2}t^{-\frac{1}{2}}dt\quad\Rightarrow\quad dt = 2t^{\frac{1}{2}}du

    Substitute: . \int t^{-\frac{1}{2}}\cdot\cos(u)\cdot\left(2t^{\frac{1}{2}  }du\right) \;= \;2\int\cos u\,du

    You can finish it now . . .



    5)\;\int \cos\theta\sin^6\theta d\theta

    We have: . \int(\sin\theta)^6(\cos\theta\,d\theta)

    Let: u = \sin\theta\quad\Rightarrow\quad du = \cos\theta\, d\theta\quad\Rightarrow\quad d\theta  = \frac{du}{\cos\theta}

    Substitute: . \int u^6\,\cos\theta\,\frac{du}{\cos\theta} \;= \;\int u^6\,du . . . got it?

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  9. #9
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    thanks you. as you can tell i am struggling with this class. is anyone willing to help me with these?

    Differentiate:

    1. y=cos^{-1}(e^{2x})
    2. h(t)=cot^{-1}(t)+cot{-1}(\frac{1}{t})

    Evaluate the integral:

    1. \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{6}{\sqrt{t-t^2}}
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  10. #10
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    Quote Originally Posted by c_323_h

    1. \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{6}{\sqrt{t-t^2}}
    Manipulate,
    \int \frac{6}{t-t^2} dt
    As, (since t\geq 0) (I factored it)

    12 \int \cdot \frac{1}{\sqrt{1-t}}\cdot \frac{1}{2\sqrt{t}} dt

    Let, u=\sqrt{t} then, \frac{du}{dt} =\frac{1}{2\sqrt{t}}.
    Thus, far we have,
    \int \frac{1}{\sqrt{t-1}} \frac{du}{dt} dt

    Since, u=\sqrt{t} then, u^2=t thus, u^2-1=t-1 thus, \sqrt{u^2-1}=\sqrt{t-1} Thus, we have,

    \int \frac{1}{\sqrt{u^2-1}} \frac{du}{dt}dt=\int \frac{1}{\sqrt{u^2-1}} du

    This is a inverse hyperbolic function which you should recognize thus,
    \cosh^{-1} u+C
    Substitute back,
    \cosh^{-1} \sqrt{t}+C
    Thus,
    \left \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{6}{\sqrt{t-t^2}}=\cosh^{-1} \sqrt{t} \right|^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}
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  11. #11
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    Quote Originally Posted by ThePerfectHacker
    x^2y'+y^2y'=2x+2yy'
    Could you tell me where you got the y' from on the right side of the equation?
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  12. #12
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    Hello again, c_323_h!

    Differentiate:

    1)\;y\:=\:\cos^{-1}(e^{2x})
    y' \;= \;\frac{-1}{\sqrt{1 - \left(e^{2x}\right)^2}}\cdot e^{2x}\cdot2 \;= \;\frac{-2e^{2x}}{\sqrt{1 - e^{4x}}}


    2)\;h(t)\:=\:\cot^{-1}(t) + \cot^{-1}\left(\frac{1}{t}\right)
    h'(t)\;=\;\frac{-1}{1 + t^2} + \frac{-1}{1 + \left(\frac{1}{t}\right)^2}\cdot(-t^{-2}) \;= \;\frac{-1}{1 + t^2} + \frac{1}{\left(1 + \frac{1}{t^2}\right)t^2}

    . . . . . =\;-\frac{1}{1 + t^2} + \frac{1}{1 + t^2} \;= \;0 **


    Evaluate the integral: . 1)\;\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{6}{\sqrt{t-t^2}}

    In the denominator, complete the square:
    t - t^2\;=\;-(t^2 - t) \;= \;-\left(t^2 - t + \frac{1}{4} - \frac{1}{4}\right) \;= -\left(\left[t - \frac{1}{2}\right]^2 - \frac{1}{4}\right) \;= \;\frac{1}{4} - \left(x - \frac{1}{2}\right)^2

    The integral becomes: . 6\int^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}} \frac{1}{\sqrt{\frac{1}{4} - (t - \frac{1}{2})^2}}\,dt

    . . Let t - \frac{1}{2}\:=\:\frac{1}{2}\sin\theta\quad \Rightarrow\quad dt\,=\,\frac{1}{2}\cos\theta\,d\theta

    . . Note that: . \sqrt{\frac{1}{4} - \left(t - \frac{1}{2}\right)^2} \;= \;\sqrt{\frac{1}{4} - \frac{1}{4}\sin^2\theta} \;=  \sqrt{\frac{1}{4}\left(1 - \sin^2\theta)}

    . . . . . . . . . . =\;\sqrt{\frac{1}{4}\cos^2\theta}\;=\;\frac{1}{2} \cos\theta

    Substitute: . 6\int\frac{1}{\frac{1}{2}\cos\theta}\cdot\frac{1}{  2} \cos\theta\,d\theta \;= \;6\int d\theta\;= \;6\theta

    Back-substitute
    We have: . \frac{1}{2}\sin\theta \,= \,t - \frac{1}{2}\quad\Rightarrow\quad \sin\theta\,=\,2t - 1\quad\Rightarrow\quad\theta \,= \,\sin^{-1}(2t-1)

    So we have: . 6\sin^{-1}(2t - 1)\,\bigg]^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}} \;= \;6\sin^{-1}\left[2\left(\frac{\sqrt{3}}{2}\right) - 1\right] - 6\sin^{-1}\left[2\left(\frac{1}{2}\right) - 1\right]

    . . . = \;6\sin^{-1}(\sqrt{3} - 1) - 6\sin^{-1}(0) \;= \;6\sin^{-1}(\sqrt{3} - 1) \;\approx\; 4.928


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **
    In #2, we have: . h(t) \:=\:\cot^{-1}(t) + \cot^{-1}\left(\frac{1}{t}\right)

    Let: \cot^{-1}(t) = \alpha . . . and: \cot^{-1}(\frac{1}{t}) = \beta

    Then \cot\alpha = t . . . and \cot\beta = \frac{1}{t}

    Since \cot\alpha = \frac{t}{1} = \frac{adj}{opp}, \alpha is in this right triangle:
    Code:
                  *
                /β|
              /   |1
            /α    |
          * - - - *
              t

    Since \cot\beta = \frac{1}{t},\;\beta is the other acute angle.

    That is: . \alpha + \beta \:=\:\frac{\pi}{2}

    Hence, the function is: . h(t)\;=\;\cot^{-1}(t) + \cot^{-1}\left(\frac{1}{t}\right) \;= \;\alpha + \beta \;= \;\frac{\pi}{2}

    No wonder the derivative is 0 . . .

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