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Math Help - Please help me check these answers

  1. #1
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    Please help me check these answers (I)

    Hi, I'm working on some problems for reviewing, but I'm not sure if the answers I'm getting are the right one. Please help me check if they're right. If they're not please let me know and I will post my work to see what I have done wrong.

    1. Find








    2. Find the equation of the line tangent to at




    3. Find the point P on the graph of closest to the point




    4. Calculate the surface area of revolution of about the x-axis over

    =
    Last edited by myngo9191; May 9th 2010 at 08:07 PM.
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  2. #2
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    Quote Originally Posted by myngo9191 View Post
    Hi, I'm working on some problems for reviewing, but I'm not sure if the answers I'm getting are the right one. Please help me check if they're right. If they're not please let me know and I will post my work to see what I have done wrong.

    1. Find






    ok

    2. Find the equation of the line tangent to at


    what is "b" ?

    3. Find the point P on the graph of closest to the point


    so far so good ... and the point is ... ?

    4. Calculate the surface area of revolution of about the x-axis over

    =

    no.
    ...
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  3. #3
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    2. Find the equation of the line tangent to at
    y=\frac{5}{13}x+ \frac{21}{13}
    3. Find the point P on the graph of closest to the point


    How do I find the point from here? Do I plug 10 in for x?
    4. Calculate the surface area of revolution of about the x-axis over
    =

    Here is my work, please let me know where I went wrong.

    y=(4-x^\frac{2}{3})^\frac{3}{2}

    y'=-x^\frac{-1}{3}(4-x^\frac{2}{3})^\frac{1}{2}

    (y')^2+1\rightarrow 1+(4-x^\frac{2}{3})^\frac{1}{2})^2=1+\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}=\frac{x^\frac{2}{3}}  {x^\frac{2}{3}}+\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}=\frac{4}{x^\frac{2}{  3}}

    u=4-x^\frac{2}{3}
    du=-\frac{2}{3}x^{-\frac{1}{3}}dx
    -3du=2x^{-\frac{1}{3}}

    2\pi\int_{2}^{3}u^{\frac{3}{2}}(-3)du=-6\pi\frac{2}{5}u^{\frac{5}{2}}\Bigg|_{2}^{3}=-\frac{12\pi}{5}(4-x^{\frac{2}{3}})^{\frac{5}{2}}\Bigg|_{2}^{3}
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  4. #4
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    Quote Originally Posted by myngo9191 View Post
    2. Find the equation of the line tangent to at
    y=\frac{5}{13}x+ \frac{21}{13}

    ok
    3. Find the point P on the graph of closest to the point


    How do I find the point from here? Do I plug 10 in for x?

    how about taking a derivative?
    4. Calculate the surface area of revolution of about the x-axis over
    =

    Here is my work, please let me know where I went wrong.

    y=(4-x^\frac{2}{3})^\frac{3}{2}

    y'=-x^\frac{-1}{3}(4-x^\frac{2}{3})^\frac{1}{2}

    (y')^2+1\rightarrow 1+(4-x^\frac{2}{3})^\frac{1}{2})^2=1+\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}=\frac{x^\frac{2}{3}}  {x^\frac{2}{3}}+\frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}=\frac{4}{x^\frac{2}{  3}}

    u=4-x^\frac{2}{3}
    du=-\frac{2}{3}x^{-\frac{1}{3}}dx
    -3du=2x^{-\frac{1}{3}}

    2\pi\int_{2}^{3}u^{\frac{3}{2}}(-3)du=-6\pi\frac{2}{5}u^{\frac{5}{2}}\Bigg|_{2}^{3}=-\frac{12\pi}{5}(4-x^{\frac{2}{3}})^{\frac{5}{2}}\Bigg|_{2}^{3}
    A = 2\pi \int_2^3 y\sqrt{1+(y')^2} \, dx

    y = (4-x^\frac{2}{3})^\frac{3}{2}

    y' = -\frac{(4-x^\frac{2}{3})^\frac{1}{2}}{x^\frac{1}{3}}

    (y')^2 = \frac{4-x^\frac{2}{3}}{x^\frac{2}{3}}

    1 + (y')^2 = \frac{4}{x^\frac{2}{3}}

    A = 2\pi \int_2^3 (4-x^\frac{2}{3})^\frac{3}{2} \cdot \frac{2}{x^\frac{1}{3}} \, dx

    A = -\frac{12\pi}{5} \left[(4-x^\frac{2}{3})^\frac{5}{2}\right]_2^3 \approx 29.658
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  5. #5
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    3. Find the point P on the graph of closest to the point


    I got this after taking the derivative

    2(x-10)+1

    Now what do I do?
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  6. #6
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    Quote Originally Posted by myngo9191 View Post
    3. Find the point P on the graph of closest to the point


    I got this after taking the derivative

    2(x-10)+1

    Now what do I do?
    have you not completed any optimization problems?

    where do extrema occur?
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  7. #7
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    I didn't even know it's an optimization problem and no I haven't done any. I went look in the book and it says I should find the critical point by solving for x. I'm not sure if that's the way to go but here it is...

     2(x-10)+1\rightarrow x=9\frac{1}{2}
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  8. #8
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    The point closest to the graph is the point where the minimum of the distance function is achieved.

    Minimum is achieved at an extremal point, so...
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  9. #9
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    go to the link ...

    http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/maxmindirectory/MaxMin.html

    read the guidelines then have a look at problem #7.
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  10. #10
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    I'm sorry but I'm really lost and I have no idea what to do with that problem.
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  11. #11
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    3. Find the point P on the graph of closest to the point


    d^2(x)'=2(x-10)+1

    \frac{-1}{2(x-10)+1} = (y-0)(x-10)

    \Big\Downarrow

    \frac{\frac{-1}{2(x-10)+1}}{y}=x-10

    \Big\Downarrow

    \frac{-1}{(2(x-10)+1)y}=x-10

    \Big\Downarrow

    \frac{-1}{(2(x-10)+1)\sqrt{x}}-x+10=0

    \Big\Downarrow

    -1-x+10=0

    \Big\Downarrow

    x=9

    \Big\Downarrow

    y=\sqrt{x}=\sqrt{9}=3

    so the point on the graph of y=\sqrt{x} closest to the point (10,0) is (9,3)
    Please let me know if this is right and if not where I have messed up at. Thank you
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  12. #12
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     <br />
D^2 = (x-10)^2 + x<br />

     <br />
\frac{d(D^2)}{dx} = 2(x-10) + 1<br />

    set the derivative equal to 0 because extrema occur at critical values ...

    2(x-10) + 1 = 0

    solve for x ...

    x = 9.5

    D^2 has a minimum value at x = 9.5 ... you can use the 1st or 2nd derivative test(s) to confirm, or you can just note that the graph of D^2 is a standard parabola with vertex at the bottom.

    so, the closest point on the curve y = \sqrt{x} to the point (10,0) is the point \left(9.5,\sqrt{9.5} \right)
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