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Math Help - Some review questions.

  1. #1
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    Some review questions.

    1. Find the limit lim as (x,y)->(0,0) {(x^2y)/(x^4+y^2)}. Does anyone see anything that doesnt make it zero easily enough? Ive tried a few, but maybe I'm missing something.

    2.Find the absolute extremes of F(x,y)=x^2+xy R={|x|<or=2, |y|<or=1}
    Someone help me set this up?

    3.Evaluate: Integral of region C Mdx + Ndy for M=e^xsiny, N=e^xcosy, C: x=t-sint, y=1-cost from (0,0) to (2pi,0).
    Looks like unit circle. I dont know how to set it up.

    Thanks
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  2. #2
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    Quote Originally Posted by dmbocci View Post
    1. Find the limit lim as (x,y)->(0,0) {(x^2y)/(x^4+y^2)}. Does anyone see anything that doesnt make it zero easily enough? Ive tried a few, but maybe I'm missing something.
    Have you tried changing this to polar co-ordinates?
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  3. #3
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    I didnt think of that. But I found that y=x^2 makes it (1/2), therefor different paths lead to different limits, so it doesnt exist. Thanks though.

    Any help on the other 2 problems?
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  4. #4
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    Quote Originally Posted by dmbocci View Post
    1. Find the limit lim as (x,y)->(0,0) {(x^2y)/(x^4+y^2)}. Does anyone see anything that doesnt make it zero easily enough? Ive tried a few, but maybe I'm missing something.
    Did you try along the path y=x^2~?
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  5. #5
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    We must have input our posts around the same time, because yea I saw what that path makes it.

    So for #2, I got critical points of (0,0) (0,1) (0,-1) (0,-2) (2,1) (2,0). But do I disregard (0,-1)&(0,-2) since they are outside my boundary?
    Either way, the values I got are 0,4,6. So would 0 be my minimum and if so in my answer would I say that the function has absolute min. at both (0,1) and (0,0). Or would 4 be my min at (2,0)?
    And 6 would be my abs. max. at (2,1)?
    Does this look right?
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  6. #6
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    #2. Max./Min. on [-2,2]x[-1,1]

    Quote Originally Posted by dmbocci View Post
    2.Find the absolute extremes of F(x,y)=x^2+xy, \ R={|x|\le 2, |y|\le 1}
    Someone help me set this up?
    Thanks
    ...

    So for #2, I got critical points of (0,0) (0,1) (0,-1) (0,-2) (2,1) (2,0). But do I disregard (0,-1)&(0,-2) since they are outside my boundary?
    Either way, the values I got are 0,4,6. So would 0 be my minimum and if so in my answer would I say that the function has absolute min. at both (0,1) and (0,0). Or would 4 be my min at (2,0)?
    And 6 would be my abs. max. at (2,1)?
    Does this look right?

    How did you find the critical points?

    The point (0, -1) is not outside your boundary. \displaystyle R is a rectangular region with verticies (2,1), (2,-1), -2, -1), and (-2, 1).
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  7. #7
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    Quote Originally Posted by dmbocci View Post
    3.Evaluate: Integral of region C Mdx + Ndy for M=e^x\sin y, N=e^x\cos y, C: x=t-\sin t, y=1-\cos t from (0,0) to (2pi,0).
    Looks like unit circle. I don't know how to set it up.

    Thanks

    \displaystyle C is a contour, so I suspect it is a contour integral that's wanted.

    The contour, \displaystyle C, is a single arch of a cycloid.

    Find \displaystyle dx and \displaystyle dy in terms of \displaystyle dt from the contour equations.

    Take \displaystyle x and \displaystyle y from the contour equations and plug them into the equations for \displaystyle M and \displaystyle N. The resulting integral looks like it would be messy.

    Alternative: Use Green's Theorem:
    \displaystyle \oint_{C_1} Mdx + Ndy=\underset{R}{\int\int}\ \left({{\partial M}\over{\partial x}}- {{\partial N}\over{\partial y}}\right)\ dA, where \displaystyle C_1 is a closed path around region, \displaystyle R.

    Break \displaystyle C_1 up into path \displaystyle C as defined above, and path \displaystyle C_0, which goes from \displaystyle (2\pi,\ 0) to \displaystyle (0,\ 0) along the \displaystyle x-axis. \displaystyle dy = 0 along \displaystyle C_0. Also, \displaystyle M=0 along \displaystyle C_0. So the integral along \displaystyle C_0 contributes nothing to the integral around \displaystyle C_0.

    Therefore, \displaystyle \oint_{C_1} Mdx + Ndy=\int_{C} Mdx + Ndy.

    Now, compute: \displaystyle \underset{R}{\int\int}\ \left({{\partial M}\over{\partial x}}- {{\partial N}\over{\partial y}}\right)\ dy\ dx.

    This will also take some work. Good Luck.
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  8. #8
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    Not thinking on #2, whoops.
    So idk if this is right, but now I have (-2,1)(-2,-1)(2,-1)(2,1)(-2,2)(2,-2)(1/2,-1)(-1/2,1) as my points to check. Does this look correct?
    If so, my values output are 6,2,-1/2, and for (-2,2)&(2,-2)= 0 . But I would exclude those last two values?
    Someone check my work?
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  9. #9
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    #2. Max./Min. on [-2,2]x[-1,1]

    Quote Originally Posted by dmbocci View Post
    2.Find the absolute extremes of F(x,y)=x^2+xy,\ R=\{|x|\le2,\ |y|\le1\}
    Someone help me set this up?

    Thanks
    ...
    Quote Originally Posted by dmbocci View Post
    Not thinking on #2, whoops.
    So idk if this is right, but now I have (-2,1)(-2,-1)(2,-1)(2,1)(-2,2)(2,-2)(1/2,-1)(-1/2,1) as my points to check. Does this look correct?
    If so, my values output are 6,2,-1/2, and for (-2,2)&(2,-2)= 0 . But I would exclude those last two values?
    Someone check my work?

    How do we find the extrema for \displaystyle F(x,\ y) over some bounded closed region, \displaystyle R\ ? Yes, look for critical points.

    Critical points occur at locations within the interior of \displaystyle R, where \displaystyle {{\partial F}\over {\partial x}}=0 and \displaystyle {{\partial F}\over {\partial y}}=0, or where these partial derivatives do not exist. They also occur along the boundary of \displaystyle R\, where directional derivatives along that boundary are zero, or don't exist. Take all of that into account, and find the minimum & maximum values of \displaystyle F at all of the critical points.

    \displaystyle {{\partial F}\over {\partial x}}=2x+y. Setting this equal to zero gives: \displaystyle y=-2x. \displaystyle {{\partial F}\over {\partial y}}=x. Setting this equal to zero gives: \displaystyle x=0. This gives a critical point at the origin.

    On the boundary \displaystyle y=1: \displaystyle F(x,\ 1)=x^2+x, so \displaystyle {{dF}\over {dx}}=2x+1=0\ \ \implies\ \  x=-{{1}\over{2}}. So, \left(-{{1}\over{2}},\ 1\right) is a critical point. Similarly, \left({{1}\over{2}},\ -1\right) is also a critical point.

    On the boundary \displaystyle x=2: \displaystyle F(2,\ y)=4+2y, so \displaystyle {{dF}\over {dy}}=2\ne0. So there are no critical points here, except at the end points.

    Critical points are thus: (0,\ 0), \left(-{{1}\over{2}},\ 1\right),\ \left({{1}\over{2}},\ -1\right),\ (2,\ 1), \ (2,\ -1),\ (-2,\ -1),\ (-2,\ 1)

    Evaluate \displaystyle F(x,\ y) at each of these points to find the extrema.
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  10. #10
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    I have what you have, since I have min of 2 at 2 points and max of 6 at 2 points, in my conclusion I would state all 4 points, correct?
    And for x=2 I exclude that output it gives because its outside the region of our graph, correct?
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