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Math Help - Some questions from old exams??

  1. #1
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    Some questions from old exams??

    1)
    a) \lim_{(x,y)\rightarrow(0,0)}\frac{sin(xy)}{x}

    b) \lim_{(x,y)\rightarrow(0,0)}\frac{xy^2}{x^4y^4}

    2) Without using distance formula, find the distance from the point (0,1,1) to the plane 4y+3z=-12

    3) A closed box to be found to have length 2ft, witdh 4ft and hight 3ft, where the measurement of each dimension ismade with a maximum possible error of +-0.02ft. The top of the box is made from material that costs only 1.50 Dollars/ ft^2. What is the maximum error involved in the computation of the cost of the box?

    4) Find an equation for the tangent plane to the surface z=e^{-(x^2+y^2)} at (0,0,1).

    5) A right circular cone is measured and is found to have base radius r=40cm and altitude h=20cm. If it is known that each measurement is accurate to within %2, what is the maximum percentage error in the measurement of the volume?

    These are five of the problems which i couldn't solve. i'll try others based on answers of these.
    Any help is appriciated. =)
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  2. #2
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    Hello, Lafexlos!

    Here's #5 . . .


    5) A right circular cone is found have base radius r = 40 cm and height h = 20 cm.
    If it is known that each measurement is accurate to within 2%,
    what is the maximum percentage error in the measurement of the volume?

    The volume of a cone is: . V \;=\;\tfrac{\pi}{3}r^2h

    Hence: . dV \;=\;\tfrac{\pi}{3}r^2dh + \tfrac{2\pi}{3}rh\,dr  \;=\;\tfrac{\pi}{3}r(r\,dh + 2h\,dr)

    The percentage error is: . E \;=\;\frac{dV}{V} \;=\;\frac{\frac{\pi}{3}r(r\,dh + 2h\,dr)}{\frac{\pi}{3}r^2h} \;=\; \frac{r\,dh + 2h\,dr}{rh} .[1]


    We are given: . \begin{Bmatrix}r \:=\: 40 & dr \:\leq\: 0.8 \\ h \:=\:20 & dh \:\leq\:0.4 \end{Bmatrix}


    Substitute into [1]: . E \;\leq\;\frac{(40)(0.4) + 2(20)(0.6)}{(40(20)} \;=\;0.06 \;=\;\boxed{6\%}

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  3. #3
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    Thanks Soroban. =)
    Still need help at* others.

    *[on, or whatever. still dont know prepositions. :A]
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  4. #4
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    For the first limit, when y=0 : the limit = 0 .. when y=2 .. the limit = 2
    so the limit does not exist.

    for the second limit, \frac{xy^2}{x^4y^4}=\frac{1}{x^3 y^2} ..

    So what will happen when (x,y) \rightarrow (0,0) ?

    Problem 4 : Shout be straightforward for you!
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  5. #5
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    Quote Originally Posted by General View Post
    for the second limit, \frac{xy^2}{x^4y^4}=\frac{1}{x^3 y^2} ..

    So what will happen when (x,y) \rightarrow (0,0) ?
    thx to our lecturer, dont know what will happen.
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  6. #6
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    Quote Originally Posted by Lafexlos View Post
    thx to our lecturer, dont know what will happen.

    When you substitute x=0 and y=0 .. what will happen to the fraction ?!
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  7. #7
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    infinite.
    Hence it doesn't exist too because of limit goes infinite?
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  8. #8
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    Quote Originally Posted by Lafexlos View Post
    infinite.
    Hence it doesn't exist too because of limit goes infinite?
    Yes ..
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  9. #9
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    ok. thanks.
    #1 & #5 solved. Try to solve #4 again. So #2 && #3 remain.
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