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Math Help - Calculus 1 Questions

  1. #1
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    Calculus 1 Questions

    1. Let a and b be differentiable.

    x^2 x<1

    ax+b x> or =1


    2. Prove lim as x approaches 3
    x^2+x=8


    3. A square plot is made at points A,B,C, and D (clockwise). The sides of the square are 100ft long. A pipeline goes from A to a point between B and C and from that point to C. The cost of each side is $20 per ft. and the cost of the pipe inside the square plot is $10 per ft. What is the minimum cost?
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  2. #2
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    For #3:

    Are you sure the costs aren't reversed. It should maybe be $20 through the square and $10 along the side.

    The point between B and C where the pipeline joins the side, call it P.

    From the drawing, this length is \sqrt{100^{2}+x^{2}}

    The distance along the side is 100-x.

    Factor in the cost:

    20\sqrt{100^{2}+x^{2}}+10(100-x)

    This is what needs minimized. Therefore, differentiate, set to 0 and solve for x.
    Last edited by galactus; November 24th 2008 at 05:39 AM.
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  3. #3
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    How did you get 8?
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  4. #4
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    Sorry, that was a typo. I fixed it.
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  5. #5
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    Quote Originally Posted by Debi
    1. Let a and b be differentiable.

    x^2 x<1

    ax+b x> or =1
    You are defining a function:

    <br />
f(x)=\left\{<br />
 \begin{array}{cc}<br />
{x^2}& {x<1}\\<br />
{ax+b}& {x \ge 1}<br />
\end{array} <br />
 \right{}<br />

    But what is the question?

    At a guess I would say you have been asked to find a and b so the f(x) is differentiable everywhere.

    f(x) is piecewise continuous and differentiable, so we need only make sure it is continuous and differentiable at the join between the pieces

    First we have to ensure that f(x) is continuous so we need

    \lim_{x \to 1_-} f(x)=\lim_{x \to 1_+}f(x),

    which is equivalent to:

    1=a+b\ \ \ \dots(1).

    Similarly we need:

    \lim_{x \to 1_-} f'(x)=\lim_{x \to 1_+}f'(x),

    which is equivalent to:

    2=a\ \ \ \dots(2).

    So s=2, and substituting back into (1) gives: b=1.

    RonL
    Last edited by CaptainBlack; July 29th 2006 at 12:26 PM.
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  6. #6
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    your way of solving the prblm isn't possible b/c the square root -99 =x
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  7. #7
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    Quote Originally Posted by CaptainBlack
    You are defining a function:

    <br />
f(x)=\left\{<br />
 \begin{array}{cc}<br />
{x^2}& {x<1}\\<br />
{ax+b}& {x \ge 1}<br />
\end{array} <br />
 \right{}<br />

    But what is the question?

    At a guess I would say you have been asked to find a and b so the f(x) is differentiable everywhere.

    RonL
    YES
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by Debi
    YES
    See the rest of that post for the solution, you have only see the first half.

    RonL
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  9. #9
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    Quote Originally Posted by Debi
    2. Prove lim as x approaches 3
    x^2+x=8
    Not sure what this question should be. The limit as x approaches
    3 is not 8

    RonL
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by galactus
    For #3:

    Are you sure the costs aren't reversed. It should maybe be $20 through the square and $10 along the side.

    The point between B and C where the pipeline joins the side, call it P.

    From the drawing, this length is \sqrt{100+x^{2}}

    The distance along the side is 100-x.

    Factor in the cost:

    20\sqrt{100+x^{2}}+10(100-x)
    Another typo, should be:


    20\sqrt{100^2+x^{2}}+10(100-x)

    RonL
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  11. #11
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    Man, I'm typoin' all over the place. Thanks for the catch, Cap'n.

    I fixed my post.
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  12. #12
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    Prove as lim x approaches 3 of the equation, x^2+x=12
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  13. #13
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    Quote Originally Posted by Debi
    Prove as lim x approaches 3 of the equation, x^2+x=12
    \lim_{x\to 3}x=3
    Proof: You need to show for any \epsilon >0 there is a \delta >0 such as, if
    0<|x-3|<\delta, \, x\in D (where D is in the domain of f(x)=x) then,
    |x-3|<\epsilon . If you take \delta = \epsilon you make this condition true.

    \lim_{x\to 3}x+1=4
    Proof: You need to show for any \epsilon >0 there is a \delta >0 such as, if
    0<|x-3|<\delta, \, x\in D (where D is in the domain of f(x)=x+1) then,
    |x+1-4|=|x-3|<\epsilon . If you take \delta = \epsilon you make the condition true.

    \lim_{x\to 3}x^2+x=12
    Proof:Factor,
    \lim_{x\to 3}x(x+1)
    Note that,
    \lim_{x\to 3}x=3--->First Paragraph
    \lim_{x\to 3}x+1=4--->Second Paragraph
    Therefore,
    \lim_{x\to 3}x(x+1)=3\cdot 4=12
    Proof Complete.
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