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Math Help - Constructing Antiderivatives and areas

  1. #1
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    Constructing Antiderivatives and areas

    1. The problem statement, all variables and given/known data
    The origin and the point (a, a) are at opposite corners of a square. Calculate the ratio of the areas of the two parts into which the curve  \sqrt{x} + \sqrt{y} = \sqrt{a} divides the square.


    2. Relevant equations
    I'm sure there will be some use of A = bh. Perhaps maybe the pythagorean theorem if the square is cut exactly in half? I'm sure I'll be using a definite integral to find the areas.



    3. The attempt at a solution
    I have no clue where to go with this.

    1. The problem statement, all variables and given/known data
    In drilling an oil well, the total cost, C, consists of fixed costs (independent of the depth of the well) and marginal costs, which depend on depth; drilling becomes more expensive, per meter, deeper into the earth. Suppose the fixed costs are 1,000,000 riyals (the riyal is the unit of currency of Saudi Arabia), and the marginal costs are C'(x) = 4000 + 10x riyals/meter, where x is the depth in meters. Find the total cost of drilling a well x meters deep.




    2. Relevant equations
    I'm using I'll use a definite integral.

    3. The attempt at a solution
    Here's what I tried.
     1,000,000 + \int (from 0 to x) of 4000 + 10x
    Evaluating that:
     1,000,000 + [4000 + 10x - (4000 + 10(0) ) ]
     1,000,000 + [4000 + 10x - 4000]
     1,000,000 + 10x riyals is the total cost.

    I'd like some direction on the first one, and confirmation with the second one.

    It's urgent, and I won't be able to look at this post for another 2 hours because I have to go on a long drive back to college. I need the help right when i get back though. Thanks in advance.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Jacobpm64 View Post
    1. The problem statement, all variables and given/known data
    The origin and the point (a, a) are at opposite corners of a square. Calculate the ratio of the areas of the two parts into which the curve  \sqrt{x} + \sqrt{y} = \sqrt{a} divides the square.
    The area below the curve in question is:

    A=\int_0^a (\sqrt{a}-\sqrt{x})^2 \,dx

    Let u=\sqrt{x}, then:

    A=\int_0^{\sqrt{a}} 2(\sqrt{a}-u)^2 u\,du

    Now the integrand may be expanded into a polynomial in u which may be integrated easily, to give:

    A=a^2/6.

    So the area in question is 1/6 of the area of the square, and the curve
    therefore divides the square into parts whose ratio is 1:5.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Jacobpm64 View Post
    1. The problem statement, all variables and given/known data
    In drilling an oil well, the total cost, C, consists of fixed costs (independent of the depth of the well) and marginal costs, which depend on depth; drilling becomes more expensive, per meter, deeper into the earth. Suppose the fixed costs are 1,000,000 riyals (the riyal is the unit of currency of Saudi Arabia), and the marginal costs are C'(x) = 4000 + 10x riyals/meter, where x is the depth in meters. Find the total cost of drilling a well x meters deep.




    2. Relevant equations
    I'm using I'll use a definite integral.

    3. The attempt at a solution
    Here's what I tried.
     1,000,000 + \int_0^x 4000 + 10x \ dx
    The total cost is:

    C(x)=1000000 + \int_0^x 4000+10 u\ du=1000000+4000x+5x^2

    RonL
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