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Math Help - 2 questions on area between curves! help please!

  1. #1
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    2 questions on area between curves! help please!

    Ahh, yikes. I just typed out all of the images to prevent any further mishaps hah.

    Find c > 0 such that the area of the region enclosed by the parabolas
    y = x^2 - c^2 and y = c^2 - x^2 is 30.


    Find the area of the region bounded by the parabola y=4x^2, the tangent line to this parabola at (3,36) and the x axis.

    Ok, so my method of solving the first one was kind of strange, and I'm fairly certain that I used a completely incorrect method, but here it is...
    I assumed that if I found the area in two quadrants then I could just multiply it by 2 and get the total area. So I did y = x^2 - c^2 = area of 15. Then I integrated, plugged in -c and c for x and solved but I came up with a wrong answer, which I think was something like 2.83.

    For the second problem I found the began by finding the derivative which was 8x then found the equation for the line by plugging in 36, 3, 8 into y=mx+b equation, which ended up being y=8x+12. Then I integrated
    S 8x+12 - 4x^2 and I'm not sure what to even use as the limits at this point, but I made multiple guesses and was wrong on all of them.

    HELP PLEASE!!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sgares View Post
    Find such that the area of the region enclosed by the parabolas and is .


    Find the area of the region bounded by the parabola , the tangent line to this parabola at and the axis.

    Ok, so my method of solving the first one was kind of strange, and I'm fairly certain that I used a completely incorrect method, but here it is...
    I assumed that if I found the area in two quadrants then I could just multiply it by 2 and get the total area. So I did = area of 15. Then I integrated, plugged in -c and c for x and solved but I came up with a wrong answer, which I think was something like 2.83.

    For the second problem I found the began by finding the derivative which was 8x then found the equation for the line by plugging in 36, 3, 8 into y=mx+b equation, which ended up being y=8x+12. Then I integrated
    S 8x+12 - 4x^2 and I'm not sure what to even use as the limits at this point, but I made multiple guesses and was wrong on all of them.

    HELP PLEASE!!
    Cannot see your images.
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  3. #3
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    Chandler, Arizona
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    Ahh, yikes. I just typed out all of the images to prevent any further mishaps hah.

    Find c > 0 such that the area of the region enclosed by the parabolas
    y = x^2 - c^2 and y = c^2 - x^2 is 30.


    Find the area of the region bounded by the parabola y=4x^2, the tangent line to this parabola at (3,36) and the x axis.

    Ok, so my method of solving the first one was kind of strange, and I'm fairly certain that I used a completely incorrect method, but here it is...
    I assumed that if I found the area in two quadrants then I could just multiply it by 2 and get the total area. So I did y = x^2 - c^2 = area of 15. Then I integrated, plugged in -c and c for x and solved but I came up with a wrong answer, which I think was something like 2.83.

    For the second problem I found the began by finding the derivative which was 8x then found the equation for the line by plugging in 36, 3, 8 into y=mx+b equation, which ended up being y=8x+12. Then I integrated
    S 8x+12 - 4x^2 and I'm not sure what to even use as the limits at this point, but I made multiple guesses and was wrong on all of them.

    HELP PLEASE!!
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  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
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    Pennsylvania
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    Quote Originally Posted by sgares View Post
    Ahh, yikes. I just typed out all of the images to prevent any further mishaps hah.

    Find c > 0 such that the area of the region enclosed by the parabolas
    y = x^2 - c^2 and y = c^2 - x^2 is 30.


    Find the area of the region bounded by the parabola y=4x^2, the tangent line to this parabola at (3,36) and the x axis.

    Ok, so my method of solving the first one was kind of strange, and I'm fairly certain that I used a completely incorrect method, but here it is...
    I assumed that if I found the area in two quadrants then I could just multiply it by 2 and get the total area. So I did y = x^2 - c^2 = area of 15. Then I integrated, plugged in -c and c for x and solved but I came up with a wrong answer, which I think was something like 2.83.

    For the second problem I found the began by finding the derivative which was 8x then found the equation for the line by plugging in 36, 3, 8 into y=mx+b equation, which ended up being y=8x+12. Then I integrated
    S 8x+12 - 4x^2 and I'm not sure what to even use as the limits at this point, but I made multiple guesses and was wrong on all of them.

    HELP PLEASE!!
    The first one is formally unanswerable, but informally

    Let c^2-x^2<x^2-c^2\quad\forall{x}\in(a,b)

    Now then you must find c by solving the following(Easy) integral equation

    \int_a^{b}\bigg[(x^2-c^2)-(c^2-x^2)\bigg]dx=30

    The tangent line is given by

    y-36=24(x-3)

    Now let g(x)=24(x-3)+36

    and f(x)=4x^2

    Now find the two values of x such that

    f(x)=g(x)

    Then test one value in each to show that either

    \forall{x}\in(a,b)\quad{f(x)}<g(x)\text{ or }g(x)<f(x)

    Then the area would be

    A=\int_a^{b}g(x)-f(x)dx

    or

    A=\int_a^{b}f(x)-g(x)dx

    respectively

    Good luck, and if you have any more questions, dont hesitate to ask.


    Mathstud
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  5. #5
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    Thank you for the help, I really do appreciate it! I'll get back to working on them now! =)
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