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Math Help - [SOLVED] calc2 major help please!

  1. #1
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    [SOLVED] calc2 major help please!

    ok so basically we are doing lengths and planes of curves.. i m lost to how the solution manual gets this..


    the get:

    f '(x) = x (x^2 + 2)^ 1/2


    = integral ( 0 to 3) x (x^2 + 2)^ 1/2

    im ok till here

    int ( 0 to 3) x (1 + {x^2 + 2} x^2)^ 1/2

    ---------------^---------------^
    my question is where did that +1 and x^2 come from how do tthey know they go there?? for other problems how would i know to get those correct values ??
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Legendsn3verdie View Post
    ok so basically we are doing lengths and planes of curves.. i m lost to how the solution manual gets this..


    the get:

    f '(x) = x (x^2 + 2)^ 1/2


    = integral ( 0 to 3) x (x^2 + 2)^ 1/2

    im ok till here

    int ( 0 to 3) x (1 + {x^2 + 2} x^2)^ 1/2

    ---------------^---------------^
    my question is where did that +1 and x^2 come from how do tthey know they go there?? for other problems how would i know to get those correct values ??
    hmm....maybe I could help if I knew what the original function was.

    --Chris
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    hmm....maybe I could help if I knew what the original function was.

    --Chris
    original function is y = (1/3) (x^2+2)^3/2
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  4. #4
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    looks like the format for an integral to calculate arc length ...

    S = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx
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    Quote Originally Posted by skeeter View Post
    looks like the format for an integral to calculate arc length ...

    S = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx
    oh wow so there is a forumal that says add 1 and square the derivative!?
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  6. #6
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    Yes. This is the standard formula for arc length.
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    hm ok then why in my original post did the book put a x^2 there instead of just squaring the derivative? instead of squaring the derivatinve the book multiplied the derivative by x^2..
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Legendsn3verdie View Post
    original function is y = (1/3) (x^2+2)^3/2
    Arc length is defined as L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx

    So we first need to find f'(x), which will then lead to [f'(x)]^2

    f'(x)=\frac{1}{2}(x^2+2)^{\frac{1}{2}}\cdot(2x)=x\  sqrt{x^2+2}

    Thus, [f'(x)]^2=x^2(x^2+2)=x^4+2x^2

    Therefore, the arc length would be L=\int_0^3\sqrt{1+(x^4+2x^2)}\,dx

    This implies that L=\int_0^3\sqrt{x^4+2x^2+1}\,dx

    It is good to note that x^4+2x^2+1 is a perfect square. So x^4+2x^2+1=(x^2+1)^2

    Thus, we have L=\int_0^3\sqrt{(x^2+1)^2}\,dx\implies L=\int_0^3 (x^2+1)\,dx

    Can you take it from here? Does this make sense?

    --Chris
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  9. #9
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    Wink

    Quote Originally Posted by Chris L T521 View Post
    Arc length is defined as L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx

    So we first need to find f'(x), which will then lead to [f'(x)]^2

    f'(x)=\frac{1}{2}(x^2+2)^{\frac{1}{2}}\cdot(2x)=x\  sqrt{x^2+2}

    Thus, [f'(x)]^2=x^2(x^2+2)=x^4+2x^2

    Therefore, the arc length would be L=\int_0^3\sqrt{1+(x^4+2x^2)}\,dx

    This implies that L=\int_0^3\sqrt{x^4+2x^2+1}\,dx

    It is good to note that x^4+2x^2+1 is a perfect square. So x^4+2x^2+1=(x^2+1)^2

    Thus, we have L=\int_0^3\sqrt{(x^2+1)^2}\,dx\implies L=\int_0^3 (x^2+1)\,dx

    Can you take it from here? Does this make sense?

    --Chris
    ty sir would kiss ya if icould!
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  10. #10
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    Quote Originally Posted by skeeter View Post
    looks like the format for an integral to calculate arc length ...

    S = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx

    imagine tryin to solve that problem not knowing that forumula! my heart sank threw the floor i was not aware at all! wow that formula is like a light from heaven
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