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Math Help - Find the length of the curve r(t) - need help integrating

  1. #1
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    Find the length of the curve r(t) - need help integrating

    Find the length of the curve  r(t)=<ln(5t), 2t , t^2> , 2\leq t \leq 3

    I am still trying to figure out how to use latex so forgive me if it is a little messed up. the integral is supposed to be from 2 to 3

    this is my solution, so far...

     r'(t) = <\frac{1}{t} , 2 , 2t >
    |r'(t)| = \sqrt{\frac{1}{t^2}+ 4 + 4t^2}

     _2\int^3 |r'(t)|dt = _2\int^3 |r'(t)| dt = _2\int^3 \sqrt{\frac{1}{t^2}+ 4 + 4t^2}  dt

    I am a bit rusty on integration and cannot get the correct answer from this point. I was thinking of using u substitution but cannot figure out how to do it correctly.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by rain21 View Post
    Find the length of the curve  r(t)=<ln(5t), 2t , t^2> , 2\leq t \leq 3

    I am still trying to figure out how to use latex so forgive me if it is a little messed up. the integral is supposed to be from 2 to 3

    this is my solution, so far...

     r'(t) = <\frac{1}{t} , 2 , 2t >
    |r'(t)| = \sqrt{\frac{1}{t^2}+ 4 + 4t^2}

     _2\int^3 |r'(t)|dt = _2\int^3 |r'(t)| dt = _2\int^3 \sqrt{\frac{1}{t^2}+ 4 + 4t^2} dt

    I am a bit rusty on integration and cannot get the correct answer from this point. I was thinking of using u substitution but cannot figure out how to do it correctly.
    \sqrt{\frac{1}{t^2}+4+4t^2}=\sqrt{\frac{4t^4+4t^2+  1}{t^2}}=\frac{\sqrt{(2t^2+1)^2}}{t}. Can you finish?
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  3. #3
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    Quote Originally Posted by rain21 View Post
    Find the length of the curve  r(t)=<ln(5t), 2t , t^2> , 2\leq t \leq 3

    I am still trying to figure out how to use latex so forgive me if it is a little messed up. the integral is supposed to be from 2 to 3

    this is my solution, so far...

     r'(t) = <\frac{1}{t} , 2 , 2t >
    |r'(t)| = \sqrt{\frac{1}{t^2}+ 4 + 4t^2}

     _2\int^3 |r'(t)|dt = _2\int^3 |r'(t)| dt = _2\int^3 \sqrt{\frac{1}{t^2}+ 4 + 4t^2} dt

    I am a bit rusty on integration and cannot get the correct answer from this point. I was thinking of using u substitution but cannot figure out how to do it correctly.

     4t^2 + 4 + \frac{1}{t^2} = \left(2t+\frac{1}{t}\right)^2 ...

    Tonio
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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    \sqrt{\frac{1}{t^2}+4+4t^2}=\sqrt{\frac{4t^4+4t^2+  1}{t^2}}=\frac{\sqrt{(2t^2+1)^2}}{t}. Can you finish?
    is the answer 20?
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by rain21 View Post
    is the answer 20?
    The answer is whaterver you get for (t^2+\ln{t})\Big|_2^3
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  6. #6
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    Quote Originally Posted by VonNemo19 View Post
    The answer is whaterver you get for (t^2+\ln{t})\Big|_2^3
    when you substitute can multiplying the denominator by 1/4 and outside the integral by 4 allow you to substitute in du?

    i dont understand how you got that
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  7. #7
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    It is not necessary to use any substitution. Just take what VonNemo19 gave you and simplify and you get something you can integrate.
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by rain21 View Post
    when you substitute can multiplying the denominator by 1/4 and outside the integral by 4 allow you to substitute in du?

    i dont understand how you got that
    Let's start here

    \int_2^3\frac{\sqrt{(2t^2+1)^2}}{t}dt=\int_2^3\fra  c{2t^2+1}{t}dt=\int_2^3(2t+\frac{1}{t})dt=(t^2+\ln  {t})\Big|_2^3
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  9. #9
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    Quote Originally Posted by VonNemo19 View Post
    Let's start here

    \int_2^3\frac{\sqrt{(2t^2+1)^2}}{t}dt=\int_2^3\fra  c{2t^2+1}{t}dt=\int_2^3(2t+\frac{1}{t})dt=(t^2+\ln  {t})\Big|_2^3
    Wow. I feel really dumb. Time to go practice Calc 1. Thanks so much.
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by rain21 View Post
    Wow. I feel really dumb. Time to go practice Calc 1. Thanks so much.
    No problem. BTW, you're not dumb.
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