# Thread: Find the length of the curve r(t) - need help integrating

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

Find the length of the curve $r(t)= , 2\leq t \leq 3$

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.

2. Originally Posted by rain21
Find the length of the curve $r(t)= , 2\leq t \leq 3$

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?

3. Originally Posted by rain21
Find the length of the curve $r(t)= , 2\leq t \leq 3$

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

4. Originally Posted by VonNemo19
$\sqrt{\frac{1}{t^2}+4+4t^2}=\sqrt{\frac{4t^4+4t^2+ 1}{t^2}}=\frac{\sqrt{(2t^2+1)^2}}{t}$. Can you finish?

5. Originally Posted by rain21
The answer is whaterver you get for $(t^2+\ln{t})\Big|_2^3$

6. Originally Posted by VonNemo19
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

7. It is not necessary to use any substitution. Just take what VonNemo19 gave you and simplify and you get something you can integrate.

8. Originally Posted by rain21
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

$\int_2^3\frac{\sqrt{(2t^2+1)^2}}{t}dt=\int_2^3\fra c{2t^2+1}{t}dt=\int_2^3(2t+\frac{1}{t})dt=(t^2+\ln {t})\Big|_2^3$

9. Originally Posted by VonNemo19
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.

10. Originally Posted by rain21
Wow. I feel really dumb. Time to go practice Calc 1. Thanks so much.
No problem. BTW, you're not dumb.