Find the length of the curve correct to four decimal places.

Hello, everyone!

My name is Nic, I'm taking Math 230 (Calculus 3) currently and it is not going well. Former math undergraduates have told me Calc 2 is the hardest.....I strongly disagree :(

So, here's where I'm at:

7. Find the length of the curve correct to four decimal places: r(t) = <t^{2}, t^{3}, t^{4}>, 0<=t<=2

Where I'm at:

L = int(a to b){|r'(t)|}

x = t^{2} => 2t

y = t^{3} => 3t^{2}

z = t^{4} => z' = 4t^{3}

L = [(2t)^{2}+(3t^{2})^{2}+(4t^{3})^{2}]^{0.5}

L = [4t^{2}+9t^{3}+16t^{4}]^{0.5 }Now, I hit a wall. Can anyone assist?

Re: Find the length of the curve correct to four decimal places.

Quote:

Originally Posted by

**nrr5094** Hello, everyone!

My name is Nic, I'm taking Math 230 (Calculus 3) currently and it is not going well. Former math undergraduates have told me Calc 2 is the hardest.....I strongly disagree :(

So, here's where I'm at:

7. Find the length of the curve correct to four decimal places: r(t) = <t^{2}, t^{3}, t^{4}>, 0<=t<=2

Where I'm at:

L = int(a to b){|r'(t)|}

x = t^{2} => 2t

y = t^{3} => 3t^{2}

z = t^{4} => z' = 4t^{3}

L = [(2t)^{2}+(3t^{2})^{2}+(4t^{3})^{2}]^{0.5}

L = [4t^{2}+9t^{3}+16t^{4}]^{0.5 }Now, I hit a wall. Can anyone assist?

Since all powers inside the root are at least 1, and t is non-negative, you can factor: [4t^{2}+9t^{3}+16t^{4}]^{0.5}= [t^{2}(4+9t+16t^{2}]^{.0.5}= t^{2}[4+ 9t+ 16t^{2}]^{0.5}.

Now complete the square inside the square root and you can use that t outside the root as part of the derivative in a substitution.

Re: Find the length of the curve correct to four decimal places.

Quote:

Originally Posted by

**HallsofIvy** [t^{2}(4+9t+16t^{2}]^{.0.5}= t^{2}[4+ 9t+ 16t^{2}]^{0.5}.

One slight correction, I presume it is a typo. The t^2 outside the square root should simply be t, not t^2.

-Dan

Re: Find the length of the curve correct to four decimal places.

Another typo - there is a mistake in the OP's last line with the powers of t - the integral should be:

$\displaystyle L = \int (4t^2 + 9t^4 + 16t^6)^{1/2}dt$

This is a bear of an integral to try to get a closed form solution for. Since the requirement is to calculate to an accuracy of 4 decimal places perhaps a numerical technique would be sufficient.

Re: Find the length of the curve correct to four decimal places.

Thanks, everyone, for the help. I spoke with my professor today and by 'correct to four decimal places' it is implying to use a calculator which can integrate, i.e. i should not solve this by hand because, as you said @ebaines, it is a bear of an integral.

Re: Find the length of the curve correct to four decimal places.

Glad to be of help! I don't have a graphing calculator, but using a simple excel spreadsheet got a result of approximately 18.68. What do you get?