Question

Vector Calculus
Find the length of the 3d curve:
r(t) = <8 + 2t, 6 - 3t, -1 +3t>
for 1<= t <=4

Answer 1

hmm, what do you know about the arc length?

Answer 2

\(\Large\rm L=\int_{a}^{b}|\vec{r}(t)|dt\)

Answer 3

So I would take the square root of the sum of each component squared first?

Answer 4

Or would I find the derivative of each component?

Answer 5

you need to find derivative of each component and then square it
the sum of all of them under radical

Answer 6

\[L= \int\limits_{1}^{4}\sqrt{(8t+t^2)^2 + (6t-(3/2)t^2)^2 + (-t+(3/2)t^2)^2}\]

Answer 7

\[L=\int\limits_{1}^{4}\sqrt{64t^2 +16t^3 +t^4 +36t^2 -18t^3 +(9/4)t^4 +t^2 -3t^3 + (9/4)t^4}\]

Answer 8

\[L=\int\limits_{1}^{4}\sqrt{101t^2 -5t^3 +(11/2)t^4}\]

Answer 9

Am I right up to here?

Answer 10

oh no, i said differentiate the components first before you square them?

Answer 11

Aha! I was backwards from the start. Thanks