Finding the bounds of integration for a vector valued function

Question

I am trying to find the arc length of $r(t) = (\sqrt{2}t)i + (\sqrt{2}t)j + (1-t^{2})k$ from $(0, 0, 1)$ to $(\sqrt{2}, \sqrt{2}, 0)$. I know how to find the integral except I am not sure what the bounds of integration should be.

I know $t$ can't be negative so $t \ge 0$, but I'm not sure what the upper bound should be. My book says it should be $1$, so is does that come from the $k$ component limiting the function first because it is the smallest component of both points? I'm not sure how to do this in three dimensions.

For which $t$ is $r(t) = (0,0,1)$? For which $t$ is $r(t) = (\sqrt{2},\sqrt{2},0)$? Each of those questions is three equations for a single unknown. — Matthew Leingang, Mar 13 '18 at 18:16

hamam_Abdallah · Accepted Answer · 2018-03-13T19:59:00.620

1

$$x (t)=y (t)=t\sqrt {2}\;\;,\;z (t)=1-t^2$$

$t=0$ gives the point $(0,0,1) $

and $t=1$ gives $(\sqrt {2},\sqrt {2},0) $.

the length is $$L=\int_0^1\sqrt {x'(t)^2+y'(t)^2+z'(t)^2}dt $$ $$=\int_0^1\sqrt {2+2+4t^2}dt $$ $$=2\int_0^1\sqrt {1+t^2}dt $$ put $t=\sinh (u)\;\;,\; dt=\cosh(u)du $

use $1+\sinh (u)^2=\cosh(u)^2$ and $$\cosh (u)^2=\frac {1+\cosh (2u)}{2} $$

edited Mar 13 '18 at 19:59

answered Mar 13 '18 at 18:39

hamam_Abdallah

62,951

You are right. I mixed up the initial function. Fixed now – dumbitdownjr Mar 13 '18 at 19:28
@dumbitdownjr Can you finish it. – hamam_Abdallah Mar 13 '18 at 20:12
I think so, although I didn't use the same substitution that you chose. I used $u = tant$. – dumbitdownjr Mar 14 '18 at 01:28

Finding the bounds of integration for a vector valued function

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