Question

What is the arc length of function r(t) = 12 years ago

Answer 1

arc length from where to where?

Answer 2

This isn't a calc 2 problem

Answer 3

The arc length of a vector valued function is defined to be:\[\large L\qquad=\qquad \int\limits |r'(t)|dt\]

So we start with,
\[\large r(t)\qquad=\qquad \left<e^{t-3},\;12,\;3t^{-1}\right>\]Taking the derivative of the components gives us,
\[\large r'(t)\qquad=\qquad \left<e^{t-3},\;0,\;-3t^{-2}\right>\]

From here we want to find the magnitude of the derivative function,
\[\large |r'(t)| \qquad=\qquad \sqrt{\left(e^{t-3}\right)^2+\left(-3t^{-2}\right)^2} \qquad=\qquad \sqrt{e^{2t-6}+9t^{-4}}\]

So that's the thing we'd integrate... hmmm..

But as experiment asked, where are we integrating from and to?
You said at t=3, but that doesn't make much sense.

Did you mean from t=0 to t=3?

Answer 4

Thanks to both @experimentX and @zepdrix ! I got it :)

Answer 5

you can express it as a function of t ... if it is what you might mean
\[ \int_t^3 |r'(t)|dt\]