Consider the vector-valued function r(f) = e^-t i+ e^-t cos t j + e^-t sin t k.
a) Find r'(t) and |r'(t)|
b) Evaluate integrate x^2 ds, where C is the graph of r(t), 0

Question

Consider the vector-valued function r(f) = e^-t i+ e^-t cos t j + e^-t sin t k.
a) Find r'(t) and |r'(t)|
b) Evaluate integrate x^2 ds, where C is the graph of r(t), 0 < t <ln3.

eliesaab · Answer

To find r'(t), find the derivative of each componnent

eliesaab · Answer

$$
r'(t)=\left\{-e^{-t},-e^{-t} \sin (t)-e^{-t} \cos (t),e^{-t} \cos (t)-e^{-t} \sin
   (t)\right\}
$$

eliesaab · Answer

$|r'(t)|$ is the length of the vector $ r'(t)$

eliesaab · Answer

You find
$$
|r'(t)|=\sqrt{e^{-2 t}+\left(-e^{-t} \sin (t)-e^{-t} \cos (t)\right)^2+\left(e^{-t} \cos
   (t)-e^{-t} \sin (t)\right)^2}
$$

eliesaab · Answer

Expand the above and simplify

eliesaab · Answer

You get
$$
|r'(t)|=\sqrt{3} \sqrt{e^{-2 t}}=\sqrt{3} e^{-t}
$$
You hsve to fill in the details

eliesaab · Answer

You need now to compute
$$
\int_0^{\ln 3} x^2 ds=\int_0^{\ln 3}  x(t)^2 |r'(t)| dt
$$
Fill in the detaila

eliesaab · Answer

details

eliesaab · Answer

$$
x(t)=e^{-t}
$$

eliesaab · Answer

The rest is easy