Question

3. (12 points) Consider the curve C defined by r(t) = (4 sint, -4 cost,0) with t€ (0,2) (a) Compute the length of the curve C. (b) Parametrize fit) with respect to are length measured from t = 0. (c) Determine the curvature of C.

Answer 1

PART A

The curve C is defined by the vector r(t) which is:

r(t) =< 4sint, -4cost, 0 >

To compute the lenght of the curve C, use the following line integral:

f(x(t), y(t), z(t)) V [r" (t)]” + [v(t)]? + [z' (t)’dt

Determine the first derivative of r(t), like follows:

(t) = 0

By substituting the corresponding values, the line integral is:

$\int_{0}^{2\pi}<4sint,-4cost,0>\sqrt{\left [ 4cost\right ]^2+\left [ 4sint \right ]^2+\left [ 0 \right ]^2}dt$

$\int_{0}^{2\pi}<4sint,-4cost,0>\sqrt{16cos^2t+16sin^2t } \ dt$$\int_{0}^{2\pi}<4sint,-4cost,0>\sqrt{16(cos^2t+sin^2t) } \ dt$

Note: Remember the following trigonometric identity.

sin(2) + cos (2) = 1

So,

$\int_{0}^{2\pi}<4sint,-4cost,0>4 \ dt$

$\int_{0}^{2\pi}16sint-16cost \ dt$

$\left [-16cost-16sint \right ]_{0}^{2\pi}$

$= (-16cos(2\pi)-16sin(2\pi))-(-16cos(0)-16sin(0))$

Therefore, the line integral for the function r(t) equals cero.

$\mathbf{\int_{0}^{2\pi}<4sint,-4cost,0>4 \ dt = 0}$

PART B

To parametrize the curve C with respect to are length, follow the steps:

r(t) =< 4sint, -4cost, 0 >

$\vec{r^{'}}(t)=<4cost,4sint,0>$

Calculate the magnitude of r'(t) like follows:

$|\vec{r^{'}}(t)|=\sqrt{\left [ 4cost\right ]^2+\left [ 4sint \right ]^2+\left [ 0 \right ]^2}$

$|\vec{r^{'}}(t)|=\sqrt{16(cos^2(t)+sin^2(t))}= \sqrt{16}=4$

$s(t)=\int_{0}^{t}4dt= 4\left [t \right ]_{0}^{t}$

$s(t)=4t$

Now, solve for t in terms of s.

$t= \frac{s}{4}$

Finally, substitute the value of t in the vector r(t).

$\mathbf{\vec{r}(t)=<4sin(\frac{s}{4}),-4cos(\frac{s}{4}),0>}$

PART C

In order to determine the curvature of the function use the following expression:

$k(t)= \frac{|T^{'}(t)|}{|r^{'}(t)|}$

Where,

$T(t)= \frac{r^{'}(t)}{|r^{'}(t)|}$

Therefore, first calculate the value of T(t):

$T(t)= \frac{<4sint,-4cost,0>}{4} = <sint,-cost,0>$

Now, the first derivate of T(t) is:

Then, determine the maginitude of T'(t):

$|T^{'}(t)|=\sqrt{cos^2(t)+sin^2(t)}= \sqrt{1}=1$

Finally, the value of the curvature is:

$\mathbf{k(t)= \frac{1}{4}}$