Particle air resistance in helix curve

Question

A particle moves around Helix curve $x = a\cos t, y=b\sin t, z = ct$

I want to calculate the work what the particle does against air $\mathbf{f}(\mathbf{r}(t)) = -k\mathbf{r'}(t)$

I know that work is $-\int_C \mathbf f\cdot \mathbf {dr}$

So far:

$0 \leq t \leq2\pi $

$x(t) = a\cos t$

$y(t) = b\sin t$

$z(t) = ct$

then:

$\mathbf r(t) = a\cos t \; \mathbf i + b\sin t \; \mathbf j + ct \; \mathbf k$

and the derivate is

$d\mathbf r = {\mathbf r'}(t) \; dt = -a\sin t \; dt \; \mathbf i + b\cos t \; dt \; \mathbf j + cdt \; \mathbf k$

$\int_0^{2\pi}\mathbf f\cdot\mathbf{dr}$

How do I calculate the integral?

Answer 1

Use the expressions below you already had in the attempt, $$\mathbf r' = -a\sin t \ \mathbf i + b\cos t \; \mathbf j + c \mathbf k$$ $$d\mathbf r = \mathbf r' dt$$

to plug into the work integral,

$$-\int_C \mathbf f\cdot d\mathbf {r}=k\int_C \mathbf r'\cdot d\mathbf {r} =k\int_0^{2\pi} [\mathbf r'(t)]^2 dt$$ $$= k\int_0^{2\pi}(a^2\sin^2t+b^2\cos ^2 t+c^2)dt $$

$$=k\pi(a^2+b^2 +2 c^2)$$

Answer 2

The force is a constant multiple of the velocity, thus, the differential of work done is given by $$-\mathbf f\cdot\mathrm d\mathbf r=-(-k\mathbf r')\cdot\mathrm d\mathbf r=k\mathbf r'\cdot \mathrm d\mathbf r.$$

You've already found what $\mathbf r'$ and $\mathrm d\mathbf r$ are, so you're there already. Just substitute and evaluate the integral.