# Calculate the work done by a particle under the influence of a force x^2 hat(i) + y^2hat(j) + z^2hat(k) along the curve x=cost, y=sint and z=e^(2t) from (1,0,1) to (1,0,4)?

Dec 18, 2016

$\frac{1}{3} \left({e}^{6 \pi} - 3\right)$

#### Explanation:

$F \left(r \left(t\right)\right) = \left({\cos}^{2} t , {\sin}^{2} t , {e}^{4 t}\right)$

$\mathrm{dr} = \left(- \sin t , \cos t , 2 {e}^{2 t}\right) \mathrm{dt}$

$F \left(r \left(t\right)\right) \cdot \mathrm{dr} = \left(- \sin t {\cos}^{2} t + \cos t {\sin}^{2} t + 2 {e}^{6 t}\right) \mathrm{dt}$

${\int}_{t = 0}^{t = \pi} \left(- \sin t {\cos}^{2} t + \cos t {\sin}^{2} t + 2 {e}^{6 t}\right) \mathrm{dt} = \frac{1}{3} \left({e}^{6 \pi} - 3\right)$

Feb 18, 2017

${\int}_{C} \setminus \vec{F} \cdot d \vec{r} = \frac{1}{3} {e}^{6 \pi} - 1 \approx 18829655$

#### Explanation:

The work done in moving a particle from the endpoints $A$ to $B$ along a curve $C$ is.

${\int}_{C} \setminus \vec{F} \cdot d \vec{r} \setminus \setminus$ where $\setminus \setminus \left.\begin{matrix}\vec{F} & = {F}_{1} \hat{i} + {F}_{2} \hat{j} + {F}_{3} \hat{k} \\ d \vec{r} & = \mathrm{dx} \hat{i} + \mathrm{dy} \hat{j} + \mathrm{dz} \hat{k}\end{matrix}\right.$

The integral is known as a line integral.

So we have:

$\vec{F} = {x}^{2} \hat{i} + {y}^{2} \hat{j} + {z}^{2} \hat{k}$

and $C$ is the arc of the parametrised curve:

$x = \cos t \setminus \setminus$, $y = \sin t \setminus \setminus$ and $\setminus \setminus z = {e}^{2 t}$

from $\left(1 , 0 , 1\right)$ to $\left(1 , 0 , 4\right)$

To evaluate the line integral we convert it to a standard integral by choosing an appropriate integration variable, In this case integrating wrt the parameter variable $t$ would seem to make sense.

On $C$, the variable $z$ varies from ${e}^{2 t} = 1 \implies t = 0$ to ${e}^{2 t} = {e}^{2 \pi} \implies t = \pi$, and these values of $t$ are also consistent with the $x$ and $y$ initial and end values. Differentiating the equations for $C$ wrt $t$ gives:

$\frac{\mathrm{dx}}{\mathrm{dx}} = - \sin t \setminus \setminus$; $\frac{\mathrm{dy}}{\mathrm{dt}} = \cos t \setminus \setminus$ and $\setminus \setminus \frac{\mathrm{dz}}{\mathrm{dt}} = 2 {e}^{2 t}$

And so we can express our vector fields in terms of $x$ alone:

$\vec{F} = {x}^{2} \hat{i} + {y}^{2} \hat{j} + {z}^{2} \hat{k}$
$\setminus \setminus \setminus \setminus = {\left(\cos t\right)}^{2} \hat{i} + \left(\sin t\right) \hat{j} + {\left({e}^{2 t}\right)}^{2} \hat{k} \setminus \setminus \setminus \setminus$ (from the equation of $C$)
$\setminus \setminus \setminus \setminus = {\cos}^{2} t \hat{i} + {\sin}^{2} t \hat{j} + {e}^{4 t} \hat{k}$

And:

$d \vec{r} = \mathrm{dx} \hat{i} + \mathrm{dy} \hat{j} + \mathrm{dz} \hat{j}$
$\setminus \setminus \setminus \setminus \setminus = \left(- \sin t \setminus \mathrm{dt}\right) \hat{i} + \left(\cos t \setminus \mathrm{dt}\right) \hat{j} + \left(2 {e}^{2 t} \setminus \mathrm{dt}\right) \hat{k} \setminus \setminus \setminus \setminus$ (from derivatives)
$\setminus \setminus \setminus \setminus \setminus = - \sin t \setminus \mathrm{dt} \hat{i} + \cos t \setminus \mathrm{dt} \hat{j} + 2 {e}^{2 t} \setminus \mathrm{dt} \hat{k}$

Hence,

${\int}_{C} \setminus \vec{F} \cdot d \vec{r} = {\int}_{C} \setminus \left({\cos}^{2} t \hat{i} + {\sin}^{2} t \hat{j} + {e}^{4 t} \hat{k}\right) \cdot \left(- \sin t \setminus \mathrm{dt} \hat{i} + \cos t \setminus \mathrm{dt} \hat{j} + 2 {e}^{2 t} \setminus \mathrm{dt} \hat{k}\right)$
$\text{ } = {\int}_{0}^{\pi} \setminus \left({\cos}^{2} t\right) \left(- \sin t\right) \mathrm{dt} + \left({\sin}^{2} t \cos t\right) \mathrm{dt} + \left({e}^{4 t} 2 {e}^{2 t}\right) \mathrm{dt}$
$\text{ } = {\int}_{0}^{\pi} \setminus - \sin t {\cos}^{2} t + {\sin}^{2} t \cos t + 2 {e}^{6 t} \setminus \mathrm{dt}$
$\text{ } = {\left[\frac{1}{3} {\cos}^{3} t + \frac{1}{3} {\sin}^{3} t + \frac{1}{3} {e}^{6 t}\right]}_{0}^{\pi}$
$\text{ } = \frac{1}{3} {\left[{\cos}^{3} t + {\sin}^{3} t + {e}^{6 t}\right]}_{0}^{\pi}$
$\text{ } = \frac{1}{3} \left\{\left({\cos}^{3} \pi + {\sin}^{3} \pi + {e}^{6 \pi}\right) - \left({\cos}^{3} 0 + {\sin}^{3} 0 + {e}^{0}\right)\right\}$
$\text{ } = \frac{1}{3} \left\{\left(- 1 + 0 + {e}^{6 \pi}\right) - \left(1 + 0 + 1\right)\right\}$
$\text{ } = \frac{1}{3} \left({e}^{6 \pi} - 3\right)$
$\text{ } = \frac{1}{3} {e}^{6 \pi} - 1$
$\text{ } = 18829655.02 \ldots$