QuestionEvaluate $\int F \cdot d r$ for $\bar{F}=x y \bar{i}-z j+x^{2} \bar{k}$ along the curve $x=t^{2}$ , $y=2t$ , $z=t^{3}$ from $t=0$ to $t=1$ . Answer: $\frac{51}{70}$ .

Studdy Solution

STEP 1

Assumptions1. The vector field is given by $\bar{}=x y \bar{i}-z \bar{j}+x^{} \bar{k}$ . . The $c$ is parameterized by $x=t^{}$ , $y=t$ , $z=t^{3}$ for $t$ in the interval [0,1].
3. We are asked to evaluate the line integral of the vector field $\bar{}$ along the curve $c$ .

STEP 2

First, we need to find the differential $d\bar{r}$ along the curve $c$ . The differential $d\bar{r}$ is given by the derivative of the position vector $\bar{r}(t)$ with respect to $t$ , multiplied by $dt$ .
$d\bar{r} = \frac{d\bar{r}}{dt} dt$

STEP 3

The position vector $\bar{r}(t)$ is given by the parameterization of the curve $c$ .
$\bar{r}(t) = x(t)\bar{i} + y(t)\bar{j} + z(t)\bar{k} = t^{2}\bar{i} +2t\bar{j} + t^{3}\bar{k}$

STEP 4

Now, differentiate the position vector $\bar{r}(t)$ with respect to $t$ .
$\frac{d\bar{r}}{dt} = \frac{dx}{dt}\bar{i} + \frac{dy}{dt}\bar{j} + \frac{dz}{dt}\bar{k} =2t\bar{i} +2\bar{j} +3t^{2}\bar{k}$

STEP 5

Multiply the derivative of the position vector by $dt$ to obtain $d\bar{r}$ .
$d\bar{r} =2t dt\bar{i} +2 dt\bar{j} +3t^{2} dt\bar{k}$

STEP 6

Now, substitute the parameterization of the curve $c$ into the vector field $\bar{}$ .
$\bar{}(t) = x(t)y(t)\bar{i} - z(t)\bar{j} + x(t)^{2}\bar{k} = t^{2}(2t)\bar{i} - t^{3}\bar{j} + (t^{2})^{2}\bar{k} =2t^{3}\bar{i} - t^{3}\bar{j} + t^{4}\bar{k}$

STEP 7

The line integral of the vector field $\bar{}$ along the curve $c$ is given by the dot product of $\bar{}$ and $d\bar{r}$ , integrated over the interval [0,1].
$\int_{c} \bar{} \cdot d\bar{r} = \int_{0}^{1} \bar{}(t) \cdot d\bar{r} dt$

STEP 8

Substitute the expressions for $\bar{}(t)$ and $d\bar{r}$ into the integral.
$\int_{0}^{1} (2t^{3}\bar{i} - t^{3}\bar{j} + t^{4}\bar{k}) \cdot (2t dt\bar{i} +2 dt\bar{j} +3t^{2} dt\bar{k}) dt$

STEP 9

Calculate the dot product inside the integral.
$\int_{}^{} (4t^{4} -2t^{3} +3t^{6}) dt$

STEP 10

Now, evaluate the integral.
$\int_{0}^{} (4t^{4} -2t^{3} +3t^{6}) dt = \left[\frac{4}{5}t^{5} - \frac{}{2}t^{4} + \frac{3}{7}t^{7}\right]_{0}^{} = \frac{4}{5} - \frac{}{2} + \frac{3}{7} = \frac{51}{70}$ The value of the line integral of the vector field $\bar{}$ along the curve $c$ is $\frac{51}{70}$ .

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