AP Calculus BC - Line Integrals of Vector Fields

Find the work done by a particle moving in a force field , moving from to on the path given by .

Explanation

The formula for work is given by

$W = \int_C F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) dt$ .

Writing our path in parametric equation form, we have

Hence

$\mathbf{r}(t) = <t,t^2>, F(\mathbf{r}(t))=<t^2,t^4>, \mathbf{r}'(t)= <1,2t>$

Plugging this into our work equation, we get

$W = \int_C F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) dt$

$= \int_{2}^{5}<t^2,t^4> \cdot <1,2t> dt$

$= \int_2^5 t^2+2t^5 dt$ .

Evaluate the integral $\int \vec{F}\cdot d\vec{r}$ on the curve $\vec{r}(t)=\left \langle t^3, t^2, t+2\right \rangle$ , where $\vec{F}=\left \langle xz, yz, xy\right \rangle$ , on the interval $0 \leq t \leq 1$

$\frac{629}{210}$

$\frac{419}{210}$

$\frac{629}{105}$

Explanation

The line integral of the vector field is equal to

$\int_{0}^{1}[\vec{F}(\vec{r}(t))\cdot \vec{r'}(t)]dt$

The parameterization (using the corresponding elements of the curve) of the vector field is

$\vec{F}(\vec{r}(t))= \left \langle t^4+2t^3, t^3+2t^2, t^5\right \rangle$

The derivative of the parametric curve is

$\vec{r'}(t)=\left \langle 3t^2, 2t, 1 \right \rangle$

Taking the dot product of the two vectors, we get

$\vec{F}(\vec{r}(t))\cdot \vec{r'(t)}= 3t^6+6t^5+2t^4+4t^3+t^5= 3t^6+7t^5+2t^4+4t^3$

Integrating this with respect to t on the given interval, we get

$\int_{0}^{1}(3t^6+7t^5+2t^4+4t^3)dt=\frac{3t^3}{7}+\frac{7t^6}{6}+\frac{2t^5}{5}+t^4\left.\begin{matrix} 1\0 \end{matrix}\right|= \frac{629}{210}$

Evaluate $\int \vec{F}\cdot d\vec{r}$ on the curve $\vec{r}(t)=\left \langle 2t, 3t, t^3\right \rangle$ , $0\leq t \leq 1$ , where $\vec{F}=\left \langle 2xz, y^2, z^2\right \rangle$ .

$\frac{164}{15}$

$\frac{32}{6}$

$\frac{32}{3}$

Explanation

The line integral of a vector field is given by

$\int \vec{F}\cdot d\vec{r} = \int [\vec{F}(\vec{r}(t))\cdot \vec{r'}(t)]dt$

So, we must evaluate the vector field on the curve:

$\vec{F}(\vec{r}(t))=\left \langle 2(2t)(t^3), 9t^2, t^6\right \rangle$

Then, we take the derivative of the curve with respect to t:

$\vec{r'}(t)=\left \langle 2, 3, 3t^2\right \rangle$

Taking the dot product of these two vectors, we get

This is the integrand of our integral. Integrating, we get

$\int_{0}^{1}(8t^4+27t^2+3t^8 )dt = \frac{8t^5}{5}+\frac{27t^3}{3}+\frac{3t^9}{9}\Big|^{1}_{0}= \frac{164}{15}$

Calculate $\int \vec{F}\cdot d\vec{r}$ on the interval $0\leq t \leq1$ , where $\vec{F}=\left \langle x, xy, z^2\right \rangle$ and $\vec{r}(t)=\left \langle 2t^2, t+2, t\right \rangle$

$\frac{25}{6}$

$\frac{25}{3}$

$\frac{5}{3}$

Explanation

To calculate the line integral of the vector field, we must evaluate the vector field on the curve, take the derivative of the curve, and integrate the dot product on the given interval.

The vector field evaluated on the given curve is

$\vec{F}(\vec{r}(t))=\left \langle 2t^2, 2t^3+4t^2, t^2\right \rangle$

The derivative of the curve is given by

$\vec{r'}(t)=\left \langle 4t, 1, 1\right \rangle$

The dot product of these is

$\vec{F}(\vec{r(t)})\cdot \vec{r'}(t)=11t^3+5t^2$

Integrating this over our given t interval, we get

$\int_{0}^{1}(10t^3+5t^2)dt=\frac{5t^4}{2}+\frac{5t^3}{3}\Big|^{1}_{0}= \frac{25}{6}$

Evaluate $\int_C \vec{F}\cdot d\vec{r}$ , where $\vec{F}(x,y,z)=10x^2y^2z^3\vec{i}+5z\vec{j}+2yz\vec{k}$ , and is the curve given by $\vec{r}(t)=t\vec{i}+t^2\vec{j}+t^3\vec{k}$ , $0\leq t \leq 1$ .