Question

Evaluate the line integral C F · dr, where C is given by the vector function r(t). F(x, y, z) = sin(x) i + cos(y) j + xz k r(t) = t4 i − t3 j + t k, 0 ≤ t ≤ 1

Answer 1

remember that $\int\limits_C {\vec{F} \cdot \vec{dr}} =\int\limits^a_b {\vec{F}(\vec{r}(t)) \cdot \frac{\vec{dr}}{dt}} \, dt$

$\vec{F}(x,y,z)=$

$\vec{r}(t)=$, or $x=t^4$, $y=-t^3$ and $z=t$

parametizing $\vec{F}$

$sin(x)=sin(t^4)$

$cos(y)=cos(-t^3)$

$xz=(t^4)(t)=t^5$

so $\vec{F}(\vec{r}(t))=$

$\frac{\vec{dr}}{dt}=\frac{d}{dt} \vec{r}(t)=$

calculate $\vec{F}(\vec{r}(t)) \cdot \frac{\vec{dr}}{dt}}$

[/tex]<sin(t^4),cos(-t^3),t^5> \cdot <4t^3,-3t^2,1>=[/tex]

$(4t^3sin(t^4))+(-3t^2cos(-t^3))+(1t^5)=$

$4t^3sin(t^4)-3t^2cos(-t^3)+t^5$

so evaluate integral

$\int\limits_C {\vec{F} \cdot \vec{dr}} =\int\limits^a_b {\vec{F}(\vec{r}(t)) \cdot \frac{\vec{dr}}{dt}} \, dt=$

$\int\limits^1_0 {4t^3sin(t^4)-3t^2cos(-t^3)+t^5} \, dt=$ (using u subsitution with u=t^4 and for cosine, u=-t^3)

$[-cos(t^4)+sin(-t^3)+\frac{1}{6}t^6]|^1_0=$

$(-cos(1^4)+sin(-(1)^3)+\frac{1}{6}(1)^6)-(-cos(0^4)+sin(-(0)^3)+\frac{1}{6}(0)^6)=$

$(-cos(1)+sin(-1)+\frac{1}{6})-(-cos(0)+sin(0)+0)=$

$-cos(1)+sin(-1)+\frac{1}{6}+1+0+0=$

$-cos(1)+sin(-1)+\frac{7}{6}$

not sure if 1 is degrees or radians