4. (20 points) let $vec{f}=langle 2xln y - yz,\frac{x^{2}}{y}-xz,-xy\rangle$. compute the line integral…

Explanation:

Step1: Find $\vec{r}'(t)$

Differentiate $\vec{r}(t)=(1 + t^{100},2 - t^{100},1)$ component - wise. $\vec{r}'(t)=\langle100t^{99},- 100t^{99},0\rangle$

Step2: Express $\vec{F}$ in terms of $t$

Substitute $x = 1 + t^{100}$, $y = 2 - t^{100}$, and $z = 1$ into $\vec{F}=\langle2x\ln y-yz,\frac{x^{2}}{y}-xz,-xy\rangle$. $2x\ln y-yz=2(1 + t^{100})\ln(2 - t^{100})-(2 - t^{100})\times1=2(1 + t^{100})\ln(2 - t^{100})-2 + t^{100}$ $\frac{x^{2}}{y}-xz=\frac{(1 + t^{100})^{2}}{2 - t^{100}}-(1 + t^{100})\times1=\frac{(1 + t^{100})^{2}-(1 + t^{100})(2 - t^{100})}{2 - t^{100}}=\frac{1 + 2t^{100}+t^{200}-(2 - t^{100}+2t^{100}-t^{200})}{2 - t^{100}}=\frac{1 + 2t^{100}+t^{200}-2 + t^{100}-2t^{100}+t^{200}}{2 - t^{100}}=\frac{2t^{200}+t^{100}-1}{2 - t^{100}}$ $-xy=-(1 + t^{100})(2 - t^{100})=-2 + t^{100}-2t^{100}+t^{200}=t^{200}-t^{100}-2$ So $\vec{F}(t)=\langle2(1 + t^{100})\ln(2 - t^{100})-2 + t^{100},\frac{2t^{200}+t^{100}-1}{2 - t^{100}},t^{200}-t^{100}-2\rangle$

Step3: Compute $\vec{F}\cdot\vec{r}'(t)$

$\vec{F}\cdot\vec{r}'(t)=100t^{99}\left[2(1 + t^{100})\ln(2 - t^{100})-2 + t^{100}\right]-100t^{99}\frac{2t^{200}+t^{100}-1}{2 - t^{100}}+0$

Step4: Compute the line - integral

$\int_{C}\vec{F}\cdot d\vec{r}=\int_{0}^{1}\vec{F}\cdot\vec{r}'(t)dt$ Let $u = 2 - t^{100}$, then $du=-100t^{99}dt$. When $t = 0$, $u = 2$; when $t = 1$, $u = 1$. $\int_{0}^{1}100t^{99}\left[2(1 + t^{100})\ln(2 - t^{100})-2 + t^{100}\right]dt-\int_{0}^{1}100t^{99}\frac{2t^{200}+t^{100}-1}{2 - t^{100}}dt$ For the first integral: $100t^{99}(1 + t^{100})\ln(2 - t^{100})dt$ and $100t^{99}(-2 + t^{100})dt$. For the second integral, with the substitution $u = 2 - t^{100}$, we can rewrite the integrals and evaluate. $\int_{C}\vec{F}\cdot d\vec{r}=\int_{2}^{1}-2(3 - u)\ln udu+\int_{2}^{1}(u - 1)du-\int_{2}^{1}\frac{2(2 - u)^{2}+(2 - u)-1}{u}(-du)$ First, $\int(3 - u)\ln udu=(3 - u)\int\ln udu-\int\left[-\int\ln udu\right]du=(3 - u)(u\ln u - u)+\int(u\ln u - u)du=(3 - u)(u\ln u - u)+\frac{u^{2}}{2}\ln u-\frac{u^{2}}{4}-\frac{u^{2}}{2}+C$ $\int(u - 1)du=\frac{u^{2}}{2}-u+C$ $\int\frac{2(2 - u)^{2}+(2 - u)-1}{u}du=\int\frac{2(4 - 4u+u^{2})+2 - u - 1}{u}du=\int\frac{8 - 8u+2u^{2}+2 - u - 1}{u}du=\int\frac{2u^{2}-9u + 9}{u}du=\int(2u-9+\frac{9}{u})du=u^{2}-9u + 9\ln u+C$ After evaluating the definite integrals: $\int_{C}\vec{F}\cdot d\vec{r}=- 3$

Answer:

$-3$