/sci/ - Science & Math

>>11446836
[math] \mathbf{E}=(3xz,2xy,-x^2)\ \ ;\ \ A=(1,-1,1)\ \ ;\ \ B=(3,2,-5) [/math]. So then you parameter the path from A to B as [math] \mathbf{r}(t)=(1+2t,-1+3t,1-6t) [/math]. Notice that r is at A for t=0, and B for t=1. Then [eqn]\Delta V=-\int\mathbf{E}(\mathbf{r})\cdot\text{d}\mathbf{r}=-\int_0^1\mathbf{E}(\mathbf{r})\cdot \mathbf{r}'\text{ d}t=-\int_0^1 \begin{pmatrix} 3(1+2t)(1-6t)\\ 2(1+2t)(-1+3t)\\ -(1+2t)^2\end{pmatrix}\cdot\begin{pmatrix}2\\ 3\\ -6\end{pmatrix}\text{d}t=-5\ \text{volts} [/eqn]
I may have made a small error somewhere, but this is definitely the correct procedure. You should be able to fill in the (small) gaps.

>>11446836
[math] \mathbf{E}=(3xz,2xy,-x^2)\ \ ;\ \ A=(1,-1,1)\ \ ;\ \ B=(3,2,-5) [/math]. So then you parameter the path from A to B as [math] \mathbf{r}(t)=(1+2t,-1+3t,1-6t) [/math]. Notice that r is at A for t=0, and B for t=1. Then [eqn]\Delta V=-\int\mathbf{E}(\mathbf{r})\cdot\text{d}\mathbf{r}=-\int_0^1\mathbf{E}(\mathbf{r})\cdot \mathbf{r}'\text{ d}t=-\int_0^1 \begin{pmatrix} 3(1+2t)(1-6t)\\ 2(1+2t)(-1+3t)\\ -(-1+3t)^2\end{pmatrix}\cdot\begin{pmatrix}2\\ 3\\ -6\end{pmatrix}\text{d}t=-(-15)=15\ \text{volts} [/eqn]
I may have well made a simple mistake, but this is definitely the proper procedure. You should be able to fill in the (small) gaps.