1)Let F =< x^4 + 3 x^2 y + z e ^{2 x} , x 3 + y , x e^{2 x} + c o s ( z ) > be a vector field...

Question:

1)Let {eq}F = < x^4 + 3x^2 y + ze^{2x}, x^3 + y, xe^{2x} + cos(z) > {/eq} be a vector field .

Determine the values of {eq}\int_{r}^{} F\cdot Ds=F(1,2,3)-F(0,0,0) {/eq} =

Vector Field:

Given: {eq}F =< x^4 + 3 x^2 y + z e ^{2 x} , x^ 3 + y , x e^{2 x} + c o s ( z ) > {/eq}

To find: {eq}F(1,2,3)-F(0,0,0) {/eq}

First find {eq}F(1,2,3) {/eq} by putting {eq}x=1\,,\,y=2\,,\,z=3 {/eq}

Find {eq}F(0,0,0) {/eq} by putting {eq}x=0\,,\,y=0\,,\,z=0 {/eq}

Answer and Explanation:

{eq}F =< x^4 + 3 x^2 y + z e ^{2 x} , x ^3 + y , x e^{2 x} + c o s ( z ) > {/eq}

{eq}F(1,2,3) =< 1^4 + 3 (1)^2 (2) + 3 e ^{2 (1)} , 1 ^3 + 2 , (1) e^{2 (1)} + c o s ( 3 ) >=\left \langle 7+3e^2,3,e^2+\cos 3 \right \rangle {/eq}

{eq}F(0,0,0) =< 0 , 0 , 1 > {/eq}

Therefore,

{eq}F(1,2,3)-F(0,0,0)=\left \langle 7+3e^2,3,e^2+\cos 3 \right \rangle-< 0 , 0 , 1 >\\ =\left \langle 7+3e^2,3,e^2+\cos 3-1 \right \rangle {/eq}


Learn more about this topic:

Loading...
Vectors: Definition, Types & Examples

from Common Entrance Test (CET): Study Guide & Syllabus

Chapter 57 / Lesson 3
24K

Related to this Question

Explore our homework questions and answers library