A blindfolded contestant makes a random selection from a bag

Mathematics

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Genrish500 [490]3 years ago

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A particle moves along line segments from the origin to the points (2, 0, 0), (2, 4, 1), (0, 4, 1), and back to the origin under

Shkiper50 [21]

The work is equal to the line integral of $\vec F$ over each line segment.

Parameterize the paths

from (0, 0, 0) to (2, 0, 0) by $\vec r_1(t)=t\,\vec\imath$ with $0\le t\le2$ ,
from (2, 0, 0) to (2, 4, 1) by $\vec r_2(t)=2\,\vec\imath+4t\,\vec\jmath+t\,\vec k$ with $0\le t\le1$ ,
from (2, 4, 1) to (0, 4, 1) by $\vec r_3(t)=(2-t)\,\vec\imath+4\,\vec\jmath+\vec k$ with $0\le t\le2$ , and
from (0, 4, 1) to (0, 0, 0) by $\vec r_4(t)=(4-4t)\,\vec\jmath+(1-t)\,\vec k$ with $0\le t\le1$

The work done by $\vec F$ over each segment (call them $C_1,\ldots,C_4$ ) is

$\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r_1=\int_0^2\vec0\cdot\vec\imath\,\mathrm dt=0$

$\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(t^2\,\vec\imath+24t\,\vec\jmath+32t^2\,\vec k)\cdot(4\,\vec\jmath+\vec k)\,\mathrm dt=\int_0^1(96t+32t^2)\,\mathrm dt=\frac{176}3$

$\displaystyle\int_{C_3}\vec F\cdot\mathrm d\vec r_3=\int_0^2(\vec\imath+(24-12t)\,\vec\jmath+32\,\vec k)\cdot(-\vec\imath)\,\mathrm dr=-\int_0^2\mathrm dt=-2$

$\displaystyle\int_{C_4}\vec F\cdot\mathrm d\vec r_4=\int_0^1((1-t)^2\,\vec\imath+2(4-4t)^2\,\vec k)\cdot(-4\,\vec\jmath-\vec k)\,\mathrm dt=-2\int_0^1(4-4t)^2\,\mathrm dt=-\frac{32}3$