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tamaranim1 [39]
3 years ago
11

A blindfolded contestant makes a random selection from a bag

Mathematics
1 answer:
Genrish500 [490]3 years ago
6 0
I’m also stuck on this question
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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

Then the total work done by \vec F over the particle's path is 46.

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The third picture!! explains it
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The slope of the points is -4x
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159,159,159,160,161,162,166,167,168,192
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Jack buys 4 / 3 pounds of<br> plums. martin buys 6 / 3 pounds of apples. who buys more fruit
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