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jarptica [38.1K]
3 years ago
14

If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

{j} \left( 1 2 \sin^{3} t \right \mathbf{k}, 0 \leq t \leq \frac{\pi}{2} and f is the radial vector field \mathbf{f} \left( x, y, z \right = x \mathbf{i} y \mathbf{j} z \mathbf{k}, compute the work done by f on a particle moving along c
Mathematics
1 answer:
jonny [76]3 years ago
6 0
With the curve C parameterized by

C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k

with 0\le t\le\dfrac\pi2, and given the vector field

\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k

the work done by \mathbf f on a particle moving on along C is given by the line integral

\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt

where

\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt

The integral is then

\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt
=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t
=269
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Is (-4,4) the solution of the system. 4x+5y>-6 and -2x+7y>20
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Answer:

  It is <em>a</em> solution, one of an infinite number of solutions.

Step-by-step explanation:

You can check to see if the given point is in the solution set:

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A student has a savings account earning 3% simple interest. She must pay $1200 for first-semester tuition by September 1 and $12
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<h3>Simple Interest</h3>

Simple interest is used when there is a single compounding per time period.

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She needs to earn $2,391.07 during the summer.

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1 year ago
2(x−23)=0 plz answer
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Your answer is A. 6b+5
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3 years ago
Read 2 more answers
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