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iogann1982 [59]
3 years ago
13

Let C = C1 + C2 where C1 is the quarter circle x^2+y^2=4, z=0,from (0,2,0) to (2,0,0), and where C2 is the line segment from (2,

0,0) to (3,3,3). Compute the work done by the force F(x,y,z) =
Mathematics
1 answer:
trapecia [35]3 years ago
5 0
Not much can be done without knowing what \mathbf F(x,y,z) is, but at the least we can set up the integral.

First parameterize the pieces of the contour:

C_1:\mathbf r_1(t_1)=(2\sin t_1,2\cos t_1,0)
C_2:\mathbf r_2(t_2)=(1-t_2)(2,0,0)+t_2(3,3,3)=(2+t_2, 3t_2, 3t_2)

where 0\le t_1\le\dfrac\pi2 and 0\le t_2\le1. You have

\mathrm d\mathbf r_1=(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1
\mathrm d\mathbf r_2=(1,3,3)\,\mathrm dt_2

and so the work is given by the integral

\displaystyle\int_C\mathbf F(x,y,z)\cdot\mathrm d\mathbf r
=\displaystyle\int_0^{\pi/2}\mathbf F(2\sin t_1,2\cos t_1,0)\cdot(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1
{}\displaystyle\,\,\,\,\,\,\,\,+\int_0^1\mathbf F(2+t_2,3t_2,3t_2)\cdot(1,3,3)\,\mathrm dt_2
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3. The backyard of a home is a rectangle 25m by 30m. A garden of uniform width is to be
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Answer:

Part A

The width of the garden is approximately 3.987 meters

Part B

Please find attached the drawing of the solution

Step-by-step explanation:

Part A

The given parameters are;

The dimensions of the backyard of the home = 25 m by 30 m

The width of the garden to be built in the yard = Uniform width

The shape of the grass left inside = Rectangular

The area of the grass at the center = The area of the garden

Let 'x' represent the width of the garden, we have;

The length of the rectangular grass area at the center, l = 30 - 2·x

The width of the rectangular area of grass at the center, w = 25 - 2·x

The area of the rectangular backyard, A = 30 m × 25 m = 750 m²

The area of the rectangular backyard, A = (The area of the garden) + (The area of the rectangle of grass inside)

The area of the rectangle of grass, GA = (30 - 2·x)·(25 - 2·x) = The area of the garden

The area of the rectangular backyard, A = 750 = (30 - 2·x)·(25 - 2·x) + (30 - 2·x)·(25 - 2·x)  = 2 × (30 - 2·x)·(25 - 2·x) = 8·x² - 220·x + 1,500

∴ 750 = 8·x² - 220·x + 1,500

8·x² - 220·x + 1,500 - 750 = 0

8·x² - 220·x + 750 = 0

x = (220 ± √(220² - 4 × 8 × 750))/(2 × 8)

x ≈ 23.513, or x = 3.987

When x = 25.513, the width of the rectangle of grass inside, w = 25 - 2 × 23.513 = -22.026, which is not a natural (physically possible)

Therefore, the possible width of the garden, x ≈ 3.987

Part B

The drawing of the solution created with MS Visio is attached

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Step-by-step explanation:


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