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LiRa [457]
4 years ago
13

Find the work done by force dield f(x,y)=x^2i+ye^xj on a particle that moves along parabola x=y^2+1 from(1,0) to (2,1)

Mathematics
1 answer:
NemiM [27]4 years ago
8 0
Call the parabola P, parameterized by \mathbf r(y)=\langle y^2+1,y\rangle with 0\le y\le 1\rangle. Then the work done by \mathbf f(x,y) along P is

\displaystyle\int_P\mathbf f(x,y)\cdot\mathrm d\mathbf r=\int_{y=0}^{y=1}\mathbf f(x(y),y)\cdot\dfrac{\mathrm d\mathbf r(y)}{\mathrm dy}\,\mathrm dy
=\displaystyle\int_0^1\langle(y^2+1)^2,ye^{y^2+1}\rangle\cdot\langle2y,1\rangle\,\mathrm dy
=\displaystyle\int_0^1(2y^5+4y^3+2y+ye^{y^2+1})\,\mathrm dy=\frac73-\frac e2+\frac{e^2}2
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Step-by-step explanation:

This is because angle two and 6 are corresponding angles. you would then subtract 180 from 45 because angle 6 and 4 are consecutive angles. This would give you 135 degrees. Then angle 1 and angle 4 are verticla agles, making them equal to each other.

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What is the value of this expression? 1/100 + 6/10<br> A.7/10<br> B.7/100<br> C.16/100<br> D.61/100
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Please answer the question below (ABOUT VECTORS AND MAGNITUDE)
coldgirl [10]

(a) <em>v</em> appears to have a fixed direction along the positive <em>x</em>-axis. If ||<em>u</em>|| = 150 N, ||<em>v</em>|| = 220 N, then when <em>θ</em> = 30°, you have

<em>u</em> = (150 N) (cos(30°) <em>i</em> + sin(30°) <em>j</em> ) ≈ (129.904 <em>i</em> + 75 <em>j</em> ) N

<em>v</em> = (220 N) (cos(0°) <em>i</em> + sin(0°) <em>j</em> ) = (220 <em>i</em> ) N

(<em>i</em> and <em>j</em> are the unit vectors in the positive <em>x</em> and <em>y</em> directions)

and their sum is

<em>u</em> + <em>v</em> ≈ (349.904 <em>i</em> + 75 <em>j</em> ) N

with magnitude

||<em>u</em> + <em>v</em>|| ≈ √((349.904)² + (75)²) N ≈ 357.851 N ≈ 357.9 N

and at angle <em>φ</em> made with the positive <em>x</em>-axis such that

tan(<em>φ</em>) ≈ (75 N) / (349.904 N)   →   <em>φ</em> ≈ 12.098° ≈ 12.1°

(b) Letting <em>θ</em> vary from 0° to 180° would make <em>v</em> a function of <em>θ</em> :

<em>u</em> = (150 N) (cos(<em>θ</em>) <em>i</em> + sin(<em>θ</em>) <em>j</em> ) = (150 cos(<em>θ</em>) <em>i</em> + 150 sin(<em>θ</em>) <em>j</em> ) N

Then

<em>u</em> + <em>v</em> = ((220 + 150 cos(<em>θ</em>)) <em>i</em> + (150 sin(<em>θ</em>)) <em>j</em> ) N

→   <em>M</em> = ||<em>u</em> + <em>v</em>|| = √((220 + 150 cos(<em>θ</em>))² + (150 sin(<em>θ</em>))²) N

<em>M</em> = √(48,400 + 66,000 cos(<em>θ</em>) + 22,500 cos²(<em>θ</em>) + 22,500 sin²(<em>θ</em>)) N

<em>M</em> = 10 √(709 + 660 cos(<em>θ</em>)) N

(c) As a function of <em>θ</em>, <em>u</em> + <em>v</em> makes an angle <em>α</em> with the positive <em>x</em>-axis such that

tan(<em>α</em>) = (150 sin(<em>θ</em>) / (220 + 150 cos(<em>θ</em>))

→   <em>α</em> = tan⁻¹((15 sin(<em>θ</em>) / (22 + 15 cos(<em>θ</em>)))

(d) Filling in the table is just a matter of evaluating <em>M</em> and <em>α</em> for each of the given angles <em>θ</em>. For example, when <em>θ</em> = 0°,

<em>M</em> = 10 √(709 + 660 cos(0°)) N = 370 N

<em>α</em> = tan⁻¹((15 sin(0°) / (22 + 15 cos(0°))) = 0°

When <em>θ</em> = 30°, you get the same result as in part (a).

When <em>θ</em> = 60°,

<em>M</em> = 10 √(709 + 660 cos(60°)) N ≈ 323.3 N

<em>α</em> = tan⁻¹((15 sin(60°) / (22 + 15 cos(60°))) = 23.8°

and so on.

4 0
3 years ago
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