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V125BC [204]
3 years ago
14

Find the work required to move an object in the force field F = e^x + y (1, 1, z) along the straight line from A(0, 0, 0) to B(-

4, 5, -5). Check to see if the force is conservative. Select the correct choice below and fill in the answer box to complete your choice. (Type an exact answer.) A. The force is not conservative. The work is ___. B. The force is conservative. The work is _____ .
Mathematics
1 answer:
DENIUS [597]3 years ago
8 0

\vec F(x,y,z)=(e^x+y)(1,1,z)

is conservative if we can find a scalar function f such that \nabla f=\vec F. This would require

f_x=e^x+y

f_y=e^x+y

f_z=(e^x+y)z

Integrating both sides of the first equation wrt x gives

f(x,y,z)=e^x+xy+g(y,z)

Differentiating both sides of this wrt y gives

f_y=x+g_y=1\implies g_y=1-x\implies g(y,z)=y-xy+h(z)

but we assumed g was a function of y and z, independent of x. So there is no such f and \vec F is not conservative.

To find the work, first parameterize the path (call it C) by

\vec r(t)=(1-t)(0,0,0)+t(-4,5,-5)=(-4t,5t,-5t)

for 0\le t\le1. Then

\vec r'(t)=(-4,5,-5)

and the work is given by the line integral,

W=\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_0^1(e^{-4t}+5t,e^{-4t}+5t,-(e^{-4t}+5t)5t)\cdot(-4,5,-5)\,\mathrm dt

W=\displaystyle\int_0^1(125t^2+5t+(25t+1)e^{-4t})\,\mathrm dt=\boxed{\frac{2207}{48}-\frac{129}{16e^4}}

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Kobotan [32]

The volume of the region R bounded by the x-axis is: \mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}

<h3>What is the volume of the solid (R) on the X-axis?</h3>

If the axis of revolution is the boundary of the plane region and the cross-sections are parallel to the line of revolution, we may use the polar coordinate approach to calculate the volume of the solid.

From the given graph:

The given straight line passes through two points (0,0) and (2,8). Thus, the equation of the straight line becomes:

\mathbf{y-y_1 = \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)}

here:

  • (x₁, y₁) and (x₂, y₂) are two points on the straight line

Suppose we assign (x₁, y₁) = (0, 0) and (x₂, y₂) = (2, 8)  from the graph, we have:

\mathbf{y-0 = \dfrac{8-0}{2-0}(x-0)}

y = 4x

Now, our region bounded by the three lines are:

  • y = 0
  • x = 2
  • y = 4x

Similarly, the change in polar coordinates is:

  • x = rcosθ,
  • y = rsinθ

where;

  • x² + y² = r²  and dA = rdrdθ

Therefore;

  • rsinθ = 0   i.e.  r = 0 or θ = 0
  • rcosθ = 2 i.e.   r  = 2/cosθ
  • rsinθ = 4(rcosθ)  ⇒ tan θ = 4;  θ = tan⁻¹ (4)

  • ⇒ r = 0   to   r = 2/cosθ
  •    θ = 0  to    θ = tan⁻¹ (4)

Then:

\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^2 (rdr d\theta )}

\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}

Learn more about the determining the volume of solids bounded by region R here:

brainly.com/question/14393123

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3 0
2 years ago
Write the slope intercept form of a line through (6, -7) and parallel to y= x - 9
AveGali [126]

Answer:

y = x - 13

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The equation of a line in slope- intercept form is

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y = x - 9 ← is in slope- intercept form

with slope m = 1

Parallel lines have equal slopes, thus

y = x + c ← is the partial equation of the parallel line

To find c substitute (6, - 7) into the partial equation

- 7 = 6 + c ⇒ c = - 7 - 6 = - 13

y = x - 13 ← equation of parallel line

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The letter P must be in the middle since this location is the vertex. Something like angle BPA is the same as angle APB. We can swap the order of the outer letters without any change to the angle itself. Effectively, this means each angle has two possible names.

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