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Korolek [52]
3 years ago
14

Write the equation of the line parallel to the line 3x + 5y = 7 and passes through the point (-1, 3) in standard form. Please he

lp
Mathematics
1 answer:
makkiz [27]3 years ago
7 0

Answer:

<h2>          3x + 5y = 12   </h2>

Step-by-step explanation:

3x + 5y = 7         {subtract 3x from both sides}

5y = -3x + 7        {divide both sides by 5}

y = -³/₅x + ⁷/₅        ←  slope-intercept form

y=m₁x+b₁   ║   y=m₂x+b₂   ⇔    m₁ = m₂

{Two lines are parallel if  their slopes are equal}

y = -³/₅x + ⁷/₅     ⇒   m₁ = -³/₅    ⇒    m₂ = -³/₅

(-1, 3)  ⇒  x₁ = -1,  y₁ = 3

The point-slope form:

y - 3 = -³/₅(x - (-1))

y - 3 = -³/₅(x + 1)           {myltiply both sides by 5}

5y - 15 = -3x - 3           {add 3x to both sides}

3x + 5y - 15 = - 3        {add 15 to both sides}

<u>3x + 5y = 12</u>

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Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find
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Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk x^2+y^2\le3, I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere x^2+y^2+z^2=3.

a. Let C denote the hemispherical <u>c</u>ap z=\sqrt{3-x^2-y^2}, parameterized by

\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k

with 0\le u\le2\pi and 0\le v\le\frac\pi2. Take the normal vector to C to be

\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k

Then the upward flux of \vec F=(2z+2)\,\vec k through C is

\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^{\pi/2}((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm dv\,\mathrm du

\displaystyle=3\int_0^{2\pi}\int_0^{\pi/2}\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du

=\boxed{2(3+2\sqrt3)\pi}

b. Let D be the disk that closes off the hemisphere C, parameterized by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath

with 0\le u\le\sqrt3 and 0\le v\le2\pi. Take the normal to D to be

\vec s_v\times\vec s_u=-u\,\vec k

Then the downward flux of \vec F through D is

\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv

=\boxed{-6\pi}

c. The net flux is then \boxed{4\sqrt3\pi}.

d. By the divergence theorem, the flux of \vec F across the closed hemisphere H with boundary C\cup D is equal to the integral of \mathrm{div}\vec F over its interior:

\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV

We have

\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2

so the volume integral is

2\displaystyle\iiint_H\mathrm dV

which is 2 times the volume of the hemisphere H, so that the net flux is \boxed{4\sqrt3\pi}. Just to confirm, we could compute the integral in spherical coordinates:

\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi

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