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Nimfa-mama [501]
2 years ago
6

Pls help (links will get reported)

Mathematics
1 answer:
joja [24]2 years ago
3 0

Answer:

-938 ft

Step-by-step explanation:

the bird will be going down in elevation. the distance between -422 and -1360

-1360+422= -938

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Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li
Sophie [7]
Consider the closed region V bounded simultaneously by the paraboloid and plane, jointly denoted S. By the divergence theorem,

\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV

And since we have

\nabla\cdot\mathbf f(x,y,z)=1

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV
=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr
=\dfrac\pi2

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by D, we have

\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS

Parameterize D by

\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k
\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k

which would give a unit normal vector of \mathbf k. However, the divergence theorem requires that the closed surface S be oriented with outward-pointing normal vectors, which means we should instead use \mathbf s_v\times\mathbf s_u=-u\,\mathbf k.

Now,

\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du
=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du
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\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2
6 0
3 years ago
On a morning of a day when the sun will pass directly​ overhead, the shadow of a 60​-ft building on level ground is 25 ft long.
AysviL [449]

Answer:

The shadow is decreasing at the rate of 3.55 inch/min

Step-by-step explanation:

The height of the building = 60ft

The shadow of the building on the level ground is 25ft long

Ѳ is increasing at the rate of 0.24°/min

Using SOHCAHTOA,

Tan Ѳ = opposite/ adjacent

= height of the building / length of the shadow

Tan Ѳ = h/x

X= h/tan Ѳ

Recall that tan Ѳ = sin Ѳ/cos Ѳ

X= h/x (sin Ѳ/cos Ѳ)

Differentiate with respect to t

dx/dt = (-h/sin²Ѳ)dѲ/dt

When x= 25ft

tanѲ = h/x

= 60/25

Ѳ= tan^-1(60/25)

= 67.38°

dѲ/dt= 0.24°/min

Convert the height in ft to inches

1 ft = 12 inches

Therefore, 60ft = 60*12

= 720 inches

Convert degree/min to radian/min

1°= 0.0175radian

Therefore, 0.24° = 0.24 * 0.0175

= 0.0042 radian/min

Recall that

dx/dt = (-h/sin²Ѳ)dѲ/dt

= (-720/sin²(67.38))*0.0042

= (-720/0.8521)*0.0042

-3.55 inch/min

Therefore, the rate at which the length of the shadow of the building decreases is 3.55 inches/min

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