Please answer quickly with little explanation

Change each of the following points from rectangular coordinates to spherical coordinates and to cylindrical coordinates.

FromTheMoon [43]

Answer and Step-by-step explanation: Spherical coordinate describes a location of a point in space: one distance (ρ) and two angles (Ф,θ).To transform cartesian coordinates into spherical coordinates:

$\rho = \sqrt{x^{2}+y^{2}+z^{2}}$

$\phi = cos^{-1}\frac{z}{\rho}$

For angle θ:

If x > 0 and y > 0: $\theta = tan^{-1}\frac{y}{x}$ ;

If x < 0: $\theta = \pi + tan^{-1}\frac{y}{x}$ ;
If x > 0 and y < 0: $\theta = 2\pi + tan^{-1}\frac{y}{x}$ ;

Calculating:

a) (4,2,-4)

$\rho = \sqrt{4^{2}+2^{2}+(-4)^{2}}$ = 6

$\phi = cos^{-1}(\frac{-4}{6})$

$\phi = cos^{-1}(\frac{-2}{3})$

For θ, choose 1st option:

$\theta = tan^{-1}(\frac{2}{4})$

$\theta = tan^{-1}(\frac{1}{2})$

b) (0,8,15)

$\rho = \sqrt{0^{2}+8^{2}+(15)^{2}}$ = 17

$\phi = cos^{-1}(\frac{15}{17})$

$\theta = tan^{-1}\frac{y}{x}$

The angle θ gives a tangent that doesn't exist. Analysing table of sine, cosine and tangent: θ = $\frac{\pi}{2}$

c) (√2,1,1)

$\rho = \sqrt{(\sqrt{2} )^{2}+1^{2}+1^{2}}$ = 2

$\phi = cos^{-1}(\frac{1}{2})$

$\phi$ = $\frac{\pi}{3}$

$\theta = tan^{-1}\frac{1}{\sqrt{2} }$

d) (−2√3,−2,3)

$\rho = \sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}+3^{2}}$ = 5

$\phi = cos^{-1}(\frac{3}{5})$

Since x < 0, use 2nd option:

$\theta = \pi + tan^{-1}\frac{1}{\sqrt{3} }$

$\theta = \pi + \frac{\pi}{6}$

$\theta = \frac{7\pi}{6}$

Cilindrical coordinate describes a 3 dimension space: 2 distances (r and z) and 1 angle (θ). To express cartesian coordinates into cilindrical:

$r=\sqrt{x^{2}+y^{2}}$

Angle θ is the same as spherical coordinate;

z = z

Calculating:

a) (4,2,-4)

$r=\sqrt{4^{2}+2^{2}}$ = $\sqrt{20}$

$\theta = tan^{-1}\frac{1}{2}$

z = -4

b) (0, 8, 15)

$r=\sqrt{0^{2}+8^{2}}$ = 8

$\theta = \frac{\pi}{2}$

z = 15

c) (√2,1,1)

$r=\sqrt{(\sqrt{2} )^{2}+1^{2}}$ = $\sqrt{3}$

$\theta = \frac{\pi}{3}$

z = 1

d) (−2√3,−2,3)

$r=\sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}}$ = 4

$\theta = \frac{7\pi}{6}$

z = 3

5 0

3 years ago

What is the area of the composite figure? _______ cm2 pls and ty

LuckyWell [14K]

The length of line in green = 5 cm
the length of line in yellow = 3 cm
the length of line in red = 6 cm
the length of line in white = 16 cm
the total area = 100 + 12 + 12 + 20 + 24 + 96 = 264 cm2

3 0

3 years ago

Read 2 more answers

How do you find angles on a rhombus

Naily [24]

Answer:

you can try adding the adding angles up to 180

or if its graphed you can count the units

Step-by-step explanation:

and know that all four sides are congruent and diagonals are perpendicular

i tried :)

6 0

2 years ago

18h=252 what is it???

Nadusha1986 [10]

Answer:

h=14

Step-by-step explanation:

18h=252

18h/18=252/18

h=14

5 0

3 years ago

A plane flying horizontally at an altitude of 3 miles and a speed of 500 mi/h passes directly over a radar station. Find the rat

konstantin123 [22]

Answer:

The rate at which the distance from the plane to the station is increasing is 331 miles per hour.

Step-by-step explanation:

We can find the rate at which the distance from the plane to the station is increasing by imaging the formation of a right triangle with the following dimensions:

a: is one side of the triangle = altitude of the plane = 3 miles

b: is the other side of the triangle = the distance traveled by the plane when it is 4 miles away from the station and an altitude of 3 miles

h: is the hypotenuse of the triangle = distance between the plane and the station = 4 miles

First, we need to find b:

$a^{2} + b^{2} = h^{2}$ (1)

$b = \sqrt{h^{2} - a^{2}} = \sqrt{(4 mi)^{2} - (3 mi)^{2}} = \sqrt{7} miles$

Now, to find the rate we need to find the derivative of equation (1) with respect to time:

$\frac{d}{dt}(a^{2}) + \frac{d}{dt}(b^{2}) = \frac{d}{dt}(h^{2})$

$2a\frac{da}{dt} + 2b\frac{db}{dt} = 2h\frac{dh}{dt}$

Since "da/dt" is constant (the altitude of the plane does not change with time), we have:

$0 + 2b\frac{db}{dt} = 2h\frac{dh}{dt}$

And knowing that the plane is moving at a speed of 500 mi/h (db/dt):

$\sqrt{7} mi*500 mi/h = 4 mi*\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{\sqrt{7} mi*500 mi/h}{4 mi} = 331 mi/h$

Therefore, the rate at which the distance from the plane to the station is increasing is 331 miles per hour.

I hope it helps you!

4 0

2 years ago