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3 years ago

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When the Olympic Games were held outside Mexico City in 1968, there was much discussion about the effect the high altitude (7340

feet) would have on the athletes. Assuming air pressure decays exponentially by 0.4% every 100 feet, by what percentage is air pressure reduced by moving from sea level to Mexico City?

1 answer:

Artyom0805 [142]3 years ago

7 0

Answer: By moving from sea level to Mexico City air pressure is reduced by 25%.

Step-by-step explanation:

Hi, to answer this we have to apply a decrease rate formula

P0 (1-r)^x

Where:

p0= initial pressure

r = decrease rate. (0.4 = 0.4/100=0.004 decimal form)

x = feet (hundredths, since the air pressure decays every 100 feet we have to divide the input value by 100)

Replacing with the values given

P0 (1- 0.4/100)^(7340/100) =

P0 (0.996)^73.4 =

P0 ( 0.75)=

The air pressure at Mexico City compared to the sea level is 75% (0.75 x 100 =75%)

To obtain the percent decrease

100% -75% =25%

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Step-by-step explanation:

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Name/ Uid:1. In this problem, try to write the equations of the given surface in the specified coordinates.(a) Write an equation

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To find:

(a) Equation for the sphere of radius 5 centered at the origin in cylindrical coordinates

(b) Equation for a cylinder of radius 1 centered at the origin and running parallel to the z-axis in spherical coordinates

Solution:

(a) The equation of a sphere with center at (a, b, c) & having a radius 'p' is given in cartesian coordinates as:

$(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=p^{2}$

Here, it is given that the center of the sphere is at origin, i.e., at (0,0,0) & radius of the sphere is 5. That is, here we have,

$a=b=c=0,p=5$

That is, the equation of the sphere in cartesian coordinates is,

$(x-0)^{2}+(y-0)^{2}+(z-0)^{2}=5^{2}$

$\Rightarrow x^{2}+y^{2}+z^{2}=25$

Now, the cylindrical coordinate system is represented by $(r, \theta,z)$

The relation between cartesian and cylindrical coordinates is given by,

$x=rcos\theta,y=rsin\theta,z=z$

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Thus, the obtained equation of the sphere in cartesian coordinates can be rewritten in cylindrical coordinates as,

$r^{2}+z^{2}=25$

This is the required equation of the given sphere in cylindrical coordinates.

(b) A cylinder is defined by the circle that gives the top and bottom faces or alternatively, the cross section, & it's axis. A cylinder running parallel to the z-axis has an axis that is parallel to the z-axis. The equation of such a cylinder is given by the equation of the circle of cross-section with the assumption that a point in 3 dimension lying on the cylinder has 'x' & 'y' values satisfying the equation of the circle & that 'z' can be any value.

That is, in cartesian coordinates, the equation of a cylinder running parallel to the z-axis having radius 'p' with center at (a, b) is given by,

$(x-a)^{2}+(y-b)^{2}=p^{2}$

Here, it is given that the center is at origin & radius is 1. That is, here, we have, $a=b=0,p=1$ . Then the equation of the cylinder in cartesian coordinates is,

$x^{2}+y^{2}=1$

Now, the spherical coordinate system is represented by $(\rho,\theta,\phi)$

The relation between cartesian and spherical coordinates is given by,

$x=\rho sin\phi cos\theta,y=\rho sin\phi sin\theta, z= \rho cos\phi$

Thus, the equation of the cylinder can be rewritten in spherical coordinates as,

$(\rho sin\phi cos\theta)^{2}+(\rho sin\phi sin\theta)^{2}=1$

$\Rightarrow \rho^{2} sin^{2}\phi cos^{2}\theta+\rho^{2} sin^{2}\phi sin^{2}\theta=1$

$\Rightarrow \rho^{2} sin^{2}\phi (cos^{2}\theta+sin^{2}\theta)=1$

$\Rightarrow \rho^{2} sin^{2}\phi=1$ (As $sin^{2}\theta+cos^{2}\theta=1$ )

Note that $\rho$ represents the distance of a point from the origin, which is always positive. $\phi$ represents the angle made by the line segment joining the point with z-axis. The range of $\phi$ is given as $0\leq \phi\leq \pi$ . We know that in this range the sine function is positive. Thus, we can say that $sin\phi$ is always positive.

Thus, we can square root both sides and only consider the positive root as,

$\Rightarrow \rho sin\phi=1$

This is the required equation of the cylinder in spherical coordinates.

Final answer:

(a) The equation of the given sphere in cylindrical coordinates is $r^{2}+z^{2}=25$

(b) The equation of the given cylinder in spherical coordinates is $\rho sin\phi=1$

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3 years ago