Assignment 9 Query

course Mth 174

??????h???????assignment #009?????????????

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Physics II

10-20-2007

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21:43:33

query problem 8.4.3 (8.3.4 3d edition) was 8.2.6) moment of 2 meter rod with density `rho(x) = 2 + 6x g/m

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RESPONSE -->

Not 8.4.3 4th edition, actually 8.4.4 which is based on 8.4.3.

The first part of the problem was to devise a Riemann sum approximating the total mass of the rod. To do this, I took the density and summed small parts of it over the length of the rod. This gave me the sum:

sum( (2 + 6x) 'dx )

I then took the limit as x->0 of the Riemann sum to find the integral of the mass:

int( (2 + 6x) 'dx, x, 0, 2)

To find the value, I simply found the antiderivative:

2x + 3x^2 over the interval 0 to 2. This turned out to be the exact value 16 grams.

Now, to find the moment. The equation to find the momemt is to take the integral and multiply it by the variable:

int( x(2 + 6x) 'dx, x, 0, 2)

I then integrated this over the same period:

x^2 + 2x^3 over 0 to 2 = 20

Then to find the center of mass I divide the moment by the mass:

20 / 16 = 1.25 m

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21:43:50

what is the moment of the rod?

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RESPONSE -->

The moment is 20 grams.

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21:44:07

What integral did you evaluate to get a moment?

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RESPONSE -->

I used the integral:

int( x(2 + 6x) 'dx, x, 0, 2)

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21:53:17

query problem 8.4.12 (was 8.2.12) mass between graph of f(x) and g(x), f > g, density `rho(x)

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RESPONSE -->

I believe the first thing to do is to find the area between the two lines by subtracting the lesser function from the greater. Since the density function varies with respect to x, the simplest thing to do is to slice the graph up with respect to x. Then the area turns into:

( f(x) - g(x) ) 'dx

We then multiply this area by the density to find the mass:

( f(x) - g(x) ) 'rho(x) 'dx

To find the overall mass, I need to take the integral of this over the period a <= x <= b:

int( ( f(x) - g(x) ) 'rho(x) 'dx, x, a, b )

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21:53:39

what is the total mass of the region?

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RESPONSE -->

I believe the total mass is equal to the integral of:

int( ( f(x) - g(x) ) 'rho(x) 'dx, x, a, b )

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21:53:49

What integral did you evaluate to obtain this mass?

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RESPONSE -->

Whoops, see prior answer.

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21:55:15

What is the mass of an increment at x coordinate x with width `dx?

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RESPONSE -->

The mass at coordinate x would be:

( f(x) - g(x) ) 'rho(x) 'dx

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21:56:14

What is the area of the increment, and how do we obtain the expression for the mass from this area?

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RESPONSE -->

The area of the increment would be:

( f(x) - g(x) ) 'dx

To obtain mass from this, we should just multiply this by the density function at that point.

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21:57:19

How to we use the mass of the increment to obtain the integral for the total mass?

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RESPONSE -->

I think we just integrate the mass function over the period a <= x <= b.

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22:21:41

query problem 8.5.13 (8.4.10 3d edition; was 8.3.6) cylinder 20 ft high rad 6 ft full of water

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RESPONSE -->

I believe the first thing I need to do here is to slice up the tank into rings so that each ring has to undergo the same amount of force to be taken out of the tank. Since the slices are circles, the formula is pretty standard:

V = 'pi (6)^2 'dh

Next I need to find the force required to move that slice. Since all the units are in SAE, the units for water density are already in weight:

Force = 62.4 ( 36 'pi ) 'dh

Now I need to find the work that will be done lifting that force over the distance from where it is in the tank to a point 10 feet above the tank. Since the depth of the tank is 20 feet, then the maximum distance that will be moved will be 30 feet. So, in terms of h, the distance moved will be:

30 - h

And:

Work = 62.4 ( 36 'pi ) 'dh (30 - h) = 2246.4 'pi (30 -h) 'dh

Next, to find the overall amount of work done I need to convert this statement into a Riemann sum:

sum( 2246.4 'pi (30 -h) 'dh, i from 1 to n )

As the width of the interval approaches 0, the sum turns into an integral:

int( 2246.4 'pi (30 -h) 'dh, h, 0, 20 )

Evaluating this numerically gives the large number:

2,822,909.5 ft-lbs of work to empty the barrell.

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22:21:53

how much work is required to pump all the water to a height of 10 ft?

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RESPONSE -->

2,822,909.5 ft-lbs of work.

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22:23:05

What integral did you evaluate to determine this work?

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RESPONSE -->

I used the integral:

int( 2246.4 'pi (30 -h) 'dh, h, 0, 20 )

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22:29:02

Approximately how much work is required to pump the water in a slice of thickness `dy near y coordinate y? Describe where y = 0 in relation to the tank.

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RESPONSE -->

The work required to move a slice of thickness 'dy near coordinate y is:

62.4 ( 36 'pi ) (30 - y) 'dy = 2246.4 'pi (30 - y) 'dy

(I previously used h for my variable)

y = 0 at 10 feet above the tank, the point where all the liquid needs to be moved to.

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22:31:41

Explain how your answer to the previous question leads to your integral.

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RESPONSE -->

Basically I took the formula for the work needed to move a tiny slice and summed it over the entire tank region. At that point, I took the limit of the function as h->0 and turned it into an integral.

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23:13:08

query problem 8.3.18 CHANGE TO 8.5.15 work to empty glass (ht 15 cm from apex of cone 10 cm high, top width 10 cm)

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RESPONSE -->

Hmm. This is neither 8.3.15 or 8.5.15. I think I can figure it out with a few assumptions. The first is that the glass is filled with water and the second is that the liquid will be moved 15cm above the top of the glass (moving the liquid at the bottom over a period of 25cm).

Before doing anything, I need to convert all of the measurements into meters:

10cm = 0.1m

15cm = 0.15m

The first thing to do is to slice up the glass. Slicing the cone into circles sounds like a good way to do it. Therefore, the volume for a slice will be:

V = 'pi (w / 2)^2 'dh

Gravitational force will then be:

F = 1000 (9.8) ( 'pi (w / 2)^2 'dh ) = 2450 'pi w^2 'dh

Work = (2450 'pi w^2 'dh) ( 0.25 - h )

Turning this into a Riemann sum:

sum( 2450 'pi w^2 'dh ( 25 - h ) ) over the interval 0 to 0.1.

Taking the limit of h leaves us with the integral:

int( (2450 'pi w^2 'dh)( 0.25 - h ), h, 0, 0.1)

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23:36:16

how much work is required to raise all the drink to a height of 15 cm?

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RESPONSE -->

Whoops, I just realized that I didn't make one part of my steps previously clear (the variable change from w to h):

To find w in terms of h, I can use comparable triangles:

w / h = 0.1 / 0.1

w / h = 1

w = h

My definite integral is then:

int( 2450 'pi h^2 (0.25 - h) 'dh, h, 0, 0.1)

Evaluated numerically: work overall = 0.44898 joules

Excellent solution; however you the apex of the cone is its 'point', which is lowest, so your .25 m should be .15 m.

You worked in Joules; the given solution here is in ergs and the final result is about 2 million ergs, about .2 Joules. The discrepancy is easily accounted for by the different limits on our integrals.

See my plausibility estimate at the bottom; raising the water to the height you use would raise the water an average of about 18 cm, which would by 18/8 = 2.25 times as far as in my estimate, giving you about 2.25 times as much work. The result is very close to what you got.

** The diameter of the top of the cone is equal to the vertical distance y from the apex to the top.

At height y the diameter of the cone is easily seen to be equal to y, so the cross-section at height y has radius .5 y and therefore area A = `pi ( .5 y ) ^ 2 = `pi / 4 * y^2.

A slice of thickness `dy at height y has approximate volume A * `dy = `pi/4 * y^2 * `dy.

This area is in cm^3 so its mass in grams is equal to the volume in cm^3 and the weight in dynes is 980 * mass = 245 `pi y^2 `dy.

This weight is raised from height y to height 15, a distance of 15 - y. So the work to raise the slice is force * distance = 245 `pi y^2 `dy * ( 15 - y ) = 245 ( 15 - y ) `pi y^2 `dy = 245 `pi ( 15 y^2 - y^3 ) `pi `dy

Slices go from y = 0 to y = 10 cm so the integral is 245 `pi ( 15 y^2 - y^3 ) `pi dy, evaluated from y = 0 to y = 10.

We get 245 `pi * (5 y^3 - y^4 / 4) evaluated between 0 and 10.

The result is 245 `pi * 2500 ergs, close to 2 million ergs.

Most of these calculations were done mentally. Check them. However I believe that the process is correct. **

A plausibility calculation follows:

The volume of a cylinder 10 cm high with top width 10 cm would be about 800 cm^3 (pi / 4 * volume of 1000 cm^3 containing cube, rounded to nearest hundredth); the volume of the cone would be 1/3 this, or around 250 cm^3 so the mass would be 250 grams; its weight in dynes would be 980 times this, or about 250,000 dynes.

The water is raised to a height 5 cm above the top of the cone; more of the mass is concentrated near the top of the cone, so we might assume that the mass is raised an ‘average’ of about 8 cm, leading us to the same estimate, about 250,000 dynes * 8 cm = 2 million ergs.

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23:36:24

What integral did you evaluate to determine this work?

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RESPONSE -->

See above.

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23:38:44

Approximately how much work is required to raise the drink in a slice of thickness `dy near y coordinate y? Describe where y = 0 in relation to the tank.

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RESPONSE -->

I used the formula:

2450 'pi h^2 (0.25 - h) 'dh

In relation to the glass, y=0 is the point 15cm above the top of the glass, the height to which the liquid should be raised.

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23:42:27

How much drink is contained in the slice described above?

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RESPONSE -->

The total volume of liquid would be the integral of the volume:

V = 'pi (h / 2)^2 'dh = 'pi / 4 h^2 'dh

int( 'pi / 4 h^2 'dh, h, 0, 0.1) = 2.518x10^(-4)

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23:43:57

What are the cross-sectional area and volume of the slice?

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RESPONSE -->

The area of the slice would be:

'pi h^2

The volume would be:

'pi h^2 'dh

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23:44:33

Explain how your answer to the previous questions lead to your integral.

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RESPONSE -->

Each function builds upon the previous functions until we find one that we can integrate to find whatever we are looking for.

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23:54:28

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

Cooler and cooler. There is some really neat stuff here, as I'd never really thought about dealing with depths, pressures, and other objects with a basic mathematical formula which can then be used to derive other chunks of useful information.

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Assignment 9 Query

You're doing great here. See my notes.