Assignment 3 Query

course Mth 174

waǐy{eĖassignment #003

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Physics II

09-17-2007

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18:11:00

query 6.5 #8 Galileo: time for unif accel object to traverse dist is same as if vel was ave of init and final; put into symbols and show why true

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RESPONSE -->

a) To find the statement in symbols, I first found the two equations, one related distance to time and the other relating velocity to time:

d = (1/2) constant * t^2 + C

v = constant * t

Then I solved for t in each equation:

t = 'sqrt( (d - C) / (1/2) constant) )

t = (v - v0) / constant

The constant term here is the constant acceleration due to whatever, such as gravity. C is another constant such as height when dropped. V is simply the velocity.

b) Solving the distance formula for the time it took an item dropped from a 100 ft building to reach the ground gave me this result:

t = 'sqrt( -100 / -16 ) = 2.5

Putting t=2.5 into the normal distance formula gave a distance of 0.

In the velocity formula, the result was -80 ft/s. Since I needed to find the mean value, the equation for that is: (0 - 80) / 2 = -40 ft/s.

Also, the acceleration constant here is due to gravity, so it is -32 ft/s.

Now that I have all those variables, I can plug and chug and see if the two equations return the same result:

t(d) = 'sqrt( (0 - 100) / -16 ) = 2.5

t(v) = -40 / 32 = 1.25

Hmm. What did I miss?

c) I believe this shows that the time it takes to traverse a particular distance under uniform acceleration is linked only to the acceleration constant, and nothing else.

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18:12:08

how can you symbolically represent the give statement?

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RESPONSE -->

I think:

'sqrt ( (d - C) / (1/2)constant ) = (v - v0) / constant

But it doesn't work out right in the end.

** Using s for the distance fallen we can translate Galileo's statement as follows:

t = s / [ (vf + v0)/2 ].

A ball dropped from rest will have velocity function

v = a * t

and position function

s = .5 a t^2.

Both of these results are easily obtained, for the given initial condition that initial velocity is zero, by a straightforward integration.

For given distance of fall s the position function gives us

time of fall t = sqrt(2 s / a).

At this time the velocity function tells us that

Final velocity = a * time of fall = a sqrt(2 s / a) = sqrt( 2 a s).

From this we calculate the average of initial and final velocities and show that at this velocity the time required to fall distance s is the same as above:

The average of initial and final velocities is

Ave of initial and final velocities = (0 + sqrt( 2 a s) ) / 2 = sqrt( a s / 2 ).

If an object travels distance s at this velocity sqrt( a / (2 s) ) then the time required is

Time required at ave of initial and final velocities

= s / sqrt( a s / 2 ) = sqrt(2 s / a),

Which is in agreement with the result obtained from the position function.

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18:16:41

How can we show that the statement is true?

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RESPONSE -->

The statement would be true if both sides returned the same value, but they don't.

Anyway, I think (to answer the question) that formula relies primarily on the constant acceleration. The two formulas differ only in how they get there, one using distance and the other using velocity.

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18:17:34

How can we use a graph to show that the statement is true?

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RESPONSE -->

Hmm. Interesting. The only thing that comes to mind is that the curve should be the same, independent of the formula used.

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18:26:19

query problem 7.1.22 (3d edition #18) integral of `sqrt(cos(3t) ) * sin(3t)

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RESPONSE -->

I found the integral to be:

- (1/3) (cos(3t))^(3/2)

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18:33:02

what did you get for the integral and how did you reason out your result?

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RESPONSE -->

Whoops, I had the wrong result:

-(2/9)(cos(3t))^(3/2) + C

First I went to the inner function, cos(3t), and set that equal to w. I then found the derivative of w:

dw = -3sin(3t) dt

That fits the equation except for the -3. For a -3 to disappear, it must be cancelled out by -(1/3).

Now we have to consider the antiderivative of 'sqrt(w). Since it is a power function (1/2), the antiderivative must be one great (3/2). Also, power functions multiply the result by the power, so to have a 1 coefficient, we must cancel out that 3/2 with a 2/3.

Result:

-(1/3)*(2/3)(cos(3t))^(3/2) = -(2/9)(cos(3t))^(3/2) + C

very good

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18:34:56

query problem 7.1.20 (3d edition #21) antiderivative of x^2 e^(x^3+1)

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RESPONSE -->

I found the derivative to be:

e^(x^3 + 1) / 3 + c

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18:35:10

what is the antiderivative?

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RESPONSE -->

e^(x^3 + 1) / 3

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18:37:46

What substitution would you use to find this antiderivative?

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RESPONSE -->

I selected the inner function, x^3 +1, and then found the derivative:

dw = 3x^2 dt

This fits the original integral except for the absence of the 3, so it must have been canceled out, leaving an antiderivative of:

e^(x^3 + 1) / 3

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19:52:31

query problem 7.1.37 (3d edition #35) antiderivative of (t+1)^2 / t^2

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RESPONSE -->

I found the antiderivative to be:

t + ln(t^2) - 1/t + C

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19:54:27

what is the antiderivative?

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RESPONSE -->

t + ln(t^2) - 1/t + C

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19:56:18

What substitution would you use to find this antiderivative?

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RESPONSE -->

Actually, I didn't find it by substitution. I instead expanded the top expression (t+1)^2, then split it up into several different functions, then found the antiderivative of each individual function.

Good. That's the way to do this one.

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20:26:51

query 7.1.64 (3d edition #60). int(1/(t+7)^2, t, 1, 3)

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RESPONSE -->

I found the antiderivative to be:

- 1 / (t + 7)

The definite integral then becomes:

1/10 - 1/8

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20:27:00

What did you get for the definite integral?

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RESPONSE -->

1/10 - 1/8

Just about right except for the sign of your result. You missed a negative sign or two.

1/8 > 1/10 so your result is negative. The function is positive over the range of integration.

The correct answer is -1/10 - (-1/8) = 1/8 - 1/10 = 1/40.

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20:27:27

What antiderivative did you use?

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RESPONSE -->

I used the antiderivative:

-1 / (t + 7) + C

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20:27:55

What is the value of your antiderivative at t = 1 and at t = 3?

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RESPONSE -->

At t = 1, the value of the antiderivative is: 1/8

At t=3, the value of the antiderivative is: 1/10

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20:32:48

query 7.1.86. World population P(t) = 5.3 e^(0.014 t).

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RESPONSE -->

a) In 1990, the model gives the population 5.3 billion. In 2000, the modeled population is 6.1 billion.

b) If I understand this correctly, this is the formula I should use:

( F(b) - F(a) ) / 2 + F(a)

Then the average population of the world between 1990 and 2000 becomes:

( 6.1 - 5.3 ) / 2 + 5.3 = 5.7 billion.

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20:33:04

What were the populations in 1990 and 2000?

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RESPONSE -->

1990: 5.3 billion.

2000: 6.1 billion.

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20:34:23

What is the average population between during the 1990's and how did you find it?

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RESPONSE -->

I found the average population to be 5.7 billion. I found it by using this formula:

( F(b) - F(a) ) / 2 + 5.3

That's the average of the initial and final populations.

Review the definition of the average value of a function:

The average value of a function over an interval is equal to its integral over that

interval, divided by the length of the interval.

If the function is linear, then the average of its initial and final values is equal to

its average value (see the above given solution to the Galileo problem).

If the function is not linear, it is very unlikely that this will be the case, and if the

function has nonzero positive or negative concavity on the interval it will never be the

case.

The exponential function is not linear, so this won't work here.

You have to integrate the function and divide by the 10-year interval of integration.

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20:41:43

What is the value of your antiderivative at t = 1 and at t = 3?

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RESPONSE -->

That's interesting. So the function P = 5.3e^(0.014t) isn't the antiderivative? In that case, I need to assume that this is a derivative and find the antiderivative:

Antiderivative: 378.6e^(0.014t)

Value at t=1: 383.94

Value at t=3: 394.84

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20:44:04

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

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RESPONSE -->

I am a bit confused about that last problem and the Galileo connection. Still, there is quite a bit of interesting and useful stuff about antiderivatives. It would be rather difficult to find a distance formula with just a rate formula without the antiderivation process.

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Right.

Excellent work, but be sure to see my notes to clarify the two points you mention.