query assignment 3

course Mth 174

界潱OxИg籁伌剤擊挟嚤胊ssignment #003

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Physics II

09-19-2006

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13:11:21

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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13:11:23

how can you symbolically represent the give statement?

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s(t) = v'(t)

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 position function s = .5 a t^2; since a = 32 ft/s^2 we have s = 16 t^2.

We assume that the time of fall is t. Then using Galilieo's assumption we show that they are consistent with the result we get from s = 16 t^2.

If an object is dropped from rest and falls for time t it will reach velocity vf = a t = 32 t. So the average of its initial and final velocities will be (vf + v0) / 2 = (32 t + 0) / 2 = 16 t.

The distance fallen is s = 16 t^2.

The time to fall distance s = 16 t^2 at average velocity 16 t is s / t = 16 t^2 / (16 t) = t, which agrees with the time the object was allowed to fall.

A numerical example for a given s:

When s = 100, then, t is the positive solution to 100 = 16 t^2. This solution is t = 2.5.

The velocity function is v = a t. We have to find v when s = 100. When s = 100, we have t = 2.5, as just seen. So v = a t = 32 * 2.5 = 80, representing 80 ft/sec.

Now we know that s = 100, vf = 80 and v0=0. Substituting into t = s / [ (vf + v0)/2 ] we get t = 100 / [ (80 + 0) / 2 ] = 100 / 40 = 2.5.

This agrees with the t we got using s = .5 a t^2.

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13:11:26

How can we show that the statement is true?

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13:11:29

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

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13:11:32

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

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13:13:31

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

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first i figure out that u would = cos(3t) and that du = -3sin(3t)dt the next thing that you need to do is solve for sin(3t)dt when you do you get sin(3t)dt = -1/3du. then if you put w back into your equation you get `sqrt(w)*sin(3t)dt since we know that sin(3t)dt is -1/3 we can rewite the equation as -1/3*w^1/2 if we simplfy further we get -1/3 * (w^3/2)/ 3/2. The you get the final answer being -2/9(cos(3t))^3/2

In a nutshell, because the v vs. t graph is linear, the average velocity is equal to the average of the initial and final velocities. Since

time of fall = displacement / average velocity,

it follows that

time of fall = displacement / (ave of initial and final vel).

This latter expression is just the time that would be required to fall at a constant velocity equal to the average of initial and final velocities.

More rigorously:

The graph is linear, so the area beneath the graph is the area of a triangle.

The base of the triangle is the time of fall, and its altitude is the final velocity.

By a simple construction we know that the area of the triangle is equal to the area of a rectangle whose length is equal to the base of the triangle, and whose width is equal to half the altitude of the triangle.

It follows that the area of the triangle is equal to half the final velocity multiplied by the time interval.

Since the initial velocity is 0, the average of initial and final velocities is half the final velocity.

So the area of the triangle is the product of the time of fall and the average of initial and final velocities

area beneath graph = time of fall * ave of init and final vel

The area beneath the graph is equal to the integral of the velocity function over the time interval, which we know is equal to the displacement. So we have

displacement = time of fall * ave of init and final vel, so that
time of fall = displacement / ave of init and final vel.

This leads to the same conclusion as above.

Also, more symbolically:

A graph of v vs. t graph is linear. Over any time interval t = t1 to t = t1 + `dt, displacement is represented by the area of the corresponding trapezoid, which is d = (v1 + v2) / 2 * `dt.

Solving for `dt we again obtain `dt = d / [ (v1+v2)/2 ].

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13:13:35

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

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13:14:44

what is the antiderivative?

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you find out that u would be x^3 + 1 then du = 3x^2dx if you solve for x^2dx you get x^2dx = 1/3du if you put u back into the equation you get e^u*x^2dx since we know that x^2dx is 1/3du we can rewrite the equation as 1/3e^u so the finial answer is 1/3e^(x^3 + 1)

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13:15:24

What substitution would you use to find this antiderivative?

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x^3 + 1

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13:18:47

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

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13:22:33

what is the antiderivative?

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U = 1 + 3t^2 du = 6tdx tdx = 1/6du. now you can rewrite the equation as 1/6(1/u) or 1/6 ln(u). the finial answer is 1/6 ln(1 + 3x^2)

expand (t+1)^2 to get t^2+2t+1.

divide by t^2 to get 1 + 2t^-1 + t^-2 dt.

Integrate to get

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

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13:23:02
What substitution would you use to find this antiderivative?

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1/6

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13:47:30
query 7.1.64 (3d edition #60). int(1/(t+7)^2, t, 1, 3)

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13:47:33
What did you get for the definite integral?

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you use u = t+7 du = dx 1/u^2 the answer being ln((t+7)^2)

This situation involves a power function. The integral of 1/u is ln(u) but the integral of 1 / u^2 is - 1 / u. We substitute u = t+7 to get du = dt; limits t = 1 and t = 3 become u = 8 and u = 10. So the integral becomes

int( u^-2, u, 8, 10).

Antiderivative can be -u^-1 or -1/u.

So definite integral is -1/10 - (-1/8) = 1/8 - 1/10 = 1/40. **

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