Assignment 3

course Mth 174

—ºÚ§–ä óÜ‚¢«Û‡àÎç´assignment #003

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Physics II

06-06-2008

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12:43:47

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 -->

(dy)(g) / t = (dy)((v + vd)/2) / t

dy = change in distance

g = acceleration

(v + vd) / 2 = average velocity

t = time

This question confused me and I know you don't realy give credit to answer that I can't explain, but my answer was pretty much a guess.

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

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

An object accelerating for time t, starting with initial velocity v0 at t = 0, has velocity function

v = v0 + a * t

and position function

s = .5 a t^2 + v0 t,

assuming that s = 0 at t = 0.

Both of these results are easily obtained, for the given initial condition that initial velocity is zero, by a straightforward integration of the acceleration function (to get the velocity function) then the velocity function (to get the position function).

For given displacement s we can solve the position equation for t. The equation is rearranged to the form of a standard quadratic:

.5 a t^2 + v0 t – s = 0, with solutions

t = (-v0 +- sqrt(v0^2 + 2 a s) ) / a ).

Substituting this into the velocity function we obtain the final velocity:

Final velocity
= v0 + a t = v0 + a * (-v0 +- sqrt(v0^2 + 2 a s) / a)
= +- sqrt(v0^2 + 2 a s) .

The average of the initial and final velocities is therefore

(initial vel + final vel) / 2 = (v0 +- sqrt(v0^2 + 2 a s)) / 2.

Traveling at this velocity for time t = (-v0 +- sqrt(v0^2 + 2 a s) ) / a ) the displacement will be

Displacement = velocity * time
= ( v0 +- sqrt(v0^2 + 2 a s)) (-v0 +- sqrt(v0^2 + 2 a s) ) / a ) / 2

The numerator is the product of the sum and the difference of two quantities and simplifies to 2 a s. The denominator is 2 a so the expression simplifies to just s.

This confirms that the distance traveled is the same as the distance that would be traveled at the average of the initial and final velocities.

Alternatively:

For uniform acceleration the velocity function can be expressed as

v0 + a t.

Integrating this function between clock times t1 and t2 we obtain 1/2 a (t2^2 – t1^2) + v0 ( t2 – t1).

Dividing the integral by the length of the interval we get the average value of the function—i.e., the average velocity. The length of the interval is t2 – t1 so

Integral / interval = (1/2 a (t2^2 – t1^2) + v0 ( t2 – t1)) / (t2 – t1) = 1/2 a ( t2 + t1) + v0.

The initial value of the velocity on the interval is v0 + a t1, the final velocity v0 + a t2, so the average of initial and final velocities is

Ave of init and final velocity = (v0 + a t1 + v0 + a t2) / 2 = v0 + 1/2 a ( t1 + t2).

Thus the average of the initial and final velocities is equal to the average velocity on the interval, and the result is proven.

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12:44:01

how can you symbolically represent the give statement?

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This was in my last ""answer""

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12:44:19

How can we show that the statement is true?

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Also in the ""answer"" two questions ago.

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12:47:35

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

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I do not understand the first answer, so I am not sure how to construct a graph. I would say that the x-axis is time. But that is as far as I got.

The average value of any linear function over an interval is equal to the average of its initial and final values over the interval.

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 which is 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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13:04:45

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

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RESPONSE -->

I have to admit that I am extremely confused at this substitution thing. My only attempt was that I found:

w = -cos(3t)

dw = -3sin(3t) * dt

And this is not very much, but I just do not know what to do from here. Even the examples confused me.

** TYPICAL INCORRECT SOLUTION: (-2/3) (cos3t)^(3/2)

INSTRUCTOR COMMENT: The derivative of cos(3t) is -3 sin(3t), wo the derivative of -2/3 (cos(3t))^(-3/2) is 3 sqrt(cos(3t) sin(3t). This solution didn't account for the Chain Rule.

To perform the integral use substitution. Very often the first substitution you want to try involves the inner function of a composite, and that is the case here. Cos(3t) is the inner function of the composite sqrt(cos(3t)), so we try using this as our substitution:

w = cos (3t) dw = -3 sin (3t)

so that sin(3t) = -dw / 3.

Thus our expression becomes

w^(1/2) * (-dw / 3).

The integral of -1/3 w^(3/2) with respect to w is -1/3 * (2/3) w^(3/2), treating w^(1/2) as a power function.

This simplifies to

-2/9 w^(3/2) or

-2/9 * (cos(3t))^(3/2).

The general antiderivative is

-2/9 * (cos(3t))^(3/2) + c,

where c is an arbitrary constant.**

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13:04:57

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

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RESPONSE -->

I didn't get an integral

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13:18:22

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

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This one I think I may have done a little better on.

w = x^3 + 1

dw = 3x^2 * dx

x = (w - 1) ^ (1/3)

dx = 1 / [3 * (w -1) ^ (2/3)]

[(w - 1) ^ (1/3)] ^ 2 * e ^ w

Good manipulations up to this point; however this doesn't get you where you need to go. See my next comment.

Antiderivative of (w - 1) ^ (2/3) * e ^ w

(w - 1) ^ (5/3) / (5/3) = (3/5) * (w - 1) ^ (5/3) * e ^ w + C

An antiderivative of (w - 1) ^ (2/3) * e ^ w is not equal to the product of an antiderivative of (w-1)^(2/3) and an antiderivative of e^w. Recall the product rule, where the derivative of f g is f ' g + g ' f. The derivative of f g is emphatically not f ' g '. Reversing this error doesn't work either. f g is not an antiderivative of f ' g '.

Derivative of supposed antiderivative: (3/5)(5/3) * (w - 1) ^ (2/3) * e ^ w

If you take the derivative of this expression, you will have to apply the product rule, and you will not get the original integrand. Nice try, but it just doesn't work out.

Try substituting for the inner function of the composite:

u = x^3 + 1 yields
du = 3 x^2 dx, so
x^2 dx = 1/3 du.

This makes the integrand 1/3 e^u du.

The antiderivative is 1/3 e^u + c, or 1/3 e^(x^3+1) + c.

If we take the derivative of this function we do indeed obtain our original integrand.

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13:18:57

what is the antiderivative?

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RESPONSE -->

I put it in my last answer.

(3/5) * (w - 1) ^ (5/3) * e ^ w + C

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13:19:21

What substitution would you use to find this antiderivative?

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I used x = (w - 1) ^ (1/3)

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13:33:45

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

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RESPONSE -->

Well, my answer seems extremely wrong, but I still think I need some extra help with this substitution thing.

I changed the equation to (t + 1)^2 * t^-2

w = t^-2

dw = -2t ^ -3 * dt

t = w ^ (-1/2)

dt = (-1/2) * w ^ (-3/2)

Substituting t into the equation:

(w ^ (-1/2) + 1) ^ 2 * (w ^ -1/2) ^ -2

(w ^ (-1/2) + 1) ^ 2 * w

(w ^ (1/4) + 2w ^ (-1/2) + 1) * (w)

w ^ (5/4) + 2w ^ (1/2) + w

I am not sure if this equation is correct, but the antiderivative of w ^ (5/4) + 2w ^ (1/2) + w is:

(4/9)w ^ (9/4) + (4/3)w ^ (3/2) + (1/2)w ^ 2

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

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

Each term of this function is just a power function. Integrate term-by-term to get

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

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13:33:48
what is the antiderivative?

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13:34:11
What substitution would you use to find this antiderivative?

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RESPONSE -->
I substituted t = w ^ (-1/2)

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13:35:13
query 7.1.64 (3d edition #60). int(1/(t+7)^2, t, 1, 3)

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RESPONSE -->

This situation involves a composite of the power function 1 / z^2 and the linear function t + 7. The latter is the ‘inner’ function of the composite.

We therefore 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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