assn 3

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** 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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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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The correct solution would have its derivative equal to the original function.

However, the derivative of

(1/3 sin(3t) * sin (3t)) + (cos (3t) * -1/3 cos (3t) ^.5 * .7t^1.5

is not

`sqrt(cos(3t) ) * sin(3t), so your solution is not correct.

To evaluate the integral use 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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If you take the derivative of this function you don't get the original expression.

Let u = x^3 + 1 so

du = 3 x^2 dx, so

x^2 dx = 1/3 du.

So the integrand is 1/3 e^u du.

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

**

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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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04:00:20

what is the antiderivative?

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RESPONSE -->

see prevous answer

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04:01:22

What substitution would you use to find this antiderivative?

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RESPONSE -->

the product rule I used in place of the quatoent rule.

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04:06:16

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

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RESPONSE -->

1 / (t+7) ^2

for the antiderivative i get 1 / (t +7t)^3 / 3

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To integrate P(t) = 5.3 e^(0.014 t) use the substitution

u = .014 t so that

du = .014 dt and

dt = du /.014 = 70 du, approximately, yielding the integral of

5.3 e^u * 70 du = 370 e^u du.

An antiderivative of 370 e^u is 370 e^u, which translates to 370 e^(.014 t).

The question should have asked for the antiderivative at t = 0 and t = 10, which would correspond to the years 1990 and 2000.

At t = 0 your antiderivative would have value 378.57.

At t = 10 your antiderivative would have value 435.4607952.

The change in the value of the antiderivative is therefore 435.46 - 378.57 = 56.89.

The change in the antiderivative is the definite integral of the function.

The average value of the function is equal to its definite integral divided by the time interval:

average value = 56.89 / 10 = 5.689.

This is quite close to, but a little less than the average of the initial and final populations.

It is a less because the exponential function is concave upward, and the population curve therefore 'dips' below the straight line connecting the initial and final points; this is so on any interval.

However the curvature on this interval is not pronounced, and the function never 'dips' far below the straight line, so the average is close to the straight-line average you gave.

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You need to review the process of integration by substitution. The expressions you obtained for your antiderivatives were generally incorrect; your antiderivatives should always be checked by taking their derivatives, to see if you got the original expression.