Assignment 3

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** 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

 

• ave.
  of init and final vel = (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 =
  average velocity * time of
  travel

= ( 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.

 

 

 

Alternative solution:

 

For uniform acceleration the

velocity function can be expressed as

 

• v0 + a t.

 

Integrating this function

between clock times t1 and t2 we obtain the displacement s:

 

• s =
  1/2 a (t2^2 – t1^2) + v0
  ( t2 – t1). 

 

Dividing the integral

(i.e., the displacement) 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

 

• displacement / interval =

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).

This is

identical to the expression obtained previously.

 

Thus the average of the

initial and final velocities is equal to the average velocity on the interval,

and the result is proven.</h3>

 

 

 

Still another

solution:

 

If an object is dropped from

rest and falls for time t it will reach velocity vf = a t.  So the average of

its initial and final velocities will be

 

• Ave of init and final
  vel =(vf + v0) / 2 = (32 t + 0) / 2 = 16 t.

The distance fallen is

known to be

 

• s = 16 t^2.

 

But this is exactly the distance

the object would travel in time t if its average velocity was 16 t.  Thus

the two quantities are identical.

 

 

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.  **

 

 

 

Another

argument:

 

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.

 

 

 

The same argument, expressed

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 derivative of cos(3t) is -3 sin(3t)

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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Note that the derivative of -2/3 (cos3t)^3/2 + C is 3 cos(3t), not cos(3t).

cos(3t) is a composite function and it's necessary to use the chain rule.

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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.

 

Each term of this function is a power function. Integrate term-by-term and add the integration constant to get the general antiderivative

 

general antiderivative = t + 2ln(t) - t^-1 +C

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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

 

integral( u^-2, u from 8 to 10).

 

The integrand is the power function u^-2, or 1 / u^2. Its simplest antiderivative can be expressed as -u^-1 or -1/u.

 

So the definite integral is

change in antiderivative =

-1/10 - (-1/8) =

1/8 - 1/10 =

1/40.

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An antiderivative of 5.3

e^(0.014 t) is found by letting u equal the ‘inner function’ of the composite:

 

u = .014 t so

du = .014 dt and

dt = du / .014.

 

Thus the expression 5.3

e^(0.014 t) dt becomes 5.3 e^u * du / .014.

 

An antiderivative of e^u

with respect to u is just e^u, so the antiderivative of the function is

 

• 5.3 e^u / .014 = 380
  e^u, approximately.  In terms of t this is

• 380 e^(.014 t).

 

At t = 0 this antiderivative

would have value about 380.

 

At t = 0 your antiderivative

would have value about 435.

 

• The change in the
  value of the antiderivative is therefore about 435 – 380 = 55.

• 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 less because

the exponential function is concave upward, and the population

curve therefore 'dips' below the straight line connecting the

initial and final points.

 

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 in your

initial solution.</h3>

The

average population seems as if it would be the average of the two, therefore

 

6.1 + 5.3/ 2 = 5.7

billion

• 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
  not linear, it is very unlikely that the average of its
  initial and final values is equal to its average value

• If
  the function has nonzero positive or negative concavity on the
  entire interval it will never
  have an average value equal to the average of its initial
  and final values.