query 3 revised

#$&*

course mth 174

4/24 11:10 pm

Query 3#$&*

course Mth 174

4/23 11:40 am

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

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Question: Section 6.5 Problem 3

6.5.3 (previously 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

How can you symbolically represent the give statement?

How can we show that the statement is true?

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Your solution:

We know that acceleration is uniform

We know that the time for an object travelling at uniform acceleration to cross a distance is equal to the average of the initial and final velocities.

Let a = uniform acceleration

Let t = time

Let v0 = initial velocity

Let v1 = final velocity

So we can begin by finding the velocity function with respect to time

V(t) = v0+ a*t

Now we have to integrate this function which will give us the position function

P(t) = v0t + 1/2at^2+c

To show the statement is true we would find the average velocity over the interval by evaluating the interval at a and b, subtracting the a from b, and dividing this difference by the length of the interval. We would set this number equal to the time for the object to travel across the distance at a uniform acceleration and make sure they are the same.

I would use t_1 and t_2 rather than a and b, mostly because a could be confused with acceleration.

&&&I can see how a and b could be confusing, in my answer below I used T1 and T2 as suggested&&&

Following this strategy, what expression would you get for the following quantities:

the two velocities

&&& the 2 velocities are t1 and t2.

Using the velocity function, v(t)=v0+a*t we get

V(t1)=v0t1+a*t1

V(t2)=v0t2+a*t2 &&&

how far the object travels between the two clock times

&&&To find how far the object travels we would use the position function which is P(t) = v0t + 1/2at^2+c where c is the initial position and could be written at P0. We would evaluate at t1 and t2 and subtract to find the distance travelled between clock times. So,

P(t1)=v0t1+(1/2at1)^2+P0

P(t2)=VoT2+(1/2at2)^2+Po

So to find the difference travelled we do

{v0t1+(1/2at1)^2+P0}-{ VoT2+(1/2at2)^2+Po} &&&

the average of the two velocities

&&&To find the average velocities we would find the velocity at clock time t1 and clock time t2 add together and divide by the length of the integral

Velocity at clock time V(t1)=v0t1+a*t1

V(t2)=v0t2+a*t2

Using the values found above for the values of the function at t1 and t2

{((V0T1+a*T1)+(V0T2+aT2))/2}*t2-t1&&&

how far the object would have traveled during the same time interval if it was traveling at the average velocity

&&&If the object was travelling at average velocity on this same time interval, we go back to the velocity function of V(t)=V0t+a*t and evaluate using what

We Have found to be the average velocity {((V0T1+a*T1)+(V0T2+aT2))/2}*t2-t1

So a function for how far the object would have travelled by average velocity uses the position function which would be evaluated at the average velocity

this expression tells us the velocity of the object now we need to go back to the position function to find out how far it travelled using the same method above of evaluating the function at T1 and T2 adding those and dividing by the length of the integral&&&

&&&I think I should have simplified my functions at some point because it seems that my answers are really long??? Am I on the right track in answering these questions or am I missing something???&&&

@&
You're definitely on the right track.

You do have one error that should be corrected:

If v(t) = v0 + a * t, which is correct, then

v(t_1) = v0 + a * t_1,

not

v0 t_1 + a * t_1.

A similar note would apply to your expression for v(t_2).

You have two expressions for the distance traveled, one known to be accurate and the other based on the premise that the average velocity is the average of initial and final velocities. Your expressions are:

{v0t1+(1/2at1)^2+P0}-{ VoT2+(1/2at2)^2+Po}

{((V0T1+aT1)+(V0T2+aT2))/2}(t2-t1)

Both perpetuate your previous error, where v_0 was multiplied by t values, but this will be very easy to correct.

Having made the corrections, you need only verify that the two expressions are the same. This will be simply an exercise in algebra, and I don't believe you'll find it too difficult.

I would suggest, however, that you be consistent with upper and lower cases. You have used both upper and lower cases for v0 and for t_1. In general variables are considered to be case-sensitive. In any event, it's easier to look at and manipulate your expressions if you are consistent with case. For these variables, lower-case is generally chosen.
*@

confidence rating #$&*: 2

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Given Solution:

If acceleration is uniform, then it follows by straightforward integration of the constant a with respect to t that

v(t) = v_0 + a t,

where a is the acceleration and v_0 is the velocity at clock time t = 0.

It also follows that the position function, found by integrating the velocity function, can be expressed as

x(t) = x_0 + v_0 t + 1/2 a t^2.

From the velocity function we can get expressions for the initial and final velocity over an interval. If we average these and multiply by the duration of the interval we get a prediction fo the distance traveled by the object.

From the position function we can calculate the two positions and subtract to determine the actual change in position.

It will turn out that the two results are identical, confirming the statement:

Between clock time t = t_1 and t = t_2, the velocity will change from

v(t_1) = v_0 + a t_1

v(t_2) = v_0 + a t_2.

The average of these velocities is

ave of init and final vel = (v(t_1) + v(t_2) ) / 2

= ( v_0 + a t_1 + v_0 + a t_2) / 2

= v_0 + a ( t_1 + t_2 ) / 2.

The time interval is `dt = t_2 - t_1. Moving a velocity equal to the average of the two velocities the object would travel through a displacement of

`dx = (ave of velocities) * `dt

= ( v_0 + a ( t_1 + t_2 ) / 2 ) * ( t_2 - t_1)

= v_0 * (t_2 - t_1) + 1/2 a ( t_2^2 - t_1^2).

Now using the position function x(t) = x_0 + v_0 t + 1/2 a t^2:

The change in position between t_1 and t_2 is

x(t_2) - x(t_1) = (x_0 + v_0 t_2 + 1/2 a t_2^2) - (x_0 + v_0 t_1 + 1/2 a t_1^2)

= v_0 ( t_2 - t_1) + 1/2 a ( t_2^2 - t_1^2).

Our two expressions agree, proving the original statement true.

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.

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 ].

Still another way of structuring the argument:

** 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.

Self-critique

###I’m not sure if my solution is correct or only partly correct. Following my chain of thinking, if my thinking is at all correct, what do I need to change on my answer????

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Self-critique rating:

See my note at the end of your solution.

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Question: **** query problem 7.1.18 integral of ` sqrt(cos(3t) ) * sin(3t)

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Your solution:

We begin with the integral of the squareroot of cos3t * sin 3t.

I rewrote this as the integral of (cox3t)^1/2 * sing 3t

We will need to use substitution to solve.

I let u = cos3t

So d’ = -3sint3t

So -1/3du=sint3t

You'll want to use the full notation, since there's a pretty specific reason for every symbol:

let u = cos3t

du / dt = -3sint3t so

du = -3 sin(3t) dt and

-1/3du=sint3t

Then sqrt( cos(3 t) ) * sin(3 t) dt becomes -1/3 u du, and the rest of the solution continues as you have written it.

So I now rewrite at -1/3 times the integral of u^1/2du

After integration I have -1/3(2/3u^3/2)

Simplifying I have -2/9(u^3/2)

Substituting back in for u I have

-2/9(cost3t)^3/2 + c

confidence rating #$&*: 3

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Given Solution:

** 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 is not consistent with 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.**

DER

22:16:28

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Self-critique (if necessary):

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Self-critique Rating:3

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Question: Section 7.1 problem 6

7.1.6 (previously 7.1.21) antiderivative of x^2 e^(x^3+1)

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Your solution:

The problem is the integral of x^2*e^x^3+1

I begin by letting u = x^3+1

So that du = 3x^2

So that 1/3du=x^2

Rewriting we have 1/3 time the integral of e^u du

So after integrating we have 1/3(e^u)

Substituting back in we have

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

confidence rating #$&*: 3

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Given Solution:

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.

An antiderivative is then

• antiderivative = 1/3 e^u, or 1/3 e^(x^3+1)

leading to the general indefinite integral

• indefinite integral = 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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Self-critique (if necessary):

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Self-critique Rating:3

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Question: Section 7.1 Problem 9

7.1.9 (previously 7.1.35) Find an antiderivative of (t+1)^2 / t^2

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Your solution:

I begin with the integral of (t+1)^2 divided by t^2

I first work out the (t+1)^2 to get (t+1)(t+1)=t^2+2t+1

So now we have the integral of (t^2+2t+1)(t^-2)

I now multiply these 2 together to get the integral of (t^0 + 2t^-1+t^-2)

Which can be rewritten to the integral of 1+2t^-1+t^-2

So we integrate getting

t + 2lnt -t^-1+c

confidence rating #$&*: 3

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Given Solution:

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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Self-critique (if necessary):

### I got the same answer but I used a different process, is this ok???

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Self-critique Rating:3

Your solution is fine. I doesn't appear to me that it differs in any significant way from process used in the given solution.

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Question: Section 7.1 Problem 13

7.1.13 (previously 7.1.60). int(1/(t+7)^2, t, 1, 3) **** What did you get for the definite integral?

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Your solution:

The problem is the definite integral from 1 to 3 of dt/(t+7)^2

I rewrite the inside as (t+7)^-2dt

So I let u = t+7

So that du = 1dt

Now I have to also rewrite my integral adding 7 to top and bottom

So now my problem is the definite integral between 8 and 10 of u^-2

After integration the inside becomes -u^-1

So I evaluate at 10 and subtract the evaluation at 8 to get the solution

-(10)^-1-(-(8)^-1))

-.1-(-.125)

.025

confidence rating #$&*: 2

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Given Solution:

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.

Amplified explanation

The given definite integral becomes the integral of 1 / u^2, integrated from u = 8 to u = 10. We can find our integral just in terms of the function u:

The definite integral is the change in the antiderivative function. The antiderivative function is - 1 / u, which between u = 8 and u = 10 changes from -1/8 to -1/10.

Subtracting -1/8 from -1/10 we get 1/40.

Alternatively, having found our antiderivative in terms of u, we could have substituted t + 7 for u and done the integral in terms of t:

Our antiderivative -1/u is equal to -1 / (t + 7).

Our original integral was form t = 1 to t = 3.

At t = 1 our antiderivative is -1/(1 + 7) = -1/8. At t = 3 our antiderivative is -1 / (3 + 7) = -1/10. Once more the change in the antiderivative is found to be 1/40.

Still another way of putting it:

If f(t) = 1 / (t + 7)^2, our antiderivative is F(t) = - 1 / (t + 7). Thus

integral ( f(t), t from 1 to 3) = F(3) - F(1) = -1/10 - (-1/8) = 1/40.

F(3) - F(1) is, of course, the change in the value of F(t) between t = 1 and t = 3; i.e., this is the change in the value of the antiderivative function.

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Self-critique (if necessary):

### I got the answer correct after remembering that I needed to change my interval I see that we added 7 to the interval and I see that u = t+7 when I do substitution, but why does the interval have to change???? Does the interval always change on u substitutions????

It usually does, but there could be cases where it doesnt.

When integrating a function of u, the limits on the integral need to be values of u.

If the limits on the original integral were on t, as in this case, we would not expect the u values to be the same as the t values. To do this as a definite integral over u, you would have to find the u values that correspond to the given t values.

Another strategy is to simply do an indefinite integral in u, then find your antiderivative as a function of t by converting the expression back to t. You can then find the definite integral, which is the change in the antiderivative function between the two limits, using the limits on t.

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Self-critique Rating:3

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Question: Section 7.1 Problem 16

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

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Your solution:

We begin with p(t) = 5.3e^0.014t

We want to find the population in 1990 and in 2000 as well as the average population over this 10 year time span.

The initial population is in 1990 which would be p(0) at time 0

So for 1990 we have p(0)=5.3e^(.014)(0) = 5.3 billion

For 2000 which is 10 years later we have p(10) = 5.3e^0.014*10 = 6.1 billion

Now we also want to find the average population over this time period

So we need to find the definite integral from 0-10 of 5.3 e^0.014t

So I need to let u = 0.014t

So that du = .014dt ( I need 5.3 out front)

378.6du=dt

If du = .014 dt, then dt = (1 / .014) du.

1 / .01 = 100, and 0 / .1 = 10, so 1 / .014 is between 10 and 100. So 1 / .014 is not 378.6.

It's best to avoid approximations in any case, though in this case since your final result will be an approximation it wouldn't really be wrong to use one.

1 / .014 is about 72, so a rough approximation would be dt = 72 du.

I would write it, though, as dt = 100 / 14 * du, probably reducing this to 50/7 du.

Your integrand would then become

5.3 e^u * 50/7 du.

This will be about 380 e^u du, but the 5.3 is not part of dt so it's not part of the resulting du. It is mulitplied by the resulting du.

Now I rewrite at 378.6 times the definite integral from 0 to 10 of e^u du

t changes during this interval from 0 to 10.

u is much smaller than t. So u doesn't change from 0 to 10.

After integrating I have 378.6*e^u

Substituting back in for u I have

378.6*e^0.014t

So now I need to find the value at 10 and then subtract 0 and then divide by 10 years

So I have (378.6*e^.014*10) - (378.6*e^0.014*0) this gives me

435-378 which is 57 which I have to divide by 10 which gives me 5.7 as the average population.

confidence rating #$&*: 2

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Given Solution:

What were the populations in 1990 and 2000?

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

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In 1990, t = 0 so the population is P(0) = 5.3 billion

In 2000, t = 0 so the population is P(10) = 6.1 billion

To find the average population we integrate the population function over the 10-year interval, then divide by the 10 years:

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.

INCORRECT IF NOT UNREASONABLE ANSWER:

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

6.1 + 5.3/ 2 = 5.7 billion

INSTRUCTOR RESPONSE:

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.

Now 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). However:

• 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.

The exponential function is not linear, and is in fact concave upward on this interval, so averaging initial and final values on the interval won't work here.

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

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COMMON INTEGRATION ERRORS: integral( x^2 * e^(x^3+1) ) = integral( x^2) * integral (e^(x^3 + 1) ) = x^3 / 3 * e^(x^3 + 1)

INSTRUCTOR RESPONSE: You have essentially assumed that the integral of a product function is equal to the product of the integrals. The integral is of the form

• integral (f * g)

with f(x) = x^2 and g(x) = e^(x^3 + 1).

However integral (f * g) is not the same as integral (f) * integral(g):

The derivative of our antiderivative function must equal the original integrand. That is, the derivative of integral(f * g) must be f * g.

However the derivative of integral (f) * integral(g) is expressed by the product rule as

(integral (f) * integral(g)) ' = ( integral (f) ) ' * integral(g) + integral (f) * (integral(g)) ' = f * integral(g) + g * integral(f).

This is much different than the original integrand f * g, so integral ( f * g) is not generally the same as integral(f) * integral(g).

Without going into the details, we'll also assert that

• integral(f / g) is not the same as integral(f) / integral(g),

which you can if you wish verify for yourself by taking the derivative of integral(f) / integral(g), using the quotient rule. What you get is nothing like the original integrand f / g.

So for example integral( (t+1)^2 / t^2 ) is not the same as integral ( t + 1) / integral (t^2).

Note also that, while the integral of e^x with respect to x is e^x + c, the integral of (e^(x^3 + 1) ) with respect to x is NOT the same as e^(x^3 + 1) + c:

• The derivative of e^(x^3 + 1) is 3 x^2 * e^(x^3 + 1), not e^(x^3 + 1).

So e^(x^3 + 1) is not an antiderivative of e^(x^3 + 1).

Self-critique

### I used a different step during the u substitution instead of solving for dt I said 378.6du=dt I am not sure if my answer is wrong or partly wrong. I followed the same logic but my answer is slightly different????

Everything you've done is fine, though I made some clarifying notes in your solution.

You converted your integrand back to t values before substituting your t limits, so your expressions are good.

Your answer differs from that of the given solution only in that you included more significant figures in yours.

The given initial population was 5.3, with only two significant figures. So the result 5.7 billion is consistent with the precision of the given information.

Your solution contains 3 significant figures, implying more precision that is possible from the given information.

I wouldn't penalize that in a calculus course, but it is something you should be aware of.

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#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#

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You're making good progress on the Galileo problem. Check my note. One fairly simple correction, then a little algebra and you should have that one.

If you get it and are sure of your solution, there's no need to resubmit. However if that's not the case, or if you have questions, or if you just want me to see your solution, please do.
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