Assignment 14

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course Mth 163

2/28/13 around 6 p.m.

014.

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Question: `q001. Note that this assignment has 7 questions

If a(n) = a(n-1) + 2^n, with a(0) = 3, then substitute in turn the values 1, 2, 3 and 4 into the equation to obtain the values a(1), a(2), a(3) and a(4).

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Your solution:

Substituting a(1) we get

A(1-1) + 2^1 = 2

A(1) = a(0) + 2 = 3+2 = 5

Substituting a(2) we get

A(2-1) + 2^2 = 5

A(2) = a(1) + 2^2 = 5+4 = 9

Substituting a(3) we get

A(3-1) + 2^3 = 8

A(3) = a(2) + 2^3 = 9 + 8 = 17

Substituting a(4) we get

A(4-1) + 2^4 = 19

A(4) = a(3) + 2^4 = 17+16 = 33

confidence rating #$&*: 3

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Given Solution:

If we substitute n = 1 into a(n) = a(n-1) + 2^n we get

a(1) = a(1-1) + 2^1 or, since 1-1 = 0 and 2^1 = 2

a(1) = a(0) + 2. Since we are given a(0) = 3 we now have

a(1) = 3 + 2 = 5.

If we substitute n = 2 into a(n) = a(n-1) + 2^n we get

a(2) = a(2-1) + 2^2 or, since 2-1 = 1 and 2^2 = 4

a(2) = a(1) + 4. Since in the previous step we found that a(1) = 5 we now have

a(2) = 5 + 4 = 9.

If we substitute n = 3 into a(n) = a(n-1) + 2^n we get

a(3) = a(3-1) + 2^3 or, since 3-1 = 2 and 2^3 = 8

a(3) = a(2) + 8. Since in the previous step we found that a(2) = 9 we now have

a(3) = 9 + 8 = 17.

If we substitute n = 4 into a(n) = a(n-1) + 2^n we get

a(4) = a(4-1) + 2^4 or, since 4-1 = 3 and 2^4 = 16

a(4) = a(3) + 16. Since in the previous step we found that a(3) = 17 we now have

a(4) = 17 + 16 = 33.

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Self-critique (if necessary): ok

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Self-critique rating: 3

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Question: `q002. If a(n) = 2 * a(n-1) + n with a(0) = 3, then what are the values of a(1), a(2), a(3) and a(4)?

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Your solution:

a(1) = 2 * (1-1) + 1 = 1-1 = 0

SO, a(1) = 2* a(0) +1

a(0) = 2 * 3 + 1 = 7

A(2) = 2*(2-1) + 2 = 2-1 = 1

So a(2) = 2*a(1) + 2

a(1) = 2*7 + 1 = 16

a(3) = 2 * (3-1) + 3 = 3-1 = 2

a(3) = 2 * a(2) + 3 = 2 *16+3 = 35

a(4) = 2 * (4-1) + 4 = 4-1 = 3

a(4) = 2 * a(3) + 4 = 2 * 35 + 4 = 74

confidence rating #$&*: 3

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Given Solution:

If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get

a(1) = 2 * a(1-1) + 1 or since 1-1 = 0

a(1) = 2 * a(0) + 1. Since we know that a(0) = 3 we have

a(1) = 2 * 3 + 1 = 7.

If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get

a(2) = 2 * a(2-1) + 2 or since 2-1 = 1

a(2) = 2 * a(1) + 2. Since we know that a(0) = 3 we have

a(2) = 2 * 7 + 2 = 16.

If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get

a(3) = 2 * a(3-1) + 3 or since 3-1 = 2

a(3) = 2 * a(2) + 3. Since we know that a(0) = 3 we have

a(3) = 2 * 16 + 3 = 35.

If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get

a(4) = 2 * a(4-1) + 4 or since 4-1 = 3

a(4) = 2 * a(3) + 4. Since we know that a(0) = 3 we have

a(4) = 2 * 35 + 4 = 74.

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Self-critique (if necessary): ok

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Self-critique rating: 3

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Question: `q003. What are the average slopes of the graph of y = x^2 + x - 2 between the x = 1 and x= 3 points, between the x = 3 and x = 5 points, between the x = 5 and x = 7 points, and between the x = 7 and x = 9 points? What is the pattern of this sequence of slopes?

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Your solution:

To find the average slopes we must first find what x equals in each case.

The coordinates we get are:

(1,0)

(3, 10)

(5, 28)

(7, 54)

(9, 88)

We then want to find the slopes between the points

We start with the slope between x = 1 & x = 3

We must use the formula y2 - y1 / x2 - x1. Plugging in the numbers we get 10 - 0 / 3 -1 = 10 / 2 = 5

For x = 3 & x = 5 we get the slope to be 9

For x = 5 & x = 7 the slope is 13

For x = 7 & x = 9 the slope is 17

The run from one point to the next is always 2. The rises are 10, 18, 26 and 34.

confidence rating #$&*: 3

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Given Solution:

At x = 1, 3, 5 , 7 and 9 we find by substituting that y = 0, 10, 28, 54 and 88.

The x = 1, 3, 5, 7 and 9 points are therefore (1,0), (3,10), (5,28), (7,54) and (9,88).

The run from one point to the next is always 2. The rises are respectively 10, 18, 26 and 34. The slopes are therefore

slope between x = 1 and x = 3: slope = rise / run = 10 / 2 = 5.

slope between x = 3 and x = 5: slope = rise / run = 18 / 2 = 9.

slope between x = 5 and x = 7: slope = rise / run = 26 / 2 = 13.

slope between x = 7 and x = 9: slope = rise / run = 34 / 2 = 17.

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Self-critique (if necessary): ok

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Self-critique rating: 3

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Question: `q004. If a solid stone sphere 4 inches in diameter weighs 3 pounds, then what would be the weight of a solid stone sphere 2 feet in diameter?

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Your solution:

We know the formula for the volume of a cube which is weight = k diameter^3

If we plug in our values for the first stone sphere we get 3lb = k4inches^3. If we solve for k we get k = .0469.

We can then plug the values in for the second stone. We must first change 2 feet to 24 inches then plug the numbers in giving us w = .0469 * 24 in ^ 3

W = 648lb

confidence rating #$&*: 3

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Given Solution:

The volume of a sphere is proportional to the cube of its diameters, and weight is directly proportional to volume so we have the proportionality

w = k d^3, where w and d stand for weight and diameter and k is the proportionality constant.

Substituting the known weight and diameter we get

3 = k * 4^3, where we understand that the weight is in pounds and the diameter in inches. This gives us

3 = 64 k so that

k = 3 / 64.

Our proportionality equation is now

w = 3/64 * d^3.

So when the diameter is 2 feet, we first recall that diameter must be in inches and say that d = 24, which we then substitute to obtain

w = 3/64 * 24^3. A simple calculation gives us the final weight

w = 648.

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Self-critique (if necessary): ok

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Self-critique rating: 3

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Question: `q005. Two boxes are each constructed of a single layer of cardboard. The first box is 12 inches by 18 inches by 24 inches and weighs 22 ounces; the second is 36 inches by 54 inches by 72 inches. Using proportionality determine the weight of the second box.

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Your solution:

We can look at the two boxes and see that each side on the second box is 3 times as big by seeing 36/12 = 3, 54/18 = 3, & 72/24 = 3

We then can use our equation again that gives us w = kx^2

We must split it and use w2 as the weight of the second box and w1 as weight of the first box and then x2 a side of the second box (i.e. 36) and x1 a side of the first box.

So we have w2/w1 = (x2/x1)^2. We already know that x2/x1 = 3 so we can plug that in and get w2/w1 = 3^2 = 9

Then we can plug that in our equation getting w2 / 22oz = 9. Multiply both sides by 22 and we get that w2 (weight of second box) = 198 oz

confidence rating #$&*: 3

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Given Solution:

The ratios 36/12, 54/18 and 72/24 of the corresponding sides are all the same and all equal to 3, so the dimensions of the sides of the second box are 3 times those of the first.

Since the thickness of the cardboard is the same on both boxes, only the dimensions of the rectangular sides change. The only thing that matters, therefore, is the surface area of the box. The proportionality is therefore of the form w = k x^2, where w is the weight of the box and x stands any linear dimension.

It follows that w2 / w1 = (x2 / x1)^2. Since as we just saw x2 / x1 = 3, we see that

w2 / w1 = 3^2 = 9. Since w1 = 22 oz, we write this as

w2 / 22 oz = 9. Multiplying both sides by 22 oz we see that

w2 = 22 oz * 9 = 198 oz.

STUDENT QUESTION

I reduced the sides of each box by dividing each side measurement by 6. Then since we were looking for weight, I used

the y = kx^3 formula to calculate the weight of the second box. I used y as the weight and x as the volume (l*w*h).

Obviously my calculation was way off. Why wouldn’t this work?

INSTRUCTOR COMMENT

Presumably the cardboard is of the same thickness for both boxes. So the amount of cardboard is determined by the surface area of the the box, not the volume.

If the box was filled with cardboard as well as being constructed of cardboard, then the proportionality with volume would be appropriate and your solution would be correct.

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Question: `q006. Between t = 1 and t = 3 the function y = .5 t^2 - 5 t + 9 has average slope -3; between t = 3 and t = 5 the average slope is -1; and between t = 5 and t = 7 the average slope is 1.

• What do you conjecture will be the average slope between t = 7 and t = 9?

• What linear function do you think gives the slope as a function of t?

• What do you think will be the slope at t = 14?

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Your solution:

For the average slope between t = 7 & t = 9 I would say it would be 3 because there is 2 in between all of the other slopes

The linear function would be ???

The slope at t = 14 would be about 8.5 or 9

confidence rating #$&*: 2

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Question: `q007. A sphere of radius .5 meters has mass 1100 kg and requires 11 ounces of paint to cover its surface. What would be the mass of a sphere of the same material whose radius is 1.3 meters, and how much paint would it take to cover its surface with a coat whose thickness was the same as that used to cover the first?

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Your solution:

I am not sure where to really start?? I would assume you would use some formula as the one before with weight, but I don’t think it will work with this problem???

confidence rating #$&*:0

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014. `query 14

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Question: `qQuery two examples and a picture ...explain the statement 'the rate of change of a quadratic

function changes at a constant rate'

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Your solution:

The rate of change of a quadratic can be calculated by looking at the gaps between lengths. The rates change at each gap and will always be the same amount. If they are the same they can be said to change at a constant rate.

confidence rating #$&*: 3

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Given Solution:

`a** We can calculate the rates of change of a quadratic function based on a series of consecutive intervals of constant

length. We find that these rates change from interval to interval, and always by the same amount. Since the rates of

change always change by the same amount, they are changing at a constant rate. **

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Self-critique (if necessary): ok

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Self-critique rating: 3

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Question: `qexplain how to get the first few members of a sequence from its recurrence relation

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Your solution:

N can be the first integer in the sequence. We will substitute this into the recurrence relation and evaluate a(n) as we have been doing. We would then substitute the next integer continuing to use a(n) until we reach the number we need

confidence rating #$&*: 3

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Given Solution:

`a** We let n be the first integer for which the value a(n) is not given, and we substitute this integer into the recurrence

relation to evaluate a(n) for this 'new' integer, using values of a(n) for previous integers. If this is not possible then we

have not been given enough information to evaluate the sequence.

We then substitute the next integer and use values of a(n) for previous integers.

We continue this process as long as necessary to get the results we need. **

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Self-critique (if necessary): ok

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Self-critique rating: 3

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Question: What are the next four numbers of the recurrence relation a(n) = 2 a(n - 1) + a(n - 2) if a(0) = 3 and a(1) = 2?

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Your solution:

A(2) = 1

A(3) = 0

A(4) = -1

A(5) = -2

@&

Your values aren't correct, and you haven't indicated how you got them.

You would apply the same reasoning you used in your solution of `q002 above.

a(n) = 2 a(n-1) + a(n-2).

You are given a(0) and a(1).

The first number you need to compute is a(2).

So you want to apply the formula with n = 2.

With n = 2 the formula is

a(2) = 2 a(2-1) + a(2-2)

which simplifies to

a(2) = 2 a(1) + a(0).

Since a(1) = 2 and a(0) = 3, we can substitute to find a(2):

a(2) = 2 * 2 + 3 = 7.

The remaining values are found using analogous reasoning.

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confidence rating #$&*:1

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Self-critique rating:

I don’t really understand this very well??? I am a little confused with the second part a(n-2). I am not sure what to do with that???

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See my notes. You solved `q002 correctly, so you pretty much understand the reasoning necessary. My notes should clarify how the a(n-2) term is taken into consideration.

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Self-critique (if necessary):

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Self-critique rating:

I don’t really understand this very well??? I am a little confused with the second part a(n-2). I am not sure what to do with that???

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See my notes. You solved `q002 correctly, so you pretty much understand the reasoning necessary. My notes should clarify how the a(n-2) term is taken into consideration.

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Self-critique (if necessary):

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Self-critique rating:

#*&!

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You should submit a revision of that last problem.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

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