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Mth 163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Questions from Inter. Prob
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I have redone some of the problems and tried to correct my errors. Could you look through these and see if I have them correct?
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Problem 10: If a cube with weight 10 has surface area 45, what is the surface area a cube with weight 2?
45 = a * 10^3
a = .045
y = .045x^3
y = .045(2^3)
y = .36
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If y was the weight and x was the length, the proportionality would be y = x^3 and this would be the weight corresponding to a length of 2 if 10 and 2 were lengths and 45 was a weight..
However 2 and 45 are surface areas, and 10 is weight.
You could begin by finding the ratio of lengths, then use this ratio to find the ratio fof volumes.
#### This assignment has been so confusing and I’m trying to wrap my head around it. I thought 10 and 2 were weights and 45 was surface area? I don’t understand how it is y = kx^3 because the problem asks for surface area which would be y = kx^2.
How then will I find the surface area of a cube with weight 2? I’m sorry I’m so confused. I don’t know why I can’t comprehend this completely. I’m really upset with myself over this….####
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You are correct that 2 is a weight, not a surface area.
In any case, as indicated in my note, you can use the ratio of lengths to find the ratio of areas.
The proportionality for volume or weight y vs. length is y = k x^3.
In this case the ratio of the lengths x would be (y2 / y1) ^(1/3) = (2/45) ^ (1/3) = .38 or so.
Now if the ratio of lengths is .38, what would be the ratio of areas?
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Don't be upset. This is challenging, and you are on the right track.
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Problem 13: By what factor does the altitude of a sandpile change if its volume changes from 5 to 2?
X2/x1 = 2/5 = .4
y2/y1 = .4^3 = .064
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If x1 and x2 are volumes with y1 and y2 being side lengths, then the proportionality would be x = k y^3, giving you ratio equation x2.x1 = (y2/y1)^3.
#### x2/x1 = 2/5 = .4
y2/y1 = .4^3 = .064
Is this correct?? ####
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If x2 / x1 = (y2 / y1)^3, then y2 / y1 = (x2 / x1)^(1/3).
Thus y2 / y1 = .4 ^(1/3), not .4^3.
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You can always check yourself by assuming the given solution. Volume is proportional to the cube of length, so the ratio of volumes is the cube of the ratio of lengths. If the ratio of lengths is .064, then the ratio of volumes would be .064^3, or about .000026.
However the ratio of volumes is .4, so your solution was not consistent with the given information.
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Problem 14: By what factor does the total surface area of a balloon change if its volume changes from 4 to 2?
X2/x1 = 2/4 = .5
y2/y1 = .5^2 = .25
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If y = k x^2, then y is the area and x is the diameter or radius or circumference.
You can find the ratio of the diameters, then use this to find the ratio of the volumes.
#### x2/x1 = 2/4 = .5 then y2/y1 = .5^3 = .125 then v2/v1 = .125^2 = .0156
Is this correct?#####
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You should clearly identify x and y, and write out the proportionality equation.
If x is diameter and y is volume then
y = k x^3
so that
y2 / y1 = (x2 / x1)^3.
You are given the two volumes, so
(x2 / x1)^3 = y2 / y1 = 2 / 4 = .5.
Solving (x2 / x1)^3 = .5, we get
x2 / x1 = .5 ^ (1/3) = .8, approx..
Then let y be surface area and x be diameter. Your proportionality is y = k x^2 so the ratio equation is
y2 / y1 = (x2 / x1)^2.
It follows that y2 / y1 = (.8)^3.
Knowing y1 you can easily find y2.
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Problem 17: If a sandpile with base diameter 3 has weight 37, what is the weight of a sandpile with base diameter 9?
37 = a(3^2)
a = 4.111
Y = 4.111x^2
y = 4.111 (3^2)
y = 36.999
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Sand occupies volume, so the weight is proportional to the cube of the base diameter.
#### This one is confusing me as well. I have done this a couple different ways on paper, but I can’t get an answer to look right??? ####
Problem 18: If a cube requiring 6 grams of paint has edge length 99, what is the edge length of a cube requiring 9 grams of paint?
99 = a 6^3
a = .4583
Y = .4583 (9^3)
y = 334.1007
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If you triple edge length you get 27 times the volume.
It appears that you mixed up the meanings of your variables.
#### These seem to be the problems that I am getting confused on! Ahh!! I am so frustrated with myself! Please help!!! And again, I sincerely apologize if I’m being a pain!! #####
Problem 19: By what factor does the volume of a sandpile change if its surface area changes from 3 to 7?
X2/x1 = 7/3 = 2.333
y2/y1 = 2.333^3 = 12.698
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That would be the ratio of volumes if the edge length increased from 3 to 7.
#### x2/x1 = 7/3 = 2.333
y2/y1 = .2333^1/2 = .483
v2/v1 = .483^3 = .1127 #####
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See my note on the previoius problem.
You need to state what each variable means, then write the proportionality equation. Having written the proportionality equation you can easily write the ratio equation.
You paint the surface of the cube. You don't fill it with paint.
For y = surface area and x = edge length the proportionality equation is
y = k x^2
so the ratio equation is
(y2 / y1) = (x2 / x1)^2
You could then either solve for proportionality constant k and use the resulting proportionality equation to answer the question, or you could use the ratio equation.
In this case you know y1 and y2, the two surface areas, and you know x1, the edge length of one of the cubes.
If you use ratios then knowing y1 and y2 you solve for (x2 / x1):
Since y2/y1 = (x2/x1)^2 you get
x2 / x1 = (y2 / y1)^(1/2).
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If you assumed that the cubes were being filled with paint, which wouldn't be the intended interpretation but would fulfill the statement of the problem, then you would use the cube and the 1/3 power. I would accept that solution.
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I can be a pain when a student doesn't make a good effort. However that isn't so in your case, and your questions are welcome.
You pretty much know what you're doing, but you're not writing the problems out in enough detail, which causes you to sometimes jump to wrong conclusions.
It you take your time and are careful, you'll get this.
Check my notes. I'll be glad to look at additional solutions and/or questions.
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