Help! Mathman!

CoyoteChris · Oct 31, 2016

When you have a calculus problem, who you gonna call? Mathman!

Spokane is semi arid but we set a one month October record this month of over six inches of rain at the airport. And counting.
I use a rain gage but some days I forget to empty it, record the reading, etc.

I set out a five gallon bucket at the beginning of the month. The bucket now has seven inches of rain in it.

The bucket is 11 inches across at the top and 10 inches at the bottom, inside diameter. It is 14 inches high.

How many inches of rain has fallen into it?

Chris who dropped out of engineering due to calculus, chemistry , and theoretical physics. :bang:

Mathman · Oct 31, 2016

6.079 inches.

The volume (of a truncated cone) is 577.726 cubic inches, with the rain collected over an area of 95.033 square inches. Divide.

By the way, problems of this sort were considered by the Egyptian mathematician-scribes of dynastic Egypt. The best surviving mathematical manuscript from this period is the Ahmes, or Rhind, Papyrus. Ahmes is the name of the scribe who copied it about 1600 BCE from an original, now lost, written about 200 years earlier. Rhind is the British tourist who bought it from local grave robbers in the 1850s CE, and later gave it to the British museum.

CoyoteChris · Nov 1, 2016

Mathman said:
6.079 inches.

The volume (of a truncated cone) is 577.726 cubic inches, with the rain collected over an area of 95.033 square inches. Divide.

By the way, problems of this sort were considered by the Egyptian mathematician-scribes of dynastic Egypt. The best surviving mathematical manuscript from this period is the Ahmes, or Rhind, Papyrus. Ahmes is the name of the scribe who copied it about 1600 BCE from an original, now lost, written about 200 years earlier. Rhind is the British tourist who bought it from local grave robbers in the 1850s CE, and later gave it to the British museum.

Thanks! I never did well in math....oh well. I can visualize the tapered sides of the bucket, the 6 inches of water falling through the upper area into the lower area, but is hard to visualize calculating the volume of the tapered bucket and compare that to the volume of water that fell. Its a good thing Mirai is doing better at probability problems in her homework than I did in engineering. Next life I am taking my slide rule and beating my Indian (dot not feather) teaching assistant in calc to death with it!

OK, maybe not. I should hug the guy. Dropping out of engineering was the best thing that ever happened to me.

We now have a new record for ANY month in the History of Spokane. 6.22" and climbing a bit before midnight. The ground soaked it all up and the aquafir really loves rain now as the snow might go away by sublimation or runoff and the summer rains, all one of them, just evaporate.

There is now for awhile an Egyptian exhibit at the local museum of fine arts with a famous human mummy, "Annie", and a pet mummy as well as lots of interesting artifacts and info. The ability of the ancients to build amazing temples and other archetecture has always interested me. Its a good thing they were better at math than I am. In the traveling exhibit, there is a copy of the Rosetta stone. Imagine NOT having found THAT baby.....stroke of luck, there, what? Or some grave robber wiping his tush with an old piece of papyrus.
Chris who finds mummy horror films too scary to watch

Manitou · Nov 1, 2016

CoyoteChris said:
Spokane is semi arid but we set a one month October record this month of over six inches of rain at the airport. And counting.
I use a rain gage but some days I forget to empty it, record the reading, etc.

I set out a five gallon bucket at the beginning of the month. The bucket now has seven inches of rain in it.

The bucket is 11 inches across at the top and 10 inches at the bottom, inside diameter. It is 14 inches high.

How many inches of rain has fallen into it?

(((10.5)^2 + 10^2) / (2 * 11^2)) * 7 = 6.081611

Mathman · Nov 1, 2016

CoyoteChris said:
I can visualize the tapered sides of the bucket, the 6 inches of water falling through the upper area into the lower area, but is hard to visualize calculating the volume of the tapered bucket and compare that to the volume of water that fell.

Actually, since the tapering is very slight, you wouldn't be too far off just to say "7 inches is 7 inches" and call it a day.

CoyoteChris · Nov 1, 2016

Manitou said:
(((10.5)^2 + 10^2) / (2 * 11^2)) * 7 = 6.081611

What? Not even an integral sign? A closed volume integral? :confused2:
Orthogonal function?
6 1/2 years of college....wasted!

Manitou · Nov 1, 2016

CoyoteChris said:
What? Not even an integral sign? A closed volume integral? :confused2:
Orthogonal function?
6 1/2 years of college....wasted!

It my times it used to be 8th grade math...

CoyoteChris · Nov 1, 2016

Mathman said:
Actually, since the tapering is very slight, you wouldn't be too far off just to say "7 inches is 7 inches" and call it a day.

Now I wish I hadnt missed those days I forgot to record my rain gage. We could indeed have received more rain here than the airport. Or less.

The tapering thing gives me an idea to guess the error. It is easy to calculate the area of the top circle, that lets the rain in.
It is easy to calculate the bottom area and the area of the seven inch mark by measuring the diameter there. With alittle thought about percentages, one may be able to come up with a WAG for
how much rain fell. Sometimes I take the inches and calculate how much rain actually fell on my property which is 15 acres.

CoyoteChris · Nov 1, 2016

Manitou said:
It my times it used to be 8th grade math...

Ouch! I didnt even understand the carrots!

Mrs. P · Nov 1, 2016

It's been a wet month indeed. It's been raining in the Yakima Valley -- but Sunshine today! YAY!

I've been thinking of taking a calculus class. For fun.

Manitou · Nov 1, 2016

CoyoteChris said:
You were 8th grade math or taught 8th grade math....?
Mathman thinks I taught spelling but it really was vocational education.

It's used to be a simple 3D geometry in where I came from and it used to be taught in the 8th grade.

Now it's different. Here it is:
1950: the lumberman sold wood for $100. The labor cost him 4/5 of that amount. How much did he make?
1990: the lumberman sold wood for $100. The labor cost him $80. How much did he make?
2000: the lumberman sold wood for $100. The labor cost him $80, so he made $20. Circle '$20'.
2016: the lumberman sold wood for $100. Color the lumberman.

Mrs. P · Nov 1, 2016

Manitou said:
It's used to be a simple 3D geometry in where I came from and it used to be taught in the 8th grade.

Now it's different. Here it is:
1950: the lumberman sold wood for $100. The labor cost him 4/5 of that amount. How much did he make?
1990: the lumberman sold wood for $100. The labor cost him $80. How much did he make?
2000: the lumberman sold wood for $100. The labor cost him $80, so he made $20. Circle '$20'.
2016: the lumberman sold wood for $100. Color the lumberman.

Okay. While I definitely will criticize the US school system for various reasons. I can assure you that my math problems -- I took algebra -- circa late 1990s involved more than the problem that you listed for 1990s -- in fact I learned that problem in elementary school -- probably around 2nd grade. Also I took geometry in 9th grade, so I wasn't THAT far of from when you took geometry.

Manitou · Nov 1, 2016

Mrs. P said:
Okay. While I definitely will criticize the US school system for various reasons. I can assure you that my math problems -- I took algebra -- circa late 1990s involved more than the problem that you listed for 1990s -- in fact I learned that problem in elementary school -- probably around 2nd grade. Also I took geometry in 9th grade, so I wasn't THAT far of from when you took geometry.

I want to keep this humorous. There is nothing to criticize...

That's the way things go today.
But seriously, when you come back to your original problem you can really see it's simple. The volume of water must stay the same both in your bucket and in the 11 inch diameter cylinder. The volume of a cylinder is equal to radius-squared times two times pi times the height you are looking for. For the bucket it's the average of top and bottom areas times the height, which is 7. Pi gets cancelled, thus the formula.

Mrs. P · Nov 1, 2016

Manitou said:
I want to keep this humorous. There is nothing to criticize... That's the way things go today.
But seriously, when you come back to your original problem you can really see it's simple. The volume of water must stay the same both in your bucket and in the 11 inch diameter cylinder. The volume of a cylinder is equal to radius-squared times two times pi times the height you are looking for. For the bucket it's the average of top and bottom areas times the height, which is 7. Pi gets cancelled, thus the formula.

Probably I think probably the worst criticism about U.S. math, which is in its ability to connect math to real life issues.

CoyoteChris · Nov 1, 2016

Mrs. P said:
It's been a wet month indeed. It's been raining in the Yakima Valley -- but Sunshine today! YAY!

I've been thinking of taking a calculus class. For fun.

I am trying not to hate you, Mrs. P.

I think the whole world is smarter than me... :slink:

CoyoteChris · Nov 1, 2016

Manitou said:
I want to keep this humorous. There is nothing to criticize... That's the way things go today.
But seriously, when you come back to your original problem you can really see it's simple. The volume of water must stay the same both in your bucket and in the 11 inch diameter cylinder. The volume of a cylinder is equal to radius-squared times two times pi times the height you are looking for. For the bucket it's the average of top and bottom areas times the height, which is 7. Pi gets cancelled, thus the formula.

See, that's exactly where I have my head stuck up my bucket. Obviously, if you double the area of a circle, it will double the water. That's an arithmetic relationship. If the bucket were make up of two straight sided sections but different diameters, each of equal height, this would be easy for me to understand. But those darned sides are slanted. And as you increase the diameter of a circle, the area changes geometrically. To put it another way, when you start with the area of the bottom of the bucket, and label it "x", then go up one inch and calculate the area of THAT circle, and call it y, then go up another inch and calculate the area of that circle, and call it z, there is a geometric relationship between your values.

What I thought calculus would do for us is make the bucket an infinate number of increasing diameter sections.

CoyoteChris · Nov 1, 2016

Mathman said:
6.079 inches.

The volume (of a truncated cone) is 577.726 cubic inches, with the rain collected over an area of 95.033 square inches. Divide.

By the way, problems of this sort were considered by the Egyptian mathematician-scribes of dynastic Egypt. The best surviving mathematical manuscript from this period is the Ahmes, or Rhind, Papyrus. Ahmes is the name of the scribe who copied it about 1600 BCE from an original, now lost, written about 200 years earlier. Rhind is the British tourist who bought it from local grave robbers in the 1850s CE, and later gave it to the British museum.

Hummmmm....starting to get visual handle.....
https://rechneronline.de/pi/truncated-cone.php

http://keisan.casio.com/has10/SpecExec.cgi?id=system/2006/1223372110

http://www.aqua-calc.com/calculate/volume-truncated-cone

Mathman · Nov 1, 2016

Manitou said:
(((10.5)^2 + 10^2) / (2 * 11^2)) * 7 = 6.081611

It should be

(((10.5)^2 + 10.5*10 + 10^2) / (3 * 11^2)) * 7 = 6.0792

Manitou said:
For the bucket it's the average of top and bottom areas times the height, which is 7. Pi gets cancelled, thus the formula.

No, it is the average in the sense of the mean value -- that is, the integral. So instead of getting (r[SUB]1[/SUB]^2 + r[SUB]2[/SUB]^2) / 2, you get

(r[SUB]1[/SUB]^3 - r[SUB]2[/SUB]^3) / (3 * (r[SUB]1[/SUB] - r[SUB]2[/SUB])) = (r[SUB]1[/SUB]^2 + r[SUB]1[/SUB]*r[SUB]2[/SUB] +r[SUB]2[/SUB]^2) / 3

Since all these r's are close, the answer is off only in the third decimal place, in this example.

Mathman · Nov 1, 2016

CoyoteChris said:
Obviously, if you double the area of a circle, it will double the water….

That is, it will double the amount of water that passes through in a given amount of time. This is the time derivative of the thing you want, not the thing itself.

...That's an arithmetic relationship. If the bucket were make up of two straight sided sections but different diameters, each of equal height, this would be easy for me to understand. But those darned sides are slanted. And as you increase the diameter of a circle, the area changes geometrically. To put it another way, when you start with the area of the bottom of the bucket, and label it "x", then go up one inch and calculate the area of THAT circle, and call it y, then go up another inch and calculate the area of that circle, and call it z, there is a geometric relationship between your values.

What I thought calculus would do for us is make the bucket an infinate number of increasing diameter sections.

This is correct. On link 3 in post 17 above, scroll down to formula 1 for the volume. This gives the result of the integration, skipping all the calculus.

CoyoteChris · Nov 1, 2016

Mathman said:
6.079 inches.

The volume (of a truncated cone) is 577.726 cubic inches, with the rain collected over an area of 95.033 square inches. Divide.

By the way, problems of this sort were considered by the Egyptian mathematician-scribes of dynastic Egypt. The best surviving mathematical manuscript from this period is the Ahmes, or Rhind, Papyrus. Ahmes is the name of the scribe who copied it about 1600 BCE from an original, now lost, written about 200 years earlier. Rhind is the British tourist who bought it from local grave robbers in the 1850s CE, and later gave it to the British museum.

Ah HA! The lightbulb came on! Here is how I pictured the problem (thanks to Mathman) so I could understand it.
The fact that the numbers made things non messy helped, along with the web calculator.
First, find the volume of the water in the bucket, then using that number, and the diameter of the top "hole", you can back shoot the height of a true cyclinder full of that volume.
The web calculator says the volume of the water in the bucket is indeed like MM says. 577 inches. V = h * π / 3 * (R² + Rr + r²)
So, that much water must have dropped through the top hole, right? So if we visualize a cylinder with a top hole diameter and bottom diameter the same, we can calculate
the true rain fall height, or height of that cylinder.
Cylinder volume is Pi times radius squared times height. We want Height so we divide both sides of the equation by Pi r squared, stick in our values, and voila! 6.07 and change inches.
This is why Mathman gets the big money!:thumbsup:

I should add that the delta r of bucket is indeed arithmetic, top to bottom. So without measuring, we know since the top hole is 5.5 inches and the bottom is 5 inches, the 7 inch point is 5.25.

Help! Mathman!

CoyoteChris

Mathman

CoyoteChris

Manitou

Mathman

CoyoteChris

Manitou

CoyoteChris

CoyoteChris

Mrs. P

Uno, Dos, twizzle!

Manitou

Mrs. P

Uno, Dos, twizzle!

Manitou

Mrs. P

Uno, Dos, twizzle!

CoyoteChris

CoyoteChris

CoyoteChris

Mathman

Mathman

CoyoteChris

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