Calculus problem (differentiation)

A reservoir of square cross-section has sides sloping at an angle
of 45◦ with the vertical. The side of the bottom is 200 feet. Find an
expression for the quantity pouring in or out when the depth of water
varies by 1 foot; hence find, in gallons, the quantity withdrawn hourly
when the depth is reduced from 14 to 10 feet in 24 hours.

No idea how to do this. Any help?

Its page 44 (pdf page 55) on this;
http://www.gutenberg.org/files/33283/33283-pdf.pdf?session_id=ab4a95f4a247411edbb1cd58a0c5f504e547700f

The answer is given but I just don't know where they got the volume from. I cannot picture what the reservoir or the sides look like because if two of the sides are at an angle, then it must be that the other two sides are not squares. In fact, I simply don't understand anything about this question. I can do the differentiation, but I just don't know what the question is asking, nor how to construct the formula.

Any help appreciated.

Reply 1

TeeEm

19

this is a problem on related rates of change (chain rule)

the shape is a frustrum of a square pyramid, although it is not clear whether it slopes inwards (sensible way) or outwards (inverted)

It is messy numbers and my brain has long stalled ... hopefully someone else please...

Reply 2

ghostwalker

17

Original post by djpailo

I just don't know where they got the volume from.

Any help appreciated.

Had a go, and got bogged down; internet to the rescue - see here.

Edit: Given the book was published in 1914, I suspect this would have been covered in the syllabii then, but not so now.

(edited 8 years ago)

Reply 3

DFranklin

18

Original post by djpailo

No idea how to do this. Any help?

Its page 44 (pdf page 55) on this;
http://www.gutenberg.org/files/33283/33283-pdf.pdf?session_id=ab4a95f4a247411edbb1cd58a0c5f504e547700f

The answer is given but I just don't know where they got the volume from. I cannot picture what the reservoir or the sides look like because if two of the sides are at an angle, then it must be that the other two sides are not squares. In fact, I simply don't understand anything about this question. I can do the differentiation, but I just don't know what the question is asking, nor how to construct the formula.

Any help appreciated.

As others have said, this is all a bit of a mess.

It would appear the intent is for the volume to be a truncated, inverted square pyramid, so at a height 'h' above the bottom of the reservoir, the cross section is a square with sides 200+2h.

But the answer in the book assumes you know the formula for the volume of such a shape, and at that point you might as well just find the two volumes to find the average flow anyhow.

Reply 4

0le

OP

21

Original post by DFranklin

As others have said, this is all a bit of a mess.

It would appear the intent is for the volume to be a truncated, inverted square pyramid, so at a height 'h' above the bottom of the reservoir, the cross section is a square with sides 200+2h.

But the answer in the book assumes you know the formula for the volume of such a shape, and at that point you might as well just find the two volumes to find the average flow anyhow.

I understand the 200 + 2h part now, but I'm still confused why they substituted h=12 into dV/dh rather than h=4 which gives the change in flow rate. From the next bit, I also don't understand where they got 6.25 from nor why they randomly decide to multiply by 4 here.

I thought dV/dh simply means the change of volume over the change in height. In other words, if the height changes by a certain amount, how much volume of water changes. So they clearly ask for a change of height of 4, why then do they substitute h=12 :s

Reply 5

ghostwalker

17

Original post by djpailo

but I'm still confused why they substituted h=12 into dV/dh rather than h=4 which gives the change in flow rate.

Depth has gone from 14 to 10 feet, so they've used h=12, the average depth to get a value for dV/dh. This is used as an approximation for dV/dh over the whole interval h=10 to h=14.

From the next bit, I also don't understand where they got 6.25 from nor why they randomly decide to multiply by 4 here.

They've gone from cubic feet to gallons - conversion factor is....

Multiply by 4 to get the total volume in 4 feet.

Reply 6

DFranklin

18

Original post by ghostwalker

They've gone from cubic feet to gallons - conversion factor is....

Multiply by 4 to get the total volume in 4 feet.

And then thank God that these days we use metric!

Reply 7

0le

OP

21

Original post by ghostwalker

Depth has gone from 14 to 10 feet, so they've used h=12, the average depth to get a value for dV/dh. This is used as an approximation for dV/dh over the whole interval h=10 to h=14.

They've gone from cubic feet to gallons - conversion factor is....

Multiply by 4 to get the total volume in 4 feet.

I still don't get why it isn't 4. :s dV/dh is the change in volume of water with height. The height hasn't changed by 12 feet. Its changed by 4 feet and dh is the change in height :s

(edited 8 years ago)

Reply 8

ghostwalker

17

Original post by djpailo

I still don't get why it isn't 4. :s dV/dh is the change in volume of water with height. The height hasn't changed by 12 feet. Its changed by 4 feet and dh is the change in height :s

dV/dh is the rate of change of volume. You need the rate of change of volume over the interval when h=14 to h=10. They've used the rate of change of volume when h is 12, being half way between 10 and 14, as an approximation for the whole interval.

Actual change in volume is approximately "rate of change for the average height (uses h=12)" times "change in height (and this is where the 4 comes in)".

Reply 9

ghostwalker

17

Original post by DFranklin

And then thank God that these days we use metric!

Certainly makes life easier now, but really throws students studying older works; and pity the americans.

Now was that an Imperial Gallon, or a US Gallon...?

And why is my space telescope (Hubble) out of focus?

Gotta laugh.

That said, in days of yore, mathematicans would have been more aware of what units they were working in, as a matter of course.

Reply 10

0le

OP

21

Original post by ghostwalker

dV/dh is the rate of change of volume. You need the rate of change of volume over the interval when h=14 to h=10. They've used the rate of change of volume when h is 12, being half way between 10 and 14, as an approximation for the whole interval.

Actual change in volume is approximately "rate of change for the average height (uses h=12)" times "change in height (and this is where the 4 comes in)".

I see. So they've evaluated the rate of change at h=12, whereas if you were to put h=4, it would be evaluating the gradient when f(h) = f(4), which would be a different part of the curve. Thanks, I'm moving onto the next parts of the book. The plan is to have this and some basic statistics done by around December and then move onto multi-variable calculus and then vector calculus.

(edited 8 years ago)

Reply 11

Mathcity23

1

Could someone explain the (200 + 2h) part?? If anyone could post a picture that'd be great. I just don't get how the area of the top of the frustum is larger than the bottom....

(edited 5 years ago)

Reply 12

ghostwalker

17

Original post by Mathcity23

Could someone explain the (200 + 2h) part?? If anyone could post a picture that'd be great. I just don't get how the area of the top of the frustum is larger than the bottom....

Seems to be well explained if you read through the thread.

And I got to wonder why your first post on TSR is to resurrect a 3 year old thread, that refers to a book what was published over 100 years ago!

Calculus problem (differentiation)

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