2nd order differential equation...anyone?

[Bravus]

Could even solve PDEs (partial differential equations) back in the day...

[Robert]

5 hours ago, David Geelan said:

 

That's right....you are the man!

[Robert]

On 6/12/2016 at 6:38 PM, Robert said:

Hint:  Use the identity, Sin2x = 2SinxCosx

y = Sin(2arcSin(1/7)) 

Let x (an angle) = arcSin(1/7) 

substituting into the above identity:  Sin(2arcSin(1/7)) = 2Sin(arcSin(1/7))Cos(arcSin(1/7))

Okay, first what is Sin(arcSin(1/7)) equal to? 

f(f^-1(x)) = x  (the function of the inverse function = x)

So, Sin(arcSin(1/7)) = 1/7

More later....

I need to get ready for work. 

[Robert]

Secondly we need to solve the other factor:   Cos(arcSin(1/7))

We need to rewrite arcSin(1/7) 

Here it is:     Sin(BAC) = BC:AB  or (1/7)/1   AB being = 1

Now we need to compute the Cos of angle BAC or Cos(BAC) = AC:1

What is AC?  AC^2 + BC^2 = 1, therefore AC = (1 - BC^2)^1/2 = (1 - (1/7)^2)^1/2 or [4(3)^1/2]/7 or .989743319

So the Cos(BAC) = [4(3)^1/2]/7

Rewriting the above we get:  arcCos([4(3)^1/2]/7) = BAC (angle) = Cos(arcSin(1/7))

So Cos(arcSin(1/7)) = Cos(arcCos([4(3)^1/2]/7)) = [4(3)^1/2]/7

Placing all this info back into  Sin(2arcSin(1/7)) = 2Sin(arcSin(1/7)Cos(arcCos[4(3)^1/2]/7) we get:

Sin(2arcSin(1/7)) = 2(1/7)([4(3)^1/2]/7) or .282783805

I need to double check my work at home....

Got to go.

 

[Robert]

7 hours ago, Robert said:

Secondly we need to solve the other factor:   Cos(arcSin(1/7))

We need to rewrite arcSin(1/7) 

Here it is:     Sin(BAC) = BC:AB  or (1/7)/1   AB being = 1

Now we need to compute the Cos of angle BAC or Cos(BAC) = AC:1

What is AC?  AC^2 + BC^2 = 1, therefore AC = (1 - BC^2)^1/2 = (1 - (1/7)^2)^1/2 or [4(3)^1/2]/7 or .989743319

So the Cos(BAC) = [4(3)^1/2]/7

Rewriting the above we get:  arcCos([4(3)^1/2]/7) = BAC (angle) = Cos(arcSin(1/7))

So Cos(arcSin(1/7)) = Cos(arcCos([4(3)^1/2]/7)) = [4(3)^1/2]/7

Placing all this info back into  Sin(2arcSin(1/7)) = 2Sin(arcSin(1/7)Cos(arcCos[4(3)^1/2]/7) we get:

Sin(2arcSin(1/7)) = 2(1/7)([4(3)^1/2]/7) or .282783805

I need to double check my work at home....

Got to go.

 

cheating with my calculator, yes, y = .282 is the answer. 

[Aliensanctuary]

Here's a practical problem to solve:

Create a formula to determine the volume of water in gallons at depth x in a cylindrical tank 8 ft wide and 20 ft long laying on its side.

This is a water tank that is used to store water that supplies three houses on some property that I own. I actually solved the problem years ago, but have since lost the solution. We use a long pole to determine the depth of the water in the tank. I don't think that differential equations are required for the solution, and maybe not even calculus. One would need to calculate the area of a sector of a circle, then compute the volume of the partial cylinder.

From what I recall, calculus can be used to determine the area under a curve in a graph, while differentials can be used to determine the instantaneous slope of a graph.

[Outta Here]

I'd start with using Google, not Calculus. I found this formula:

[Outta Here]

Then I found an Excel template that I would use for my practical answer.

[Bravus]

Hmm, think I'll head home from the office now, but this is a fun kind of problem. I'll be back.

[Aliensanctuary]

21 hours ago, Aliensanctuary said:

From what I recall, calculus can be used to determine the area under a curve in a graph, while differentials can be used to determine the instantaneous slope of a graph.

Oops, I meant integration can be used to determine the area under a graph. Rather than sector, it should be segment of a circle. I should take my brain in to get a lube job due to the build up of rust.

Quote

20 hours ago, Aubrey said:

I'd start with using Google, not Calculus. I found this formula:

 

According to this formula, the terms between the outside brackets must be the area of the segment of the end of the cylinder. Not sure, but this may show the area of the sector minus the area of the triangle above the segment, as shown below.

http://regentsprep.org/Regents/math/geometry/GP16/CircleSectors.htm

Quote

Problem: Find the area of a segment of a circle with a central angle of 120 degrees and a radius of 8  Express answer to nearest integer.

Solution: Start by finding the area of the sector: Now, find the area of the triangle.  Dropping the altitude forms a 30-60-90 degree triangle.  Using trig. (or the 30-60-90 rules), find the altitude, which is 4, and the other leg, which is 6.92820323 (or ). square units

What a person might do is to calculate predetermined depths, say, 1ft, 2ft, 3ft, etc., and have a little cheat sheet with this info to approximate the amount of water remaining in the tank. A little twist in the practical application is that the tank is several inches out of level, being lower at the intake\outgo hole in the bottom.

Now, if the segment were placed on a graph, flat side down, a person could determine the area using integration. I'm not sure what the function would be, maybe related to pi r squared.

[phkrause]

17 minutes ago, Aliensanctuary said:

I should take my brain in to get a lube job due to the build up of rust.

Don't we all! 

[Robert]

On 6/16/2016 at 6:01 AM, Aliensanctuary said:

Here's a practical problem to solve:

Create a formula to determine the volume of water in gallons at depth x in a cylindrical tank 8 ft wide and 20 ft long laying on its side.

So 8 ft diameter by 20 ft in length?

 

[Robert]

1 hour ago, Robert said:

So 8 ft diameter by 20 ft in length?

 

 

 

 

[Robert]

Found mistake

 

 

[Robert]

Okay, found mistake.  Here's the correct formula:

[Robert]

Close up of formula:

[Robert]

Note that if x = 0 Volume = 0

Note that if x =4 (the radius or the diameter divided by 2) the volume = 1/2 pi*(r^2)20 or 1/2 the volume of a cylinder (V=pi(r^2)h) 

The answer is in cubic feet.  1 gallon = 0.133681 cubic ft

So the answer must be multiplied by (1/0.133681)

[Bravus]

I'm imagining it as a cylinder lying on its side, so the 'height' of the cylinder (in this case the length) is 20 ft, and the diameter is 8 ft (so the radius is 4).

My imaginary way of calculating it (I'm a physicist, not a mathematician, so I start by trying to understand the problem first and then using the math as a tool) is as a series of rectangles of increasing area - 20 ft long by (something) wide) - and constant thickness. The thicker the slices the easier the calculation, the thinner the slices the more accurate the calculation. In the limit of an infinite number of slices of infinite thinness is where calculus lives.

We only need to do the bottom half (the first 4 ft), because the top is just the mirror image of that. 

We need an equation for the length of a chord in a circle of radius 4, in terms of how far the chord is from the centre.

The expression is of the form: V = 20 (integral from 4 to 0 (formula for chord length in terms of r).dr

I found the formula 

Chord length

=

2

√

r

2

−

d

2

Where r is the radius and d is the perpendicular distance of the chord from the centre. In our case, r is a constant, 4 ft. We'll make the formula in terms of d.

(Just saw Robert's solution but deliberately didn't look closely: I'll carry on doing it my way and see if we arrive near the same place)

V = 40 (integral from 4 to 0 (sqrt(16-d^2)).dd)

OK, taking it to the paper: this forum is wonderful for many things, but not for math markup! 

 

[Bravus]

Hmm, ended up having to go to Wolfram Integrator online for the integral, because my skills are pretty rusty too, but it's pretty ugly. 

I feel like I'm missing something: I know my expression above simply gives the volume for the bottom half of the tank, which could be got at much more simply:

V = 1/2 * pi *r^2 * h = 1/2 * pi * 16 * 20 = 160 pi ~= 503 cubic feet. I'll leave the translation into gallons for those familar with archaic measurement systems. :-p

I think the integral needs to be, not from 4 to 0 feet, but from 4 to (4-depth) feet, but that just makes it uglier.

[Bravus]

Yes, Robert has done it correctly. It's *inherently* ugly.

[Robert]

On 6/16/2016 at 6:01 AM, Aliensanctuary said:

From what I recall, calculus can be used to determine the area under a curve in a graph, while differentials can be used to determine the instantaneous slope of a graph.

No need of calculus here....

[Bravus]

The way I'd solve it in real life is a lookup table: if the water is x ft deep the volume is y. That could be created empirically, or by doing the calculations (which you'd have a computer do) for a number of values.

Just for funsies, I might see what would happen if I used inch-thick slices instead of calculus: how inaccurate would my answers be?

V = 40 (sqrt(16-d^2)) where d is in feet, but can be measured in inches = feet * 12. That is, if it's 2 feet it's 24 inches.

For the case when d, the distance from the centre, is 0 (i.e. the tank is half full) we get 40 x sqrt(16) = 40 x 4 = 160... we seem to be missing a factor of pi.

[Bravus]

Anyway, I came to the office on a weekend to get some work done, so I'd better do that.

[Robert]

Here's the formula simplified and complete with conversion from cubic feet to gallons:

[Outta Here]

I'd just use this: http://mathcentral.uregina.ca/QQ/database/QQ.09.08/h/tank_calculator.xlsx I'm telling you, it'll save a lot of time!