Math problem where graphing calculator may slow you down - part II.

[CR Haeger]

Hello - here is a problem that should be fairly easily solvable by hand and with a trig table:

Assume f(x) = max(sin(x), cos(x)) using radians. If the area under the curve f(x) = 4.0 and the lower limit of x = 0.0, what is the upper limit of integration?

What do your machines and/or brains come up with?

Note - I got this to work on the HP Prime using the AREA() function in Function APP, but had problems using solve() or fsolve() properly. Num-Solv on a TI36x had to think about this a looong time.

[Marcus von Cube]

(09-12-2014 12:43 PM)CR Haeger Wrote:  If the area under the curve f(x) = 4.0

Is the area positive if the function is negative? Or are we just talking about the definite integral?

EDIT: If the area is always positive, the answer 2π. I split the interval at π/4, π, π+π/4, π+π/2, 2π. All the square roots cancel each other.

[CR Haeger]

(09-12-2014 04:33 PM)Marcus von Cube Wrote:  

(09-12-2014 12:43 PM)CR Haeger Wrote:  If the area under the curve f(x) = 4.0

Is the area positive if the function is negative? Or are we just talking about the definite integral?

EDIT: If the area is always positive, the answer 2π. I split the interval at π/4, π, π+π/4, π+π/2, 2π. All the square roots cancel each other.

Hi Marcus, I should have stated that its a definite integral where area is negative when f(x) is negative. Your solution and abs(f(x)) are interesting too - thanks!

[Thomas Klemm]

My solution by hand: \(2\pi+\arccos(3\sqrt{2}-4)\)
This is about: 7.6089
I used the HP-15C with this program to calculate the function:

Code:

LBL A
1
->R
x<y
x<>y
RTN

And then another program to calculate the integral which I finally used with the solver.

Cheers
Thomas

[Bunuel66]

By hand and a bit of brain....
f(x) is periodic with a 2pi period.
from 0 to pi/4 f(x)=cos(x)
from pi/4 to 5pi/4 f(x)=sin(x)
from 5pi/4 to 2pi f(x)=cos(x)

Integrating is straightforward and gives 4/sqrt(2) over 0 to 2pi
going up to 9pi/4 gives 5/sqrt(2).
Then the value searched for is between 9pi/4 and 9pi/4+pi, in other words on the second interval of the definition of f(x).
This equivalent to search for X with: sum from pi/4 to X of sin(x) dx=4-5/sqrt(2).
Integrating analytically and solving lead to: X=acos(6/sqrt(2)-4)
The solution is then 9pi/4+X-pi/4~7.6089

Nothing more necessary than any standard scientific calculator...
BTW, I'm cheating slightly, I checked the value of the integral with my own python software ;-)
which uses a HP like algorithm....

[Thomas Klemm]

(09-13-2014 09:12 PM)Bunuel66 Wrote:  Integrating is straightforward and gives 4/sqrt(2) over 0 to 2pi

We should make sure that the integral is always < 4. Thus we calculate it up to the first zero of the function \(f(x)\) which occurs at \(x=\pi\):
\[
\int_{0}^{\pi}\max(\sin(x),\cos(x))dx=1+\sqrt{2}\approx 2.41421
\]

Cheers
Thomas

[Bunuel66]

Actually I made the check for the first maximum, I wanted just not to be too long with the explanations, just giving the general idea.

Regards

[Gilles]

I used the numeric solver of the 50G in FIX 4



FIX 6 takes more time and returns
X: 7.608894

This help to get the exact answer with the 50G CAS :
'X=2*PI+ACOS(3*SQRT(2)-4)'

[CR Haeger]

Thanks Giles,

Has anyone tried solving this using the HP Prime CAS, Home or Function APP? It seems to me there are only a couple of ways to find a numeric solution. I have not found any exact solution using the device either.

Screenshot with CAS settings Exact, Complex and Use i unchecked.
   

Graphically, it worked out pretty well, once I figured out AREA() syntax. Note that using integral from template did not work for me.
   
   

[Gilles]

Hi CR Haeger, I tried with the Prime both in CAS and with the SOLVE APPS without success.

I agree with you about the plotter of the Prime : just fabulous !
I like very much to zoom instantly with 2 fingers

[lrdheat]

I'm amazed at the TI 36 Pro. It took 30+ minutes (in numsolve), but it came up with the result!

[lrdheat]

I sped things up to under 15 minutes by doing a couple of experimental integrations to get an idea of where the integral would be ~4, then a quick table to zoom in on an appropriate guess, and then enter a good guess into numsolve.

[lrdheat]

...which leads me to...how to accomplish this on the WP-34S? How do I solve for an integral ?

[Thomas Klemm]

(11-29-2014 03:31 AM)lrdheat Wrote:  how to accomplish this on the WP-34S?

It's similar to what you do using the HP-15C:

Code:

LBL'FX'
1
->REC
MAX
RTN

Code:

LBL'IGR'
0
x<>y
∫'FX'
4
-
RTN

RAD
FIX 4
7.6
7.61
SLV'IGR'

7.6089

Cheers
Thomas

[CR Haeger]

(11-29-2014 12:49 AM)lrdheat Wrote:  I sped things up to under 15 minutes by doing a couple of experimental integrations to get an idea of where the integral would be ~4, then a quick table to zoom in on an appropriate guess, and then enter a good guess into numsolve.

Try turning off pretty print mode and setting the integrals resolution to say 0.1. May speed things up.

I agree - the TI36x is pretty straightforward and capable for these type if solver problems.

[lrdheat]

Thanks Thomas,

It works...couldn't remember how to have an integral in solve. Rather slow on this sort of problem!

[Thomas Klemm]

(11-30-2014 01:33 AM)lrdheat Wrote:  Rather slow on this sort of problem!

For this specific problem the derivative is trivial:

\[
\begin{align}
F(x)&=\int_{0}^{x}f(t)dt-4 \\
F'(x)&=f(x) \\
\end{align}
\]

Thus we can use Newton's method:
\[
\begin{align}
x'&=x-\frac{F(x)}{F'(x)} \\
&=x-\frac{\int_{0}^{x}f(t)dt-4}{f(x)} \\
&=x+\frac{\int_{x}^{0}f(t)dt+4}{f(x)} \\
\end{align}
\]

Instead of starting the integration from \(0\) over and over again we can reuse the result of \(F(x)\) from the previous loop:
\[
F(x')=F(x)+\int_{x}^{x'}f(t)dt
\]
This value is saved in register 01.

Code:

LBL'FX'
# 001
→REC
MAX
RTN

Code:

LBL'NWT'
# 004
STO 01
CLx
STO 00
R↓
LBL 00
ENTER↑
x<> 00
∫'FX'
RCL+ 01
STO 01
RCL 00
XEQ'FX'
/
x≈0?
SKIP 002
RCL+ 00
GTO 00
RCL 00
END

This will speed up the calculation.

Cheers
Thomas

[lrdheat]

Excellent, clear, concise as usual.