Learning How to Use the Prime G2 - Hallway Pole Problem - In Three Parts
09-02-2019, 11:42 PM (This post was last modified: 09-03-2019 06:30 AM by jlind.)
Post: #3
 jlind Junior Member Posts: 35 Joined: Aug 2019
Learning How to Use the Prime G2 - Hallway Pole Problem - Part III of III
Note:
There’s a limit to the number of images allowed in a posting. I’ve had to break this up into three parts to cope with that . . .
This is Part III of III

At the end of Part II, I had solved for the root of the first derivative and found the angle.

From there, it's a simple matter to type in F1(X) on the entry line.

This uses the angle, which is now stored in "X", it finds the longest pole length, c ~= 20.81 feet, which matches what was found using the first two methods.

While it's a certainty this is the answer from all that's been calculated and graphed thus far, showing that it's a local minima needs the second derivative, F''(X). To find it, the same process as the first derivative is used on it to get the second derivative.

The second derivative is quite messy, but that's OK as it's only needed with the angle found to see if it's positive, negative or zero. I believe some of that messiness is using degrees instead of radians. It will be a "plug and chug" calculation:
In using the second derivative test:
• A postiive number shows a local minimum
• A negative number shows a local maximum
• Zero shows an inflection point, which is neither a minimum nor a maximum, but a point at which the direction of curvature changes direction.

As with F'(x) before, F''(X) is imported from CAS into the Symbolic view, and it's stored in F3(X).

Now I can go back to the CAS view and enter in F3(X) to show that it's positive for the critical point found. I can also put in F2(X) to show the slope at the local minimum. It should theoretically be zero, but ~2.1E-12 is close enough, likely due to (cumulative) rounding errors with the approximation of the local minimum.

Finally, all three functions, F(x) in blue, F'(X) in red, and F''(X) in green graphed with the F(x) minimum marked. The curve for F'(X) shows the intersection, albeit muddled with the X-axis tick marks. Looking at it almost obviates bothering with F''(X) test as its slope is always increasing; it never changes direction even though its slope is very shallow around the F(X) minimum.

Epilogue:
Finally, a word about the ceiling and tilting the pole. Doing so changes the problem from moving a line around the corner to moving a vertical triangle around the corner. The pole is now a hypotenuse on that triangle. Its horizontal leg is the hypotenuse, "c" in the problem presented. For a uniform height ceiling (I put it at the common eight feet found in many homes), the vertical leg never changes. So, if one is able to tilt the pole, with an eight foot high ceiling, the longest pole becomes:

p^2 = c^2 + 8^2
p = (~20.82^2 + 64)^0.5
p ~= 22.3 feet

Only about 18 inches longer. Less than a 10% increase; not as much as most would think. The Doubting Thomas types are invited to work it with an equation that tilts the pole from floor to ceiling (which I've done) to verify it's a trivial add-on.

Had to wander in the wilderness of the Prime's documentation for a while to get a handle on how to use it for problems like this one. Reminded me of how much I hate PDFs for documentation. Give me a printed book. Much easier to for me use. Spent a considerable amount of time experimenting with various menus and the tools residing in them. The Zeros function ran me around in circles for a while trying to figure out how to limit the domain it would use to 0-90 degrees (0 to Pi/2). If there's a disappointment, it's in HP's documentation and the lack of real problem examples. Contains a ton of information, but found it quite difficult to use. Hope walking through the Hallway Pole problem and how I worked it is of help to someone else coping with the HP Prime's learning curve.

Normally I'd work a problem like this using the angle in radians, not in degrees, and leave the calculator set to radians, but most people relate better to angles in degrees, without having to remember 45 degrees is roughly 0.785 radians.

John
Footnote:
I've proofread this several times but I'm not perfect. If I find an error, I'll do an edit to fix it.

John

Pickett: N4-ES, N600
TI: 58, 30-III, 30x Pro MathPrint, 36x Solar, 85, 86, 89T, Voyage 200, Nspire CX II CAS
HP: 50g, Prime G2, DM42
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 Messages In This Thread Learning How to Use the Prime G2 - Hallway Pole Problem - In Three Parts - jlind - 09-02-2019, 11:32 PM Learning How to Use the Prime G2 - Hallway Pole Problem - Part II of III - jlind - 09-02-2019, 11:38 PM Learning How to Use the Prime G2 - Hallway Pole Problem - Part III of III - jlind - 09-02-2019 11:42 PM RE: Learning How to Use the Prime G2 - Hallway Pole Problem - In Three Parts - teerasak - 09-03-2019, 07:23 AM RE: Learning How to Use the Prime G2 - Hallway Pole Problem - In Three Parts - JoJo1973 - 09-03-2019, 06:00 PM RE: Learning How to Use the Prime G2 - Hallway Pole Problem - In Three Parts - jlind - 09-04-2019, 02:42 AM RE: Learning How to Use the Prime G2 - Hallway Pole Problem - In Three Parts - Albert Chan - 09-04-2019, 12:53 AM RE: Learning How to Use the Prime G2 - Hallway Pole Problem - In Three Parts - teerasak - 09-04-2019, 04:10 AM RE: Learning How to Use the Prime G2 - Hallway Pole Problem - In Three Parts - jlind - 09-04-2019, 09:17 AM RE: Learning How to Use the Prime G2 - Hallway Pole Problem - In Three Parts - jlind - 09-12-2019, 04:03 PM RE: Learning How to Use the Prime G2 - Hallway Pole Problem - In Three Parts - teerasak - 09-13-2019, 04:16 PM RE: Learning How to Use the Prime G2 - Hallway Pole Problem - In Three Parts - Thomas Klemm - 07-12-2022, 12:29 AM RE: Learning How to Use the Prime G2 - Hallway Pole Problem - In Three Parts - jte - 08-12-2022, 10:07 PM

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