Off-Beat 2/11/2018 6:19 AM
Any Maths wizards in the house?
Im helping my son with some engineering maths at home before he starts his course. (when i say help i basically try to learn it then teach him as I suck at maths) cosh^2(x)+5sinh(x)+6=1+sinh^2(x) Solve for x Trying to have a go at this altho Im struggling. From using microsoft maths it shows the answer as x = 0.864726118559. But I cant seem to get to that answer. Any help would be much appreciated!
 
J M Fahey 2/11/2018 7:50 AM
My Calculus is very rusty, learnt it 48 years ago an slowly lost it, same as German and Russian languages, for lack of everyday use. But in any case: neither MM nor any other software will help him, except as an after the fact results checker. "A number" is NOT enough, even if accurate, Teachers want him to go through all the hoops and explain them how he reached the result.
 
Off-Beat 2/11/2018 7:54 AM
Yeah thats it, Im trying to do the working out however i keep getting it wrong, so im wondering if anyone on here could figure it out. Maybe i should go back to school ha!
 
Enzo 2/11/2018 10:20 AM
Read his text. I was a physics major in college, took all the calculus and other math, had the algebra and trig and such in high school. I remember liking diffy-Q (differential equations), but don;t remember a thing about it. College was over 50 years ago. Never used the stuff once I left college. Wish I could help.
 
SoulFetish 2/12/2018 7:21 PM
For what it's worth, you (or he) might find this book helpful: [IMG]https://s3.amazonaws.com/images.addoway.com/items/3448/4645726/3448_1_66625d.png[/IMG] From the Preface: "Quick Calculus should teach you the elementary techniques of differential and integral calculus with a minimum of wasted effort on your part; it is designed for you to study by yourself Since the best way for anyone to learn calculus is to work problems, we have included many problems in this book. You will always see the solution to your problem as soon as you have finished it, and what you do next will depend on your answer. A correct answer generally sends you to new material, while an incorrect answer sends you to further explana*tions and perhaps another problem."
 
The Dude 2/12/2018 8:00 PM
[QUOTE=Off-Beat;479815]...... cosh^2(x)+5sinh(x)+6=1+sinh^2(x)......[/QUOTE] Looks like authentic frontier gibberish to me. ;)
 
Gnobuddy 2/12/2018 9:50 PM
[QUOTE=Off-Beat;479815] cosh^2(x)+5sinh(x)+6=1+sinh^2(x) Solve for x [/QUOTE] There is actually no need for calculus to solve this, you just need a little algebra, and a couple of mathematical identities connecting hyperbolic trigonometric functions. Here we go: Step 1) Bring sinh^2(x) to left side, move +6 to right side: => [cosh^2(x) - sinh^2(x)] + 5 sinh(x) = 1-6 The term in square brackets is equal to 1 (this is a mathematical identity, [cosh^2(x) - sinh^2(x)] = 1, see here: [url]https://brilliant.org/wiki/hyperbolic-trigonometric-functions/[/url] ) So now we have: 1 + 5 sinh(x) = -5 Step 2) Move the +1 to the right hand side => 5 sinh(x) = -5-1 => 5 sinh(x) = -6 Step 3) Divide both sides by 5 => sinh(x) = -6/5 => sinh(x) = -1.2 Sinh(x) is a well-behaved function, and therefore has a well-defined inverse function. So all we have to do operate on both sides of this last equation with sinh[SUP]-1[/SUP](x). Step 4) Take sinh[SUP]-1[/SUP]() of both sides: sinh[SUP]-1[/SUP][sinh(x)] = sinh[SUP]-1[/SUP](-1.2) By definition of the inverse function, the left hand side of the equation is now just (x), so we have the result we want: Step 5: [COLOR="#006400"][SIZE=3]x = sinh[SUP]-1[/SUP](-1.2)[/SIZE][/COLOR] If I was your son's teacher, I would happily accept this answer and give him full marks, but just in case his actual teacher wants a numerical answer, we can take this a little further. Step 6: The sinh[SUP]-1[/SUP](x) function can be written explicitly as ( see [url=http://wwwf.imperial.ac.uk/metric/metric_public/functions_and_graphs/hyperbolic_functions/inverses.html]Hyperbolic Functions: Inverses[/url] ): sinh[SUP]-1[/SUP](x) = ln[x + square_root(x[SUP]2[/SUP] + 1)] (This forum doesn't seem to support a square-root sign, so I wrote "square_root"; and "ln" is the natural log, i.e. log to the base "e") So we can now use this form of sinh[SUP]-1[/SUP](x) to evaluate the equation we found in step 5: [COLOR="#006400"][SIZE=3] x = ln{-1.2 + square_root[(-1.2)[SUP]2[/SUP] + 1]}[/SIZE][/COLOR] Everything on the right hand side can be found with an ordinary scientific calculator (or the Internet). Solving, I get: [COLOR="#006400"][SIZE=4]x = -1.015973134[/SIZE][/COLOR] Hope that helps! -Gnobuddy P.S. I don't get the same answer as Microsoft Maths, so your son still has some detective work to do. If I made an algebra mistake, maybe he can find and fix it. But he should be able to follow the same solution method I used.
 
SoulFetish 2/13/2018 12:50 PM
[QUOTE=Gnobuddy;479990] ...a couple of mathematical identities connecting hyperbolic trigonometric functions. .[/QUOTE] Took the words right out of my mouth. ;) What I want to know is, and how did you format the text To display postscripts & subscripts?
 
Gnobuddy 2/13/2018 4:42 PM
[QUOTE=SoulFetish;480035] What I want to know is, and how did you format the text To display postscripts & subscripts? [/QUOTE] If you click the "Quote" and then "Go Advanced" buttons, you get little icons that let you turn a bit of highlighted text into a subscript or a superscript (see attached screen capture). I didn't know they were there until last night, either! :) -Gnobuddy
 
Steve A. 2/15/2018 12:50 PM
[QUOTE=Gnobuddy;479990]There is actually no need for calculus to solve this, you just need a little algebra, and a couple of mathematical identities connecting hyperbolic trigonometric functions.[/quote] I would trust your work more than Microsoft Maths whatever that may be. I would imagine that a decent scientific calculator would be able to handle most if not all of the computations. FWIW I was just looking at calculator apps yesterday and found some that can emulate "real" calculators from HP or TI although you do need to download the ROM from an actual calculator as THAT is protected by copyrights (the emulators are perfectly legal.) On a 7 or 8 inch tablet those emulators should really rock! Steve A. EDIT here is a full-size image of the screen capture you posted on subscripts and superscripts... Thanks for educating us! [img]http://music-electronics-forum.com/attachments/47089d1518565348-subscripts_superscripts.png[/img] P.S. I found those two options after pressing "Go Advanced"...
 
Gnobuddy 2/15/2018 2:03 PM
[QUOTE=Steve A.;480252] I would trust your work more than Microsoft Maths [/quote] Well, I trust Microsoft about as far as I could throw the proverbial elephant. :D But I didn't check my answer (easy enough to do, simply back-substitute the number I got into the original equation, online, or with a calculator that does include hyperbolic trig functions.) So it's always possible I made a mistake. [QUOTE=Steve A.;480252] I would imagine that a decent scientific calculator would be able to handle most if not all of the computations. [/quote] My trusty old Casio fx260 Solar doesn't include any hyperbolic trig functions, but I wouldn't be surprised if some newer and more powerful calculators do. Sharp has some very nice engineering-oriented calculators that have become standards in many university engineering courses here in British Columbia. The web link I included in my answer lets you evaluate sinh[SUP]-1[/SUP](x) using log, square, and square root functions, which even the most basic scientific calculators do have. :) -Gnobuddy
 
potatofarmer 2/15/2018 2:22 PM
Wolfram Alpha makes for a pretty great answer check too: [url]https://www.wolframalpha.com/[/url] [url]https://www.wolframalpha.com/input/?i=cosh%5E2(x)%2B5sinh(x)%2B6%3D1%2Bsinh%5E2(x[/url])
 
Gnobuddy 2/15/2018 10:27 PM
[QUOTE=potatofarmer;480262] Wolfram Alpha makes for a pretty great answer check too: [url]https://www.wolframalpha.com/[/url] [/QUOTE] Thanks for the reminder! I've used Alpha before. I also remember when it was new, and Stephen Wolfram published a book that more or less claimed that his new software basically made every other mathematical approach ever used on earth obsolete and worthless. :) Wolfram's hyperbole aside, it is a great tool. And it seems Wolfram Alpha agrees with my step 5 equation (screenshot attached.) -Gnobuddy