Little Math Question

Nil_IQNil_IQ Join Date: 2003-04-15 Member: 15520Members
edited January 2004 in Off-Topic
<div class="IPBDescription">Exam tommorow :O</div> Yes I know there are probably better places to ask this (in fact I did try a Yahoo maths chatroom, only to have my question responded to by "A/S/L" "You male or female?" "A/S/L"), but I know you are all a (mostly) intellegent group of people, so you should find this one a doddle.

It's more of a general question than a specific one, I need to know how the differential of an equation relates to it's minimum value. For example,

A ship is to make a voyage of 100km at a constant speed of vkm/h.

The running cost of the ship is £(0.8v^2 + 2000/v) per hour.

(I) Write down the time taken to go 100km at vkm/h (100/v, easy)

(II) Hence write down the total cost of travelling 100km at vkm/h (80v + 200,000/v^2)

(III) By considering dC/dV, find the speed which keeps the cost of the journey to a minimum (to nearest km/h)

So... how do I do it? If I did it right, dC/dv is 80-400,000^-3 (I took 200,000/v^2 to be 200,000v^-2, is that right?)

But the value of 80-400,000^-3 is just 80, which is NOT the answer.

Please help a maths nub.

*edit* I'm starting to realise the error lies in the fact that I took 200,000/v^2 to be the same as 200,000v^-2. I thought the rules were 1/x = x^-1, and 2/x = 2x^-1 ? Am i wrong? *edit*

(Feel free to delete this once I get a correct answer, and I apologise for using the word "doddle")

Comments

  • Phoenix_SixPhoenix_Six Join Date: 2003-11-10 Member: 22442Members
    edited January 2004
    the 80 should be a 1.6v if I'm not mistaken.

    EDIT: Beh, I was differentiating the initial cost equation, and looked at your answer further down, assuming it was the same equation - I thought you had differentiated the term 0.8v^2 as 80.
    <!--emo&;)--><img src='http://www.unknownworlds.com/forums/html/emoticons/wink.gif' border='0' style='vertical-align:middle' alt='wink.gif'><!--endemo-->
  • Nil_IQNil_IQ Join Date: 2003-04-15 Member: 15520Members
    edited January 2004
    <!--QuoteBegin--Phoenix Six+Jan 13 2004, 03:09 PM--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> (Phoenix Six @ Jan 13 2004, 03:09 PM)</td></tr><tr><td id='QUOTE'><!--QuoteEBegin--> the 80 should be a 1.6v if I'm not mistaken. <!--QuoteEnd--></td></tr></table><span class='postcolor'><!--QuoteEEnd-->
    Um... huh? You differentiate 80v and you get 80 surely? Could you explain <i>which</i> 80 and why it should be 1.6v?

    *edit* The "^" symbol means "to the power of"
  • BogglesteinskyBogglesteinsky Join Date: 2002-12-24 Member: 11488Members
  • MadCowMadCow Join Date: 2003-02-10 Member: 13398Members
    I'll take a crack at this.... gimme a minute
  • MadCowMadCow Join Date: 2003-02-10 Member: 13398Members
    2/X is NOT X^-2 it's 2X^-1
  • Nil_IQNil_IQ Join Date: 2003-04-15 Member: 15520Members
    edited January 2004
    <!--QuoteBegin--Z.X. Bogglesteinsky+Jan 13 2004, 03:43 PM--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> (Z.X. Bogglesteinsky @ Jan 13 2004, 03:43 PM)</td></tr><tr><td id='QUOTE'><!--QuoteEBegin--> 2/X is X^-2 <!--QuoteEnd--></td></tr></table><span class='postcolor'><!--QuoteEEnd-->
    No it isn't. You try it.

    2/10 = 0.2

    10^-2 = 0.01

    2/X is 2x^-1 afaik.

    *edit* Thank you madcow. Please don't confuse me on the night before my exam boggle <!--emo&:)--><img src='http://www.unknownworlds.com/forums/html/emoticons/smile.gif' border='0' style='vertical-align:middle' alt='smile.gif'><!--endemo--> *edit*
  • BogglesteinskyBogglesteinsky Join Date: 2002-12-24 Member: 11488Members
    edited January 2004
    so it is.

    thanks for clearing that one

    (how could i be so stupid...)
  • MadCowMadCow Join Date: 2003-02-10 Member: 13398Members
    edited January 2004
    OK first:
    I believe when you typed this out you put down the wrong answer for #2. From your work on #3 it seems like you got the right idea though.
    I believe that should be 80v + 200,000/(v^2) or 80v + 200,000v^-2.

    Ok for number 3.

    C = 80v + 200,000v^-2

    Differentiate that and we get:

    dC/dv = 80 - 400,000v^-3

    I'm thinking that the lowest cost is the zero of this function, which is 17.099759 or about 17.
    I'm pretty sure that this is right now. I think you differentiated 400,000v^-2 and got 400,000^-3. The v needs to stay in that equtation <!--emo&::nerdy::--><img src='http://www.unknownworlds.com/forums/html/emoticons/nerd.gif' border='0' style='vertical-align:middle' alt='nerd.gif'><!--endemo-->
  • MrPinkMrPink Join Date: 2002-05-28 Member: 678Members
  • ZigZig ...I am Captain Planet&#33; Join Date: 2002-10-23 Member: 1576Members
  • BogglesteinskyBogglesteinsky Join Date: 2002-12-24 Member: 11488Members
    <!--QuoteBegin--MadCow+Jan 13 2004, 09:55 PM--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> (MadCow @ Jan 13 2004, 09:55 PM)</td></tr><tr><td id='QUOTE'><!--QuoteEBegin--> I'm thinking that the lowest cost is the zero of this function, which is 17.099759 or about 17.
    I'm pretty sure that this is right now. I think you differentiated 400,000v^-2 and got 400,000^-3. The v needs to stay in that equtation <!--emo&::nerdy::--><img src='http://www.unknownworlds.com/forums/html/emoticons/nerd.gif' border='0' style='vertical-align:middle' alt='nerd.gif'><!--endemo--> <!--QuoteEnd--> </td></tr></table><span class='postcolor'> <!--QuoteEEnd-->
    Don't you need to identify the turning points and stuff?
  • chis1chis1 Join Date: 2004-01-13 Member: 25281Members
    i think ill get a course in maths <!--emo&:(--><img src='http://www.unknownworlds.com/forums/html/emoticons/sad.gif' border='0' style='vertical-align:middle' alt='sad.gif'><!--endemo--> i cant even do the simplest of sums
  • PithlitPithlit Join Date: 2003-05-07 Member: 16120Members, NS1 Playtester, Constellation
    Will me get Pei for the answer?


    pei
  • MadCowMadCow Join Date: 2003-02-10 Member: 13398Members
    <!--QuoteBegin--Z.X. Bogglesteinsky+Jan 13 2004, 04:20 PM--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> (Z.X. Bogglesteinsky @ Jan 13 2004, 04:20 PM)</td></tr><tr><td id='QUOTE'><!--QuoteEBegin--> <!--QuoteBegin--MadCow+Jan 13 2004, 09:55 PM--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> (MadCow @ Jan 13 2004, 09:55 PM)</td></tr><tr><td id='QUOTE'><!--QuoteEBegin--> I'm thinking that the lowest cost is the zero of this function, which is 17.099759 or about 17.
    I'm pretty sure that this is right now.  I think you differentiated 400,000v^-2 and got 400,000^-3.  The v needs to stay in that equtation  <!--emo&::nerdy::--><img src='http://www.unknownworlds.com/forums/html/emoticons/nerd.gif' border='0' style='vertical-align:middle' alt='nerd.gif'><!--endemo--> <!--QuoteEnd--></td></tr></table><span class='postcolor'><!--QuoteEEnd-->
    Don't you need to identify the turning points and stuff? <!--QuoteEnd--> </td></tr></table><span class='postcolor'> <!--QuoteEEnd-->
    The zero of the differentiated function IS the turning point of the original function.
  • Nil_IQNil_IQ Join Date: 2003-04-15 Member: 15520Members
    edited January 2004
    <!--QuoteBegin--MadCow+Jan 13 2004, 04:28 PM--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> (MadCow @ Jan 13 2004, 04:28 PM)</td></tr><tr><td id='QUOTE'><!--QuoteEBegin--> <!--QuoteBegin--Z.X. Bogglesteinsky+Jan 13 2004, 04:20 PM--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> (Z.X. Bogglesteinsky @ Jan 13 2004, 04:20 PM)</td></tr><tr><td id='QUOTE'><!--QuoteEBegin--> <!--QuoteBegin--MadCow+Jan 13 2004, 09:55 PM--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> (MadCow @ Jan 13 2004, 09:55 PM)</td></tr><tr><td id='QUOTE'><!--QuoteEBegin--> I'm thinking that the lowest cost is the zero of this function, which is 17.099759 or about 17.
    I'm pretty sure that this is right now.  I think you differentiated 400,000v^-2 and got 400,000^-3.  The v needs to stay in that equtation  <!--emo&::nerdy::--><img src='http://www.natural-selection.org/forums/html/emoticons/nerd.gif' border='0' style='vertical-align:middle' alt='nerd.gif'><!--endemo--> <!--QuoteEnd--></td></tr></table><span class='postcolor'><!--QuoteEEnd-->
    Don't you need to identify the turning points and stuff? <!--QuoteEnd--></td></tr></table><span class='postcolor'><!--QuoteEEnd-->
    The zero of the differentiated function IS the turning point of the original function. <!--QuoteEnd--></td></tr></table><span class='postcolor'><!--QuoteEEnd-->
    He's right you know. Mad Cow, you are right in every respect.

    Yes, the answer to (II) should have been 80v + 200,000/v^2

    Yes, I did forget the v. I wrote it pretty small and didn't notice it next to all those zeros....

    And finally, yes, the answer is 17.

    Want to have a go at the last part?

    (IV) Find the minimum cost of the voyage.

    *edit* Arg, ok, so I have 400,000v^-3 = 80, then I can divide to get v^-3 = 0.0625, but if I take the -3rd root, I don't get 17, I get -0.3968. What am I doing wrong? *edit*
  • MadCowMadCow Join Date: 2003-02-10 Member: 13398Members
    <!--QuoteBegin--Nil_IQ+Jan 13 2004, 04:35 PM--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> (Nil_IQ @ Jan 13 2004, 04:35 PM)</td></tr><tr><td id='QUOTE'><!--QuoteEBegin--> *edit* Arg, ok, so I have 400,000v^-3 = 80, then I can divide to get v^-3 = 0.0625, but if I take the -3rd root, I don't get 17, I get -0.3968. What am I doing wrong? *edit* <!--QuoteEnd--> </td></tr></table><span class='postcolor'> <!--QuoteEEnd-->
    80/400,000 = .0002

    .0002^-1 = 5000

    5000^(1/3) = 17.0997

    That's the way I worked it at least.
  • Nil_IQNil_IQ Join Date: 2003-04-15 Member: 15520Members
    I probably won't have time to read your reply tonight, but why did you do 0.0002^-1?

    Surely that would just give v^-4 = 2^-5?

    And where did the ^(1/3) come from?

    I know you're right, I just can't see the logic behind it.
  • MadCowMadCow Join Date: 2003-02-10 Member: 13398Members
    edited January 2004
    Alright, I'll do both sides of the equation, that should make it more clear.

    400,000v^-3 = 80

    v^-3 = 80/400,000

    v^-3 = .0002

    (v^-3)^-1 = .0002^-1

    v^3 = 5000

    (v^3)^(1/3) = 5000^(1/3)

    v = 17.0997

    I raised everything to the -1 to get rid of the negative power, when working problems like this I always try to get rid of negatives because they usually end up messing me up somewhere. This is just the way I think and therefore the way that I solved the problem. Raising a number to 1/3 power is like finding the 3rd root, just like raising a number to the 1/2 power is like finding the square root.

    <!--QuoteBegin--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> </td></tr><tr><td id='QUOTE'><!--QuoteEBegin--> I probably won't have time to read your reply tonight, but why did you do 0.0002^-1?
    Surely that would just give v^-4 = 2^-5?<!--QuoteEnd--></td></tr></table><span class='postcolor'><!--QuoteEEnd-->

    From the above you can see that I did (v^-3)^-1. To simplify this you multiply the exponents so we have -3 * -1 = 3.
  • AngelusAngelus Join Date: 2002-12-14 Member: 10810Members, Constellation
    from what i can see, the (^1/3) was there to remove the (^3) part, so you got v on its own....

    Doesnt finding the minimum invole the second differential... or am i thinking about whether its a min or max???

    this actually got me thinking, but i dont know how close this is to the maths exam im doing... What country you in (if UK what yr/lvl/module you doing)... spose i better get some sleep... i got 2 maths exams tomorrow..... :/
  • ZaphodZaphod Join Date: 2003-12-30 Member: 24882Members
    Second derivative would be needed to figure out whether the point you found was a maximum or minimum. The 2nd derivative tells you concavity. So if the value of the 2nd derivative at a point is positive, then it is concave up (like a bowl) and the point would be a minimum (bottom of the bowl). If the value is negative, then it is concave up (hill) and the point is a maximum (top of the hill). Anywho..

    So for the last part of your prob, you found the speed to mimimize cost (17.1 km/hr), and you just plug it back into your equation for total cost ie:
    C(v) = 80*v + 200,000*v^-2 = 80*(17.1) + 200,000*(17.1)^-2 = 2051 (i think, i dont have a decent calculator around right now)
  • Har_Har_the_PirateHar_Har_the_Pirate Join Date: 2003-08-10 Member: 19388Members, Constellation
    thats freaking awsome, im gonna get u guys to do my homework sometime!!
  • MulletMullet Join Date: 2003-04-28 Member: 15910Members, Constellation
  • Phoenix_SixPhoenix_Six Join Date: 2003-11-10 Member: 22442Members
    <!--QuoteBegin--Mullet+Jan 14 2004, 12:41 AM--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> (Mullet @ Jan 14 2004, 12:41 AM)</td></tr><tr><td id='QUOTE'><!--QuoteEBegin-->math blows<!--QuoteEnd--></td></tr></table><span class='postcolor'><!--QuoteEEnd-->
    <!--QuoteBegin--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> </td></tr><tr><td id='QUOTE'><!--QuoteEBegin--><i>Mathematics is the supreme intellectual achievement and the most original creation of the human spirit</i> - Morris Kline.<!--QuoteEnd--></td></tr></table><span class='postcolor'><!--QuoteEEnd-->
    <!--QuoteBegin--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> </td></tr><tr><td id='QUOTE'><!--QuoteEBegin--><i>Mathematics is the Queen and servant of the Sciences</i> - Karl Friedrich Gauss<!--QuoteEnd--></td></tr></table><span class='postcolor'><!--QuoteEEnd-->
    <!--QuoteBegin--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> </td></tr><tr><td id='QUOTE'><!--QuoteEBegin--><i>Mathematics is just more and more complicated ways of saying 'Pigs is pigs'</i> - Bertrand Russell<!--QuoteEnd--></td></tr></table><span class='postcolor'><!--QuoteEEnd-->
    To each his own I guess <!--emo&:D--><img src='http://www.unknownworlds.com/forums/html/emoticons/biggrin.gif' border='0' style='vertical-align:middle' alt='biggrin.gif'><!--endemo-->
  • NuketheplaceNuketheplace Join Date: 2002-09-02 Member: 1266Members
    <!--QuoteBegin--MrPink+Jan 13 2004, 01:15 PM--></span><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td><b>QUOTE</b> (MrPink @ Jan 13 2004, 01:15 PM)</td></tr><tr><td id='QUOTE'><!--QuoteEBegin--> The answer is 69 <!--QuoteEnd--> </td></tr></table><span class='postcolor'> <!--QuoteEEnd-->
    Wrong the answer is 42.

    I'm still waiting for one of my math problems to equal 42 or 133.7 or 13.37 or 1.337... you get the picture. I will be a very happy man when I notice I have a l33t answer.

    Oh and yes I should just post my math homework here some time. It would get things done much faster. Better yet we need a homework forum. It would be fun for all.
  • MadCowMadCow Join Date: 2003-02-10 Member: 13398Members
    I don't think "doing" homework is a good idea, you youngsters need to learn. (youngsters, gees I'm 21 and I feel ancient) But if someone gets stumped on a problem I'll gladly lend a hand. <!--emo&::nerdy::--><img src='http://www.unknownworlds.com/forums/html/emoticons/nerd.gif' border='0' style='vertical-align:middle' alt='nerd.gif'><!--endemo-->
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