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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1367.0. "Questions" by SQGUK::PAW () Mon Jan 07 1991 09:08

    
    Hi folks,
    
    My name is Chico and i've been with Digital for six months and only
    just realised that you have a maths conference - FANTASTIC.
    
    Here are two questions , the first one is easy and the second is very
    involved. See if you can answer them:
    
    Q1. I am thinking of a number that has four digits. The first digit
        multiplied by the second gives the third. The second multiplied
    	by the fourth gives the third. The sum of the (first and third)
    	and the (second and fourth) are equal. The total sum of all the
    	digits is 20. What is the number........
    
    Q2. Given that Z1 = 3y(sqr) + 6xy +x(sqr) + 8       Z1=0 when x=2
        calculate the value for which y is satisfed in the eqaution and
    	work out the eqaution dy/dx. Calculate the value of x at this point
    	when y=5.
T.RTitleUserPersonal
Name		DateLines
1367.1GUESS::DERAMODan D'EramoMon Jan 07 1991 09:293
        Q1. 1991
        
        Dan
1367.2GUESS::DERAMODan D'EramoMon Jan 07 1991 09:3915
>>    Q2. Given that Z1 = 3y(sqr) + 6xy +x(sqr) + 8       Z1=0 when x=2
>>        calculate the value for which y is satisfed in the eqaution and
>>    	  work out the eqaution dy/dx. Calculate the value of x at this point
>>    	  when y=5.
        
        Z1 = 0, x = 2 ==> y = -2
        
        If 3y^2 + 6xy + x^2 + 8 = constant then 6y dy/dx + 6y +
        6x dy/dx + 2x = 0, or dy/dx = - (6y + 2x)/(6y + 6x) or
        dy/dx = - (3y + x)/(3(y + x)).
        
        Z1 = 0, y = 5 ==> x = -15 +/- sqrt(142)  It isn't clear
        that this is what the last sentence asked for.
        
        Dan
1367.3Solutions?SQGUK::PAWTue Jan 08 1991 04:5738
    Hi there Dan,
    
    I see you're another maths lover. Anyway the answer to question 1 is
    correct. Easy wasn't it. I got that from teletext sunday morning. The
    second question I made up and I don't agree with your answer.
    From what I can see you have done:
    
    	6ydy/dx + 6y + 6xdy/dx + 2x = 0
    
    	which leads to dy/dx = - (3y + x)/(3(y + x))
    
    	but then when you put y=5 surely you get x=15.
    
    	When I did the question my-self this is what I did:
    
    	First work out dz/dy and treat x as a constant
    
    	Then work out dz/dx and treat y as a constant
    
    	then apply  dy/dx = dy/dz * dz/dx
    
    	This then gave me:
    
    	dz/dx = 6y + 2x
    
    	dz/dy = 6y + 6x
    
    	Therefore dy/dx = (6y + 2x)/(6y+6x)
    		  dy/dx = (3y + x)/(3(x+y))
    
    	This then gave me the x= -15.
    
    	Perhaps you can explain the sign difference. To me both methods
    	seem fairly feasable.
    
    	Have you got any differential problems that I can have a go at!
    
    	Chico...
1367.4GUESS::DERAMODan D'EramoTue Jan 08 1991 11:349
        re .3,
        
>>    	which leads to dy/dx = - (3y + x)/(3(y + x))
>>    
>>    	but then when you put y=5 surely you get x=15.
        
        No, when you put in y=5 you get dy/dx = - (15 + x) / (15 + 3x)
        
        Dan
1367.5MistakeSQGUK::PAWWed Jan 09 1991 05:4714
    
    	Hi again,
    
    	I agree with what you have written. I realised my mistake.
    	
    	I was taking dy/dx to be equal to zero which is why I got
    
    	x = 15. I was wrong to assume that at y=5 we have a turning point.
    
    	Thanks for the correction.......
    
    	Do you have a differential question that I can have a go at!
    
    	chico
1367.6GUESS::DERAMODan D'EramoWed Jan 09 1991 16:5312
        re .5,
        
>>    	Do you have a differential question that I can have a go at!
        
        	dy        1
        	-- + (x + -) y = 1
        	dx        x
	
        Either for the general case or with the initial condition
        y(2) = 1.
        
        Dan