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HelpMath

Member

Registered: 2009-02-02

Posts: 3

Differientiation help

Hi all,

I had been stuck with this question for almost 2 weeks and really need your expertise.

If q = ax + by and r = bx - ay, determine the value of:

charismath

Member

Registered: 2009-01-23

Posts: 78

Re: Differientiation help

ok lemme try to help u..But i'm in class at the moment so I dnt have time to go through the latex guide..so plz adjust..Copy on paper,ul understand better..

q = ax + by    (1)                   r = bx - ay   (2)
where a and b are constants

lets start with the 1st part, the easiest one., partial diferential (dq/x)y
Look at equ(1). assume that y is a constant. n u surely know that differential of a constant is 0.

Hence (dq/dx)y=a

Coming to the second part (dx/dq)r

you will notice that we now need to xpress x in terms of q and r
To do this take (1) and make y the subject of formula.
y = (q-ax) / b

Now replace in (2)

r = bx -a(q-ax) / b
Multiply the whole thing by b..    br=b^2.x - aq+a^2.x

a^2.x + b^2.x = br + aq

Make x the subjet of formula..
you'l obtain     x=br /(a^2+b^2) + aq/(a^2+b^2)

this equ..  Look at the firs term.. it consists of only constants a,b and variable r.. Since we r diferentiating with respect to q, we take r as constant.. hence differential of the first term willl be 0...

Now no spoon feeding for u.. differentiate yourself the second term with respet to q. you'l get the value for (dx/dq)r

and multiply with (dq/dx)y to get your final answer;)
Try it.. if u still cant do it, i'll give u the final answer..

Last edited by charismath (2009-02-02 20:14:08)

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HelpMath

Member

Registered: 2009-02-02

Posts: 3

Re: Differientiation help

Thank you for your guidance!
Just to clarify my ans to (dx/dq)r = a/a^2 + b^2

so multiple the two will get: a^2 / a^2 + b^2

Am I right to say that?

charismath

Member

Registered: 2009-01-23

Posts: 78

Re: Differientiation help

U got it right Helpmath!

Oh.. I'm glad I could help u.. am a newbie here like u..
Try more similar problems.. For practice..;)

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Rob Miller

Guest

Re: Differientiation help

Cool site