x^(2/3) + y^(2/3) = a^(2/3) , 0<x<a and 0<y
wat is y' and y"?
Originally posted by ctstalin:x^(2/3) + y^(2/3) = a^(2/3) , 0<x<a and 0<y
wat is y' and y"?
y' is dy/dx
y'' is d2y/dx2
Originally posted by dkcx:y' is dy/dx
y'' is d2y/dx2
That one i know lah
Just that my answer is very very far off from the standard answer and i can't find out whats wrong
Which makes me wonder....Is "a" a constant?
Originally posted by ctstalin:That one i know lah
Just that my answer is very very far off from the standard answer and i can't find out whats wrong
Which makes me wonder....Is "a" a constant?
Oh ps, see wong.
Yes a is a constant since your question shows it as a value.
Originally posted by dkcx:Oh ps, see wong.
Yes a is a constant since your question shows it as a value.
I also thought that its a constant....but then hor the standard answer has "a" in both the first and second derivatives.... which don't make any sense cos differentiating a constant will give u nothing...
Help anyone?
Originally posted by ctstalin:
I also thought that its a constant....but then hor the standard answer has "a" in both the first and second derivatives.... which don't make any sense cos differentiating a constant will give u nothing...
Help anyone?
When you convert the question to in terms of y, its (a^2/3 - x^2/3)^3/2 and not seperately so your answer will still have the a term.
Is y' = (a^2/3 - x^2/3)^1/2?
Originally posted by dkcx:When you convert the question to in terms of y, its (a^2/3 - x^2/3)^3/2 and not seperately so your answer will still have the a term.
Is y' = (a^2/3 - x^2/3)^1/2?
you're quite close.... so just express the equation in terms of y can liao?
maybe i think too much lol
thanks
anyway here's the solution
Originally posted by ctstalin:you're quite close.... so just express the equation in terms of y can liao?
maybe i think too much lol
thanks
anyway here's the solution
ok, i just 4got the -. You can rearrange the equation i wrote and get this answer.