Results 1 to 7 of 7

Math Help - dy/dx=0

  1. #1
    Newbie
    Joined
    Dec 2006
    Posts
    8

    dy/dx=0

    hello.. how would i go about solving dy/dx=0?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by tummbler View Post
    hello.. how would i go about solving dy/dx=0?
    as in solve for y?

    wow, that's a weird question, there's got to be a trick to it that i'm not seeing, i'm going to go ahead and say

    dy/dx = 0
    => dy = 0*dx
    => int{1}dy = int{0}
    => y = C where C is a constant
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by tummbler View Post
    hello.. how would i go about solving dy/dx=0?
    If y'=0 on I some open interval.

    Then, y=C for some real number C.

    Proof: Let a<b be in I. Then consider [a,b]. The function y is differenciable fully on [a,b] because it is differenciable on I. Thus, y is continous on [a,b] and as said on (a,b). This tells us that y satisfies Lagrange's Mean Value Theorem. Hence there exists c in (a,b) so that,
    f'(c)=(f(b)-f(a))/(b-a)
    But f'(c)=0 because f'(x)=0 for all x in I.
    Thus,
    0=(f(b)-f(a))/(b-a)
    Thus,
    f(b)=f(a).
    This tells us that f is a konstant function on I.
    Q.E.D.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ThePerfectHacker View Post
    If y'=0 on I some open interval.

    Then, y=C for some real number C.

    Proof: Let a<b be in I. Then consider [a,b]. The function y is differenciable fully on [a,b] because it is differenciable on I. Thus, y is continous on [a,b] and as said on (a,b). This tells us that y satisfies Lagrange's Mean Value Theorem. Hence there exists c in (a,b) so that,
    f'(c)=(f(b)-f(a))/(b-a)
    But f'(c)=0 because f'(x)=0 for all x in I.
    Thus,
    0=(f(b)-f(a))/(b-a)
    Thus,
    f(b)=f(a).
    This tells us that f is a konstant function on I.
    Q.E.D.
    I guess that's why you're MHF's Number Theorist!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Jhevon View Post
    I guess that's why you're MHF's Number Theorist!
    That has nothing to do with number theory. I am far superior in Number/Group/Field Theory than in Analysis. But that proof is a standard. You should know it as well as you know your own name.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by ThePerfectHacker View Post
    That has nothing to do with number theory. I am far superior in Number/Group/Field Theory than in Analysis. But that proof is a standard. You should know it as well as you know your own name.
    well yeah, but i was more refering to our different approaches to the same problem...or did you respond to the problem using a proof because of what i said? (about it being a trick question)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2006
    Posts
    8
    i should i stated the whole problem. im sorry. it went as follows:

    For what value(s) of x in this equation does y attain its minimum value if a, b, and c are all positive constants?

    y=[(a^2)+(b-x)^2]^(1/2) + [(c^2)+(x^2)]^(1/2)

    and someone else told me:
    Put a=1, b=2, c=3. Then plot the curve. Is the minimum at x=2 or x=0?
    -its not.

    so, This curve is smooth and y can be minimised by solving dy/dx=0 in the usual manner. I just dont know what the "usual manner" is..
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum