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  • halp mee wit mi maths houmwurk ples

    If x = 5t^2 and y = 10t, find dy/dx and d^2y/dx^2 in terms of t.

    whoever finds it AND shows the work will get 20k in game money. Good Luck!
     

    DreadySemicolon

    Bishop of the umi order
    What are "dy" and "dx"? If they're something like sine and cosine you should say that, it makes it easier for non-native English speakers to understand the meaning of generic terms *^*
     

    DreadySemicolon

    Bishop of the umi order
    I'm sorry for trying to be nice and answer an actual question without any indication of it being a joke aside from the title, smartbottom "-_-"
    Addendum: I don't understand the hostility's raison d'être.
     
    sheesh it's 7th grade write-up excuses all over again. "you write me up for being nice and asking for a pencil". xD, Umm Semi, im sorry for whatever, i, did, idk. ...
     
    Two ways to work out dy/dx:

    Alt 1.
    t = y/10
    x = 5(y/10)^2
    x/5 = (y^2)/100
    20x = y^2
    d(y^2)/dx = d(20x)/dx
    d(y^2)/dy . dy/dx= 20
    2y . dy/dx = 20
    dy/dx = 10/y
    dy/dx = 10/10t
    dy/dx = 1/t

    Alt 2.
    dt/dy . dx/dt = dx/dy
    dx/dy = d(y/10)/dy . d(5t^2)/dt
    dx/dy = 1/10 . 10t
    dx/dy = t
    dy/dx = 1/(dx/dy)
    dy/dx = 1/t

    Two ways to work out d^2y/dx^2
    Alt 1.
    dy/dx = 10/y
    d^2y/dx^2 = d(10/y)/dx
    d^2y/dx^2 = d(10/y)/dy . dy/dx
    d^2y/dx^2 = (-10/y^2) . 10/y
    d^2y/dx^2 = -100/y^3
    d^2y/dx^2 = -100/y^3
    d^2y/dx^2 = -1/10t^3

    Alt 2.
    dy/dx = 1/t
    d^2y/dx^2 = d(1/t)/dx
    d^2y/dx^2 = (d(1/t)/dt)/(dx/dt)
    d^2y/dx^2 = (-1/t^2)/(d(5t^2)/dt)
    d^2y/dx^2 = (-1/t^2)/(10t)
    d^2y/dx^2 = -1/10t^3

    Therefore:
    dy/dx = 1/t
    d^2y/dx^2 = -1/10t^3

    I will have my 20k now thx.
     
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