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Thread: Calculus with reward

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    Default Calculus with reward

    *Reward 1 pin for most got it right before tommor 8am est*
    Last edited by Main; 03-09-2010 at 10:40 PM.
    Oh Hai Dar

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    Answer: 4 รท 0 ... OH SH*T!

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    this is serious man, please help me out here.
    Oh Hai Dar

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    well lets see...for the first one, take the derivative, and set it to 0, then solve for x

    man, this stuff is lame without just having learned it...can't remember much of it.
    Proud owner of "Efferator" my totally boted main account!
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    for the first one i get 0 as the horizontal tangent

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    0 is one of the horizontal tangents, but just from my glance I would say there is at least one more.
    Proud owner of "Efferator" my totally boted main account!
    "You see, sometimes, science is not a guess" -Xiaobing Zhou (my past physics professor, with heavy Chinese accent)

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    im just going by my crappy calculus software...the one thing canadian i am not proud of :P

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    what are you in uwo for? life sci?
    Oh Hai Dar

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    who's in uwo? im at mac for eng

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    I'm at UWO for eng. Currently life sci though.

    Close though
    Last edited by Nava2; 03-10-2010 at 01:07 PM.
    Writing an SRL Member Application | [Updated] Pascal Scripting Statements
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    13:46 <@BenLand100> <SourceCode> @BenLand100: what you have just said shows you 
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    Quote Originally Posted by Main_Ftw View Post
    *Reward 1 pin for most got it right before tommor 8am est*
    I don't have the time to solve them sorry.

    A) you will take the derivative of each. d[x^3y^2] + d[4(3x-2y)^3]=d[4y]-d[6x]

    d=derivative, you will need to use the product rule on the first, chain rule on the second and the other side is simple.

    B) The same thing as A, take the derivative of each operand.

    I would have to review for the rest, sorry. I Just don't have that time right now.

    "Failure is the opportunity to begin again more intelligently" (Henry Ford)


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    I havent' checked question 2 and I've got the working if you want it (its rather messy though).

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    can you show me your steps ?
    Oh Hai Dar

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    Ok..

    1. Horizontal tangents when dy/dx = 0,
    Find the first derivative function.. simplies to this:
    SCAR Code:
    dy/dx = 4^(4x+3).x^2.(3 + 4ln(4).x)

    Set dy/dx to 0 and solve for x; only ways to get 0 is if x^2 = 0 or (3 + 4ln(4).x) = 0 as an exponential function can never equal 0.

    Hence,
    SCAR Code:
    x = 0, x = -3/(4ln(4))

    Then you substitute those values into your initial function to get y and since slope = 0, there is no x value in those equations just the y value.

    2. Implicitly differentiate them... too hard to explain online tbh =/

    3. Again implicitly differentiate and rearrange to get dy/dx = everythingelse. You're giving the value of x = 1, so subsitute that into your y = equation to get the y values, then substitute the sets of coords (you get two y values: (1, 6) and (1, -4)) into your dy/dx function to get your gradients at those points. Then its a simple matter of finding your tangent equations.

    4. Key Point: If two lines are perpendicular then their gradients multiply to -1, ie
    SCAR Code:
    m1.m2 = -1.
    You have two functions, you can differentiate both of them in terms of x to get your gradient function in terms of x. Then multiply them together (the two functions) and set that equal to -1. Then solve the equation for x.

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    ty. First one was the one i didn't get, i got 0 but when i graphed it, it was totally different.
    Oh Hai Dar

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