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Homework Help: Two questions - (a) Differentiating inverse trig function and (b) vector

  1. Nov 23, 2008 #1
    Struggling with these two questions, any ideas?

    1. The problem statement, all variables and given/known data
    Differentiate with respect to x

    sin-1 2x - 4 cos-1[tex]\frac{x}{2}[/tex]

    2. Relevant equations
    The equations I have in my notes are identical to those derived here...

    3. The attempt at a solution
    I got

    [tex]\frac{1}{1 - 2x}[/tex] + [tex]\frac{8}{2 - x}[/tex]

    but I am totally unconvinced by myself!

    1. The problem statement, all variables and given/known data

    Resolve the vector x= (1,4,-3) along and perpendicular to a = (-2,3,1).

    2. Relevant equations


    3. The attempt at a solution

    I dont understand what the question wants me to do...
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 23, 2008 #2
    For the first part, [tex]arcsin(2x)[/tex], you got the derivative a little wrong.
    The derivative of [tex]arcsin(x)[/tex] is [tex]\frac{1}{\sqrt{1 - x^{2}}}[/tex], but x is actually 2x in this case so replace the x in the denominator of the derivative with 2x, don't forget to square it, and don't forget the chain rule! The derivative of arccos(x) is the opposite of the derivative of arcsin(x), [tex]\frac{-1}{\sqrt{1 - x^{2}}}[/tex]. You can multiply the -4 through after you find the derivative of arccos(x/2), but don't forget the chain rule here either.
  4. Nov 23, 2008 #3


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    I strongly suspect that, using the rules you refer to, you got
    [tex]\frac{1}{\sqrt{1- 4x^2}}[/itex]
    and reduce that to 1/(1-2x).

    The first is correct, the second is wrong: [itex]\sqrt{a^2- b^2}[/itex] is NOT equal to a- b!
  5. Nov 23, 2008 #4
    [tex]\sqrt{1 - 4x^{2}} \neq 1 - 2x[/tex] You just said [tex]\sqrt{a^{2} - b^{2}} \neq a - b[/tex] but you went ahead and did it anyway for the first part of the problem.

    [tex]\frac{d}{dx}(arcsin(2x)) = \frac{1}{\sqrt{1 - (2x)^{2}}} * \frac{d}{dx}(2x) = \frac{2}{\sqrt{1 - 4x^{2}}}[/tex]
    That is as reduced as it goes unless you want to rationalize the denominator which is unnecessary.
  6. Nov 23, 2008 #5
    For the second question, you need to break the vector down into projection and perpendicular components.

    You need to project the vector x on vector a, then take the projection of a away from x. You are then left with the perpendicular component (the same as resolving x and y components of a force vector).

    The projection is given by [tex]\frac{\vec{x}.\vec{a}}{\left|\vec{a}\right|^{2}}.\vec{a}[/tex]

    which gives:

    [tex]\vec{x}.\vec{a} = (1*-2) + (4*3) + (-3*1) = 7[/tex]
    [tex]\left|\vec{a}\right|^{2} = ((-2)^{2}+3^{2}+1^{2}) = 14[/tex]
    [tex]\frac{\vec{x}.\vec{a}}{\left|\vec{a}\right|^{2}}.\vec{a} = (7/14)*(-2,3,1) = (1/2).(-2,3,1) [/tex]

    Then take that away from x to get the perpendicular:

    [tex] (1,4,-3) - (1/2)*(-2,3,1) = (1/2) * (4,5,-7) [/tex]

    The full vector is then:

    [tex] x = (1/2) * (-2,3,1) + (1/2) * (4,5,-7) [/tex]

    Have a look at this:

  7. Nov 24, 2008 #6
    Thanks for the help, I eventually did get that last answer, but I did rationalise the denominator.

    Ta muchly
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