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Math Help - vectors and differentiation

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    vectors and differentiation

    The vector a depends on a parameter t, i.e. a=a(t)=a_x(t)i +a_y(t)j +a_z(t)k.. it satisfies the equation da/dt= j (Vector product) a
    show that d^2a_x/dt^2 =-a_x , da_y/dt=0 and d^2a_z/dt^2 =-a_z.

    For the vector a, find its value for t=pi if at t=0 a(0)=i+j and da/dt(0)=0k

    i have absolutely no idea how to start...
    Last edited by Emma L; January 9th 2010 at 02:14 PM.
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    Senior Member Shanks's Avatar
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    can you use latex to restate your problem?
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    Hope that's a bit clearer sorry, wasnt sure how to do the vector product in latex
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Emma L View Post
    Hope that's a bit clearer sorry, wasnt sure how to do the vector product in latex
    I'm sorry. This may be a stupid question, but what is "vector product"? Is it the cross product or the dot product? Both involve vectors.

    EDIT: It's obvious by context that it is cross product.

    \times is given by \times
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    ye sorry, it's the cross product
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    Quote Originally Posted by Emma L View Post
    The vector a depends on a parameter t, i.e. a=a(t)=a_x(t)i +a_y(t)j +a_z(t)k.. it satisfies the equation da/dt= j (Vector product) a
    show that d^2a_x/dt^2 =-a_x , da_y/dt=0 and d^2a_z/dt^2 =-a_z.

    For the vector a, find its value for t=pi if at t=0 a(0)=i+j and da/dt(0)=0k

    i have absolutely no idea how to start...
    So the problem is that you are given \vec{a} such that \frac{d\vec{a}}{dt}= \vec{j}\times \vec{a} and you are asked to show that, in that case, \frac{d^2a_x}{dt^2}= -a_x, \frac{da_y}{dt}= 0, and \frac{d^2a_z}{dt^2}= -a_z.

    Okay, go ahead and calculate \vec{j}\times \vec{a}. That's easy, it is just a_z\vec{i}- a_x\vec{k}. Setting \frac{d\vec{a}}{dt}= \frac{da_x}{dx}\vec{i}+ \frac{da_y}{dt}\vec{j}+ \frac{da_z}{dt}\vec{k} equal to that gives you three equations: \frac{da_x}{dt}= a_z, \frac{da_y}{dt}= 0, and \frac{da_z}{dt}= -a_x.

    differentiating the first of those, with respect to t, gives \frac{d^2 a_x}{dt^2}= \frac{d^a_z}{dt}= -a_x. Get the point? I'll leave the others to you now.

    For the last part you need to solve those equations. \frac{da_y}{dt}= 0 is easy: a_y is a constant and the last part tells us that that constant is 1.

    To solve the other two, use the fact, that you have now shown, that \frac{d^2a_x}{dt^2}= -a_x and \frac{d^2a_z}{dt^2}= -a_z. Solve those differential equations using the initial values a_x(0)= 1, a_x'(0)= 0, a_z(0)= 0, and a_z'(0)= 0.
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