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Find the tangential and normal components of the acceleration vector. (r)= (t^2 +1)I +t^3 j, t>=0?

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  • NCS
    Lv 7
    5 months ago
    Favorite Answer

    r(t) = t² i^+ t³ j^

    v(t) = 2t i^ + 3t² j^

    a(t) = 2 i^ + 6t j^

    According to the citation,

    tangential a_t = a(t)•v(t) / |v(t)|

    a_t = (2*2t + 6t*3t²) / √[(2t)² + (3t²)²]

    a_t = (4t + 18t³) / √[t²(4 + 9t²)]

    a_t = 2(2 + 9t²) / √(4 + 9t²)

    and

    normal a_n = |v(t) x a(t)| / |v(t)|

    where x represents the cross product

    and we already know |v(t)|

    a_n = |(2t i^ + 3t² j^) x (2 i^ + 6t j^)| / √[t²(4 + 9t²)]

    a_n = |(2t*6t - 3t²*2) k^| / √[t²(4 + 9t²)]

    a_n = |(12t² - 6t²) k^| / √[t²(4 + 9t²)]

    a_n = |6t² k^| / √[t²(4 + 9t²)]

    a_n = |6t k^| / √(4 + 9t²)

    a_n = 6t / √(4 + 9t²)

    which I don't think can be further simplified

    That's the process. Hopefully I haven't made any calculation errors.

    If you find this helpful, please award Best Answer!

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  • 5 months ago

    Vectors:

    r=(t^2+1)i+(t^3)j, t>=0 is the position vector of a point on

    the curve.

    r'=2ti+(3t^2)j=vT, where v=tsqr(4+9t^2)

    r"=v'T+(v^2)cN, where c=the curvature at the point.

    v'=(4+18t^2)/sqr(4+9t^2)

    c=|r'xr|/|r'|^3=|-(t^2+3)(t^2)k|/|[(4+9t^2)^1.5]t^3|

    =>

    c=(t^2+3)/[t(4+9t^2)^1.5], where 1/c follows the sign

    convention.

    cv^2=t(t^2+3)/sqr(4+9t^2)

    Thus,

    the tangential component

    =(4+18t^2)/sqr(4+9t^2);

    the normal component

    =t(t^2+3)/sqr(4+9t^2)

    at time t.

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  • Anonymous
    5 months ago

    You have my approval to go ahead and do it then.

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