Find the tangential and normal components of the acceleration vector. (r)= (t^2 +1)I +t^3 j, t>=0?
- NCSLv 75 months agoFavorite 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)
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.
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- PinkgreenLv 75 months ago
r=(t^2+1)i+(t^3)j, t>=0 is the position vector of a point on
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.
c=(t^2+3)/[t(4+9t^2)^1.5], where 1/c follows the sign
the tangential component
the normal component
at time t.
- Anonymous5 months ago
You have my approval to go ahead and do it then.