VC2.12

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Given,{\frac  {d^{2}r}{dt^{2}}}=6ti-24t^{2}j+4\sin tk\, [Equation 1]

with {\frac  {dr}{dt}}=-i-3k\, at t=0 [Equation 2]

and r=2i+j\, at t=0.[Equation 3]

Integrating (1),{\frac  {dr}{dt}}=3t^{2}i-8t^{3}j-4\cos tk+c\, [4] where c is an arbitrary constant vector.

Putting t=0 in (4) and using (2),we have

-i-3k=-4k+c\, so that c=-i+k\, [5]

Using (5),(4) reduces to {\frac  {dr}{dt}}=3t^{2}i-8t^{3}j-4\cos tk-i+k=(3t^{2}-1)i-8t^{3}j+(1-4\cos t)k\, [6]

Integrating (6),r=(t^{3}-t)i-2t^{4}j+(t-4\sin t)k+d\, [7] where d is a constant vector.

Putting t=0 in (7) and using(3),we have d=2i+j\, [8]

Using (8),(7)reduces to r=(t^{3}-t)i-(2t^{4}j+(t-4\sin t)k+2i+j\,

On simplification,we get

r=(t^{3}-t+2)i+(1-2t^{4})j+(t-4\sin t)k\,

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