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iren [92.7K]
3 years ago
13

I am not sure how to figure this out.

Mathematics
1 answer:
umka2103 [35]3 years ago
3 0
Change them all into the same unit of measurement and you will see which is the furthest out by default. 
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Find the solution r(t)r(t) of the differential equation with the given initial condition: r′(t)=⟨sin9t,sin6t,9t⟩,r(0)=⟨4,6,3⟩
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\mathbf r'(t)=\langle\sin9t,\sin6t,9t\rangle


\mathbf r(t)=\displaystyle\int\mathbf r'(t)\,\mathrm dt


\mathbf r(t)=\left\langle\displaystyle\int\sin9t\,\mathrm dt,\int\sin6t\,\mathrm dt,\int9t\,\mathrm dt\right\rangle


\displaystyle\int\sin9t\,\mathrm dt=\frac19\cos9t+C_1

\displaystyle\int\sin6t\,\mathrm dt=\frac16\cos6t+C_2

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With the initial condition \mathbf r(0)=\langle4,6,3\rangle, we find


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\mathbf r(t)=\left\langle\dfrac19\cos9t+\dfrac{35}9,\dfrac16\cos6t+\dfrac{35}6,\dfrac92t^2+3\right\rangle

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