How to solveu→¨=u→×k^

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Matronola3zw6 · Accepted Answer

Step 1You should perhaps more explicitly mention that k is the third canonical basis vector.Then it is immediately clear that the right side is zero in z direction and acts as a&amp;nbsp;−90∘&amp;nbsp;rotation in the xy-plane,(xi+yj)×k=-xj+yi.This rotation has 2 square roots, the rotations by&amp;nbsp;−45∘&amp;nbsp;and by&amp;nbsp;135∘.Interpreting the xy-plane as complex plane, one getsx¨+iy¨=−i(x+iy)=121−i2(x+iy)so thatx+iy=c1e1−i2t+c2e−1+i2t,≈c1,2∈BC

Abdullah Avery · Accepted Answer

Step 1So writing&amp;nbsp;u=(ui), we haveu¨1=u2,&amp;nbsp;u¨2=−u1,&amp;nbsp;u¨3=0Hence,&amp;nbsp;u3(t)=at+b&amp;nbsp;for some constants a,b, the other two equations give usddu¨1=−u1This has the characteristic equation&amp;nbsp;λ4+1=0, giving usλ1,2=1±i2,λ3,4=−1±i2Hence, we haveu1(t)=exp⁡(t2)(csin⁡(t2)+dcos⁡(t2))+exp⁡(−t2)(esin⁡(t2)+fcos⁡(t2))for some constants c,d,e,f. By using&amp;nbsp;u2=u¨1, you can calculate the missing component.

How to solve \(\displaystyle\ddot{{\vec{{{u}}}}}=\vec{{{u}}}\times\hat{{{k}}}\)

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