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Author Question: Solve the initial-value problem = k r with initial condition r(0) = i + 2j + 3k. Describe the solution curve. (Read 62 times)

amal

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on: May 27, 2021
Solve the  initial-value problem   = k × r  with initial condition  r(0) = i + 2j + 3k.  Describe the solution curve.
◦ r(t) = i +   j +  3 k,  
a circle with centre  (0, 0, 3)  and radius  5  in the plane  z = 3.
◦ r(t) = i +   j +  3 k,  
a circle with centre  (0, 0, 3)  and radius  5  in the plane  z = 3.
◦ r(t) = i -   j +  3 k,  
a circle with centre  (0, 0, 3)  and radius  5  in the plane  z = 3.
◦ r(t) = i +   j +  3 k,  
a circle with centre  (0, 0, 3)  and radius    in the plane  z = 3.
◦ r(t) = i -   j +  3 k,  
a circle with centre  (0, 0, 3)  and radius    in the plane  z = 3.
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Marked as best answer by amal on May 27, 2021

alexanderhamilton

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Reply #1 on: May 27, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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amal

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Reply 2 on: May 27, 2021
YES! Correct, THANKS for helping me on my review

Chelseyj.hasty

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Reply 3 on: Yesterday
:D TYSM

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