where i⃗,j⃗,k⃗\vec{i}, \vec{j}, \vec{k}i,j,k are unit vectors.
Finding a derivative:
dp⃗dt=d(sint)dti⃗+d(cost)dtj⃗+dtdtk⃗=(cost)i⃗−(sint)j⃗+k⃗\frac{d\vec{p}}{dt}=\frac{d(\sin{t})}{dt}\vec{i}+\frac{d(\cos{t})}{dt}\vec{j}+\frac{dt}{dt}\vec{k}=(\cos{t})\vec{i}-(\sin{t})\vec{j}+\vec{k}dtdp=dtd(sint)i+dtd(cost)j+dtdtk=(cost)i−(sint)j+k
Finding a module for an arbitrary vector
is found by the formula:
Finding a module:
∣dp⃗dt∣=(cost)2+(−sint)2+12=2|\frac{d\vec{p}}{dt}|=\sqrt{(\cos{t})^2+(-\sin{t})^2+1^2}=\sqrt{2}∣dtdp∣=(cost)2+(−sint)2+12=2
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