p ⃗ = ( sin t ) i ⃗ + ( cos t ) j ⃗ + t k ⃗ \vec{p}=(\sin{t})\vec{i}+(\cos{t})\vec{j}+t\vec{k} p = ( sin t ) i + ( cos t ) j + t k
where i ⃗ , j ⃗ , k ⃗ \vec{i}, \vec{j}, \vec{k} i , j , k are unit vectors.
Finding a derivative:
d p ⃗ d t = d ( sin t ) d t i ⃗ + d ( cos t ) d t j ⃗ + d t d t k ⃗ = ( cos t ) i ⃗ − ( sin t ) j ⃗ + k ⃗ \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} d t d p = d t d ( s i n t ) i + d t d ( c o s t ) j + d t d t k = ( cos t ) i − ( sin t ) j + k
Finding a module for an arbitrary vector
a ⃗ = a x i ⃗ + a y j ⃗ + a z k ⃗ \vec{a}=a_x\vec{i}+a_y\vec{j}+a_z\vec{k} a = a x i + a y j + a z k is found by the formula:
∣ a ⃗ ∣ = a x 2 + a y 2 + a z 2 |\vec{a}|=\sqrt{a_x^2+a_y^2+a_z^2} ∣ a ∣ = a x 2 + a y 2 + a z 2
Finding a module:
∣ d p ⃗ d t ∣ = ( cos t ) 2 + ( − sin t ) 2 + 1 2 = 2 |\frac{d\vec{p}}{dt}|=\sqrt{(\cos{t})^2+(-\sin{t})^2+1^2}=\sqrt{2} ∣ d t d p ∣ = ( cos t ) 2 + ( − sin t ) 2 + 1 2 = 2
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