Question #105718
Let p= sin t i + cost j + tk, What is |dp/dt|?
1
Expert's answer
2020-03-16T14:47:00-0400
p=(sint)i+(cost)j+tk\vec{p}=(\sin{t})\vec{i}+(\cos{t})\vec{j}+t\vec{k}


where i,j,k\vec{i}, \vec{j}, \vec{k} are unit vectors.


Finding a derivative:

dpdt=d(sint)dti+d(cost)dtj+dtdtk=(cost)i(sint)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}


Finding a module for an arbitrary vector


a=axi+ayj+azk\vec{a}=a_x\vec{i}+a_y\vec{j}+a_z\vec{k}

is found by the formula:


a=ax2+ay2+az2|\vec{a}|=\sqrt{a_x^2+a_y^2+a_z^2}

Finding a module:

dpdt=(cost)2+(sint)2+12=2|\frac{d\vec{p}}{dt}|=\sqrt{(\cos{t})^2+(-\sin{t})^2+1^2}=\sqrt{2}




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