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Sagot :
The velocity vector of the given path r(t) = (9cos2(t), 3t - t^3, 2t) is [tex]v = 9sint\hat{i}+(3-3t^{2} )\hat{j} +2\hat{k}[/tex].
According to the given question.
We have a path
r(t) = (9cos2(t), 3t - t^3, 2t)
So, the vector form of the above vector form can be written as
[tex]r(t) = 9cos2(t)\hat{i}+ (3t - t^{3} )\hat{j} + 2t\hat{k}[/tex]
As, we know that the rate of change of position of an object is called velocity vector.
Therefore, the velocity vector of the given path r(t) = (9cos2(t), 3t - t^3, 2t) is given by
[tex]v = \frac{d(r(t))}{dt}[/tex]
[tex]\implies v = \frac{d(9cost\hat{i}+(3t-t^{3})\hat{j}+2t\hat{k} }{dt}[/tex]
[tex]\implies v = \frac{d(9cost\hat{i})}{dt} +\frac{d(3t-t^{3})\hat{j} }{dt} +\frac{d(2t)}{d(t)}[/tex]
[tex]\implies v = 9sint\hat{i}+(3-3t^{2} )\hat{j} +2\hat{k}[/tex]
Hence, the velocity vector of the given path r(t) = (9cos2(t), 3t - t^3, 2t) is [tex]v = 9sint\hat{i}+(3-3t^{2} )\hat{j} +2\hat{k}[/tex].
Find out more information about velocity vector here:
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