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## Q. 2.2

Describe the curve given by the vector equation $\mathbf{f}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi$.

## Verified Solution

We first see that for every $t,|\mathbf{f}(t)|=1$ since $|\mathbf{f}(t)|=\sqrt{\cos ^{2} t+\sin ^{2} t}=1$. Moreover, if we write the curve in its parametric representation, we find that

$x=\cos t, \quad y=\sin t$                                                           (5)

and since $\cos ^{2} t+\sin ^{2} t=1$, we have

$x^{2}+y^{2}=1$

which is, of course, the equation of the unit circle. This curve is sketched in Figure 1 . Note that in the sketch the parameter t represents both the length of the arc from (1,0) to the endpoint of the vector and the angle (measured in radians) the vector makes with the positive x-axis. The representation $x^{2}+y^{2}=1$ is called the Cartesian equation of the curve given by (5). 