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## Q. 4.3

Calculate a unit normal vector to the curve f(t ) = (cos t)i + (sin t)j at t = π/4.

## Verified Solution

First, we calculate $\mathbf{f}^{\prime}(t)=-(\sin t) \mathbf{i}+(\cos t) \mathbf{j}$, and since $\left|\mathbf{f}^{\prime}(t)\right|=1$, we find that

$\mathbf{T}(t)=\frac{\mathbf{f}^{\prime}(t)}{\left|\mathbf{f}^{\prime}(t)\right|}=-(\sin t) \mathbf{i}+(\cos t) \mathbf{j}$.

Then

$\mathbf{T}^{\prime}(t)=-(\cos t) \mathbf{i}-(\sin t) \mathbf{j}=\mathbf{n}(t)$

since $\left|\mathrm{T}^{\prime}(t)\right|=1$, so that at $t=\pi / 4$,

$\mathbf{n}=-\frac{1}{\sqrt{2}} \mathbf{i}-\frac{1}{\sqrt{2}} \mathbf{j}$.

This is sketched in Figure 1 [remember that $\mathbf{f}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}$ is the parametric equation of the unit circle]. The reason the vector $\mathbf{n}$ points inward is that it is the negative of the position vector at $t=\pi / 4$. 