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## Q. 4.4

Calculate a unit normal vector to the curve f(t) = [(t³/3) – t]i + t²j at t = 3.

## Verified Solution

Here $\mathbf{f}^{\prime}(t)=\left(t^{2}-1\right) \mathbf{i}+2 t \mathbf{j}$ and
$\left|\mathbf{f}^{\prime}(t)\right|=\sqrt{\left(t^{2}-1\right)^{2}+4 t^{2}}=\sqrt{t^{4}-2 t^{2}+1+4 t^{2}}=\sqrt{t^{4}+2 t^{2}+1}=t^{2}+1$.

Thus

$\mathbf{T}(t)=\frac{\mathbf{f}^{\prime}(t)}{\left|\mathbf{f}^{\prime}(t)\right|}=\frac{t^{2}-1}{t^{2}+1} \mathbf{i}+\frac{2 t}{t^{2}+1} \mathbf{j}$.

Then

$\mathbf{T}^{\prime}(t)=\frac{d}{d t}\left(\frac{t^{2}-1}{t^{2}+1}\right) \mathbf{i}+\frac{d}{d t}\left(\frac{2 t}{t^{2}+1}\right) \mathbf{j}=\frac{4 t}{\left(t^{2}+1\right)^{\mathbf{2}}} \mathbf{i}+\frac{2-2 t^{2}}{\left(t^{2}+1\right)^{2}} \mathbf{j}$.

Finally,

\begin{aligned}\left|\mathrm{T}^{\prime}(t)\right| &=\left\{\left[\frac{4 t}{\left(t^{2}+1\right)^{2}}\right]^{2}+\left[\frac{2-2 t^{2}}{\left(t^{2}+1\right)^{2}}\right]^{2}\right\}^{1 / 2}=\frac{1}{\left(t^{2}+1\right)^{2}}\left(16 t^{2}+4-8 t^{2}+4 t^{4}\right)^{1 / 2} \\&=\frac{1}{\left(t^{2}+1\right)^{2}} \sqrt{4 t^{4}+8 t^{2}+4}=\frac{2}{\left(t^{2}+1\right)^{2}} \sqrt{t^{4}+2 t^{2}+1} \\&=\frac{2\left(t^{2}+1\right)}{\left(t^{2}+1\right)^{2}}=\frac{2}{t^{2}+1^{\prime}}\end{aligned}

so that

$\mathbf{n}(t)=\frac{\mathbf{T}^{\prime}(t)}{\left|\mathbf{T}^{\prime}(t)\right|}=\frac{t^{2}+1}{2}\left[\frac{4 t}{\left(t^{2}+1\right)^{2}} \mathbf{i}+\frac{2-2 t^{2}}{\left(t^{2}+1\right)^{2}} \mathbf{j}\right]=\frac{2 t}{t^{2}+1} \mathbf{i}+\frac{1-t^{2}}{t^{2}+1} \mathbf{j}$.

At $t=3, \mathbf{T}(t)=\frac{4}{5} \mathbf{i}+\frac{3}{5} \mathbf{j}$ and $\mathbf{n}=\frac{3}{5} \mathbf{i}-\frac{4}{5} \mathbf{j}$. Note that $\mathbf{T}(3) \cdot \mathbf{n}(3)=0$. This is sketched in Figure 2.