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For $f\left(t\right)=\left(\mathrm{sin}t,\frac{{\mathrm{cos}}^{2}t}{t}\right)$ what is the distance between $f\left(\frac{\pi }{4}\right)$ and $f\left(\pi \right)$
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Step 1
Here,
$f\left(t\right)=\left(\mathrm{sin}t,\frac{{\mathrm{cos}}^{2}t}{t}\right)$
$⇒f\left(\frac{\pi }{4}\right)=\left(\mathrm{sin}\left(\frac{\pi }{4}\right),\frac{{\mathrm{cos}}^{2}\left(\frac{\pi }{4}\right)}{\frac{\pi }{4}}\right)=\left(\frac{1}{\sqrt{2}},\frac{\frac{1}{2}}{\frac{\pi }{4}}\right)=\left(\frac{1}{\sqrt{2}},\frac{2}{\pi }\right)$
Similarly,
$f\left(\pi \right)=\left(\mathrm{sin}\pi ,\frac{{\mathrm{cos}}^{2}\pi }{\pi }\right)=\left(0,\frac{{\left(-1\right)}^{2}}{\pi }\right)=\left(0,\frac{1}{\pi }\right)$
Distance Formula:
If $A\left({x}_{1},{y}_{1}\right)\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}B\left({x}_{2}.{y}_{2}\right),$ , then
$d\left(A,B\right)=\sqrt{{\left({x}_{1}-{x}_{2}\right)}^{2}+{\left({x}_{2}-{y}_{2}\right)}^{2}}$
$⇒d\left(f\left(\frac{\pi }{4}\right),f\left(\pi \right)\right)=\sqrt{{\left(\frac{1}{\sqrt{2}}-0\right)}^{2}+{\left(\frac{2}{\pi }-\frac{1}{\pi }\right)}^{2}}=\sqrt{\frac{1}{2}+\frac{1}{{\pi }^{2}}}$