For $f\left(t\right)=(\mathrm{sin}t,\frac{{\mathrm{cos}}^{2}t}{t})$ what is the distance between $f\left(\frac{\pi}{4}\right)$ and $f\left(\pi \right)$

Cristopher Knox
2022-06-30
Answered

For $f\left(t\right)=(\mathrm{sin}t,\frac{{\mathrm{cos}}^{2}t}{t})$ what is the distance between $f\left(\frac{\pi}{4}\right)$ and $f\left(\pi \right)$

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Hayley Mccarthy

Answered 2022-07-01
Author has **19** answers

Step 1

Here,

$f\left(t\right)=(\mathrm{sin}t,\frac{{\mathrm{cos}}^{2}t}{t})$

$\Rightarrow f\left(\frac{\pi}{4}\right)=(\mathrm{sin}\left(\frac{\pi}{4}\right),\frac{{\mathrm{cos}}^{2}\left(\frac{\pi}{4}\right)}{\frac{\pi}{4}})=(\frac{1}{\sqrt{2}},\frac{\frac{1}{2}}{\frac{\pi}{4}})=(\frac{1}{\sqrt{2}},\frac{2}{\pi})$

Similarly,

$f\left(\pi \right)=(\mathrm{sin}\pi ,\frac{{\mathrm{cos}}^{2}\pi}{\pi})=(0,\frac{{(-1)}^{2}}{\pi})=(0,\frac{1}{\pi})$

Distance Formula:

If $A({x}_{1},{y}_{1})\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}B({x}_{2}.{y}_{2}),$ , then

$d(A,B)=\sqrt{{({x}_{1}-{x}_{2})}^{2}+{({x}_{2}-{y}_{2})}^{2}}$

$\Rightarrow d(f\left(\frac{\pi}{4}\right),f\left(\pi \right))=\sqrt{{(\frac{1}{\sqrt{2}}-0)}^{2}+{(\frac{2}{\pi}-\frac{1}{\pi})}^{2}}=\sqrt{\frac{1}{2}+\frac{1}{{\pi}^{2}}}$

Here,

$f\left(t\right)=(\mathrm{sin}t,\frac{{\mathrm{cos}}^{2}t}{t})$

$\Rightarrow f\left(\frac{\pi}{4}\right)=(\mathrm{sin}\left(\frac{\pi}{4}\right),\frac{{\mathrm{cos}}^{2}\left(\frac{\pi}{4}\right)}{\frac{\pi}{4}})=(\frac{1}{\sqrt{2}},\frac{\frac{1}{2}}{\frac{\pi}{4}})=(\frac{1}{\sqrt{2}},\frac{2}{\pi})$

Similarly,

$f\left(\pi \right)=(\mathrm{sin}\pi ,\frac{{\mathrm{cos}}^{2}\pi}{\pi})=(0,\frac{{(-1)}^{2}}{\pi})=(0,\frac{1}{\pi})$

Distance Formula:

If $A({x}_{1},{y}_{1})\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}B({x}_{2}.{y}_{2}),$ , then

$d(A,B)=\sqrt{{({x}_{1}-{x}_{2})}^{2}+{({x}_{2}-{y}_{2})}^{2}}$

$\Rightarrow d(f\left(\frac{\pi}{4}\right),f\left(\pi \right))=\sqrt{{(\frac{1}{\sqrt{2}}-0)}^{2}+{(\frac{2}{\pi}-\frac{1}{\pi})}^{2}}=\sqrt{\frac{1}{2}+\frac{1}{{\pi}^{2}}}$

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