Using the following notations:
θ = independent scalar variable
r(θ) = position vector depending on the variable θ
tˆ = unit tangent vector
nˆ = unit normal vector
show that
nˆ ⊥ tˆ
Given,
"\\theta=" Independent scalar variable
"r(\\theta)=" Position vector
"\\hat{t}=" unit tangent vector
"\\hat{n}=" unit normal vector
"r(\\theta)=r\\cos\\theta \\hat{i}+r\\sin\\theta\\hat{j}"
"\\hat{n}=\\frac{\\overrightarrow{r(\\theta)}}{|r(\\theta)|}=\\frac{r\\cos\\theta \\hat{i}+r\\sin\\theta\\hat{j}}{r}"
"=\\cos\\theta \\hat{i}+\\sin\\theta\\hat{j}"
unit tangent vector"(\\hat{t})=\\frac{r'(\\theta)}{|r'|}"
"\\hat{t}=\\frac{-r\\sin\\theta\\hat{i}+r\\cos\\theta\\hat{j}}{r}=-\\sin\\theta \\hat{i}+r\\cos\\theta\\hat{j}"
Now,
"\\hat{n}.\\hat{t}=(\\cos\\theta \\hat{i}+\\sin\\theta\\hat{j})(-\\sin\\theta \\hat{i}+r\\cos\\theta\\hat{j})"
"=-\\cos\\theta . \\sin\\theta+\\sin\\theta.\\cos\\theta"
"=0"
Hence, it is prove that "\\hat{n}" is perpendicular to "\\hat{t}" .
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