Question

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

Answer 1

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 $\hat{t}$ .