Short Answer

Find the unit tangent vector and the principal unit normal vector at the specified value of t.
$r (t) = (t, t^{2}), t = 1$

$\begin{array}{l} T (1) = <\frac{\sqrt{5}}{5}, \frac{2 \sqrt{5}}{5}> and \\ N (1) = <\frac{- 2 \sqrt{5}}{5}, \frac{\sqrt{5}}{5}> \end{array}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step1. Given Information

$Consider r (t) = <t, t^{2}>, t = 1 The objective is to find the unit tangent vector and the principle unit normal vector to r (t) at t = 1 The unit tangent vector of r (t) denoted by T (t) and the principal unit normal vector of r (t) denoted by N (t) are given by T (t) = \frac{r^{'} (t)}{∥r^{'} (t)∥} and N (t) = \frac{T^{'} (t)}{∥T^{'} (t)∥} Consider r (t) = <t, t^{2}> \begin{array}{l} r^{'} (t) = \frac{d}{d t} <t, t^{2}> \\ = ⟨ 1, 2 t ⟩ \\ ∥r^{''} (t)∥ = ∥ ⟨ 1, 2 t ⟩ ∥ \\ = \sqrt{1 + 4 t^{2}} \end{array}$

Step2. Continue

$The unit tangent vector to r (t) is \begin{array}{l} T (t) = \frac{r^{'} (t)}{∥r^{'} (t)∥} \\ = \frac{⟨ 1, 2 t ⟩}{\sqrt{1 + 4 t^{2}}} \\ A t t = 1, T (t) = T (1) = \frac{⟨ 1, 2 (1) ⟩}{\sqrt{1 + 4 (1)^{2}}} = \frac{⟨ 1, 2 ⟩}{\sqrt{5}} = <\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}> \\ = <\frac{\sqrt{5}}{5}, \frac{2 \sqrt{5}}{2}>, (R a t i o n a l i z e t h e d e n o m i n a t i o r) \\ W e h a v e \\ T (t) = <\frac{1}{\sqrt{1 + 4 t^{2}}}, \frac{2 t}{\sqrt{1 + 4 t^{2}}}> \\ The Quotient rule is used to find T^{'} (t) : \\ \begin{array}{c} T^{'} (t) = (<\frac{- 4 t}{{(1 + 4 t^{2})}^{\frac{3}{2}}}, \frac{2}{{(1 + 4 t^{2})}^{\frac{3}{2}}}>) \\ ∥T^{'} (t)∥ = \sqrt{\frac{16 t^{2}}{{(1 + 4 t^{2})}^{3}}} + \frac{4}{{(1 + 4 t^{2})}^{3}} \\ = \sqrt{\frac{4 (1 + 4 t^{2})}{{(1 + 4 t^{2})}^{3}}} \\ = \frac{2}{1 + 4 t^{2}} \end{array} \end{array}$

Step3. Continue

$N o w \begin{array}{l} N (t) = \frac{T^{'} (t)}{∥T^{'} (t)∥} \\ = \frac{1 + 4 t^{2}}{2} <\frac{- 4 t}{{(1 + 4 t^{2})}^{\frac{3}{2}}}, \frac{2}{{(1 + 4 t^{2})}^{\frac{3}{2}}}> \\ = <\frac{- 2 t}{\sqrt{1 + 4 t^{2}}}, \frac{1}{\sqrt{1 + 4 t^{2}}}> \\ A t t = 1, N (t) = N (1) = <\frac{- 2 (1)}{\sqrt{1 + 4 (1)^{2}}}, \frac{1}{\sqrt{1 + 4 (1)^{2}}}> = <\frac{- 2}{\sqrt{5}}, \frac{1}{\sqrt{5}}> \\ = (\frac{- 2 \sqrt{5}}{5}, \frac{\sqrt{5}}{5}) \\ T h u s \\ \begin{array}{l} T (1) = <\frac{\sqrt{5}}{5}, \frac{2 \sqrt{5}}{5}> and \\ N (1) = <\frac{- 2 \sqrt{5}}{5}, \frac{\sqrt{5}}{5}> \end{array} \end{array}$

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Find the unit tangent vector and the principal unit normal vector at the specified value of t.
$r (t) = (t, t^{2}), t = 1$

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Find the unit tangent vector and the principal unit normal vector at the specified value of t. r(t)=(t,t2),t=1

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Find the unit tangent vector and the principal unit normal vector at the specified value of t.
$r (t) = (t, t^{2}), t = 1$