Chapter 11: Q. 18 (page 898)

$r (t) = (t, t^{2})$

Short Answer

Expert verified

Thus $a_{T} = \frac{4 t}{\sqrt{1 + 4 t^{2}}}$ and $a_{N} = \frac{2}{\sqrt{1 + 4 t^{2}}}$

Step by step solution

Introduction

Consider $r (t) = <t, t^{2}>$

The goal is to determine the tangential and normal components of acceleration. $r (t)$ .

Given information

The tangential and normal components of Acceleration

$r (t)$ is a twice - differentiable vector function with $r^{'} (t) = v (t)$ and $r^{''} (t) = a (t)$ .

The tangential component of acceleration, $a_{T} = \frac{v \cdot a}{‖ v ‖}$ . The normal component acceleration,

a_{N} = \frac{‖ v \times a ‖}{‖ v ‖} r (t) = <t, t^{2}> v (t) = r^{'} (t) = ⟨ 1, 2 t ⟩ ‖ v ‖ = ‖ ⟨ 1, 2 t ⟩ ‖ = \sqrt{1 + 4 t^{2}} v \cdot a = v (t) \cdot a (t) = ⟨ 1, 2 t ⟩ \cdot ⟨ 0, 2 ⟩ = 1 (0) + 2 t (2) = 4 t

Explanation

$v \times a = v (t) \times a (t)$

$= |\begin{matrix} i j k \\ 12 t 0 \\ 0 & 20 \end{matrix}|$

$= i (0) - j (0) + k (2)$

$= ⟨ 0, 0, 2 ⟩$

$‖ v \times a ‖ = \sqrt{2^{2}} = 2$

The tangential component of acceleration,

a_{T} = \frac{v \cdot a}{‖ v ‖} a_{r} = \frac{4 t}{\sqrt{1 + 4 t^{2}}}

The normal component of acceleration,

a_{N} = \frac{‖ v \times a ‖}{‖ v ‖} = \frac{2}{\sqrt{1 + 4 t^{2}}} Thus a_{T} = \frac{4 t}{\sqrt{1 + 4 t^{2}}} and a_{N} = \frac{2}{\sqrt{1 + 4 t^{2}}}

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$r (t) = (t, t^{2})$

Short Answer

Step by step solution

Introduction

Given information

Explanation

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r(t)=t,t2

Short Answer

Step by step solution

Introduction

Given information

Explanation

One App. One Place for Learning.

Most popular questions from this chapter

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$r (t) = (t, t^{2})$