Chapter 11: Q. 26 (page 880)

For each of the vector-valued functions, find the unit tangent vector.
$r (t) = (t, t^{2}, t^{3})$

Short Answer

Expert verified

$\frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 1}} <1, 2 t, 3 t^{2}>$

Step by step solution

Step1. Given Information

$Consider r (t) = <t, t^{2}, t^{3}> The objective is to find the unit tangent vector to r (t) The unit tangent vector The unit tangent vector of r (t) denoted by T (t) is defined as T (t) = \frac{r^{'} (t)}{∥r^{'} (t)∥} Consider r (t) = <t, t^{2}, t^{3}> First we compute r^{'} (t) : \begin{array}{l} r^{'} (t) = \frac{d}{d t} <t, t^{2}, t^{3}> \\ = <\frac{d}{d t} (t), \frac{d}{d t} (t^{2}), \frac{d}{d t} (t^{3})> \\ = <1, 2 t, 3 t^{2}> \end{array}$

Step2. Continue

$\begin{array}{l} ∥r^{'} (t)∥ = ∥<1, 2 t, 3 t^{2}>∥ \\ = \sqrt{1^{2} + (2 t)^{2} + {(3 t^{2})}^{2}} \\ = \sqrt{1 + 4 t^{2} + 9 t^{4}} \\ The unit tangent vector to r (t) is \\ \begin{array}{l} T (t) = \frac{r^{'} (t)}{∥r^{''} (t)∥} \\ = \frac{<1, 2 t, 3 t^{2}>}{\sqrt{1 + 4 t^{2} + 9 t^{4}}} \\ Thus the unit tangent vector to the given vector - valued function is \\ \frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 1}} <1, 2 t, 3 t^{2}> \end{array} \end{array}$

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For each of the vector-valued functions, find the unit tangent vector.
$r (t) = (t, t^{2}, t^{3})$

Short Answer

Step by step solution

Step1. Given Information

Step2. Continue

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For each of the vector-valued functions, find the unit tangent vector.r(t)=(t,t2,t3)

Short Answer

Step by step solution

Step1. Given Information

Step2. Continue

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Applied Mathematics

Probability and Statistics

Logic and Functions

Discrete Mathematics

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Study anywhere. Anytime. Across all devices.

For each of the vector-valued functions, find the unit tangent vector.
$r (t) = (t, t^{2}, t^{3})$