Q. 26 For each of the vector-valued fu... [FREE SOLUTION]

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

$Consider r (t) = <t, t^{2}, t^{3}> The objective is to find the unit tangent vector to r (t) The unit tangent vector The unit tangent vector of r (t) denoted by T (t) is defined as T (t) = \frac{r^{鈥�} (t)}{鈭�r^{鈥�} (t)鈭�} Consider r (t) = <t, t^{2}, t^{3}> First we compute r^{鈥�} (t) : \begin{array}{l} r^{鈥�} (t) = \frac{d}{d t} <t, t^{2}, t^{3}> \\ = <\frac{d}{d t} (t), \frac{d}{d t} (t^{2}), \frac{d}{d t} (t^{3})> \\ = <1, 2 t, 3 t^{2}> \end{array}$

Step2. Continue

$\begin{array}{l} 鈭�r^{鈥�} (t)鈭� = 鈭�<1, 2 t, 3 t^{2}>鈭� \\ = \sqrt{1^{2} + (2 t)^{2} + {(3 t^{2})}^{2}} \\ = \sqrt{1 + 4 t^{2} + 9 t^{4}} \\ The unit tangent vector to r (t) is \\ \begin{array}{l} T (t) = \frac{r^{鈥�} (t)}{鈭�r^{鈥测赌�} (t)鈭�} \\ = \frac{<1, 2 t, 3 t^{2}>}{\sqrt{1 + 4 t^{2} + 9 t^{4}}} \\ Thus the unit tangent vector to the given vector - valued function is \\ \frac{1}{\sqrt{9 t^{4} + 4 t^{2} + 1}} <1, 2 t, 3 t^{2}> \end{array} \end{array}$

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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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91影视

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

Decision Maths

Pure Maths

Calculus

Discrete Mathematics

Logic and Functions

Probability and Statistics

Study anywhere. Anytime. Across all devices.

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