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

Discrete Mathematics

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Theoretical and Mathematical Physics

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Calculus

Study anywhere. Anytime. Across all devices.

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