Chapter 11: Q. 13 (page 901)

Unit tangent vectors: Find the unit tangent vector for the given function at the specified value of t.
$r (t) = <t, t^{3}>, t = 2$

Short Answer

Expert verified

Ans: Thus the unit tangent vector to $<t, t^{3}>$ at $t = 2$ is $<\frac{1}{\sqrt{145}}, \frac{12}{\sqrt{145}}>$

Step by step solution

Step 1. Given information:

$r (t) = <t, t^{3}>, t = 2$

Step 2. Simplifying the Unit tangent vectors :

Consider $r (t) = <t, t^{3}>$

First, we compute

$r^{'} (t) = \frac{d}{d t} <t, t^{3}> = <\frac{d}{d t} (t), \frac{d}{d t} (t^{3})> = <1, 3 t^{2}> ||r^{'} (t)|| = ||<1, 3 t^{2}>|| = \sqrt{(1)^{2} + {(3 t^{2})}^{2}} = \sqrt{1 + 9 t^{4}}$

Step 3. Finding the Unit tangent vectors:

The unit tangent vector to $r (t)$ is

$T (t) = \frac{r^{'} (t)}{||r^{''} (t)||} = \frac{<1, 3 t^{2}>}{\sqrt{1 + 9 t^{4}}}$

At $t = 2$ , the unit tangent vector is

$T (2) = \frac{<1, 3 (2)^{2}>}{\sqrt{1 + 9 (2)^{4}}} = \frac{⟨ 1, 12 ⟩}{\sqrt{145}} = <\frac{1}{\sqrt{145}}, \frac{12}{\sqrt{145}}>$

Thus the unit tangent vector to $<t, t^{3}>$ at $t = 2$ is $<\frac{1}{\sqrt{145}}, \frac{12}{\sqrt{145}}>$

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Unit tangent vectors: Find the unit tangent vector for the given function at the specified value of t.
$r (t) = <t, t^{3}>, t = 2$

Short Answer

Step by step solution

Step 1. Given information:

Step 2. Simplifying the Unit tangent vectors :

Step 3. Finding the Unit tangent vectors:

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Unit tangent vectors: Find the unit tangent vector for the given function at the specified value of t.r(t)=t,t3,t=2

Short Answer

Step by step solution

Step 1. Given information:

Step 2. Simplifying the Unit tangent vectors :

Step 3. Finding the Unit tangent vectors:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

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Study anywhere. Anytime. Across all devices.

Unit tangent vectors: Find the unit tangent vector for the given function at the specified value of t.
$r (t) = <t, t^{3}>, t = 2$