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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

01

Step 1. Given information:

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

02

Step 2. Simplifying the Unit tangent vectors :

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

First, we compute

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

03

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

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

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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Most popular questions from this chapter

Imagine that you are driving on a twisting mountain road. Describe the unit tangent vector, principal unit normal vector, and binomial vector as you ascend, descend, twist right, and twist left.

Find parametric equations for each of the vector-valued functions in Exercises 26鈥�34, and sketch the graphs of the functions, indicating the direction for increasing values of t.

$r (t) = (1 + \sin t, 3 - \cos 2 t), f o r t 鈭� [0, 2 蟺]$

Let $f (t)$ be a differentiable scalar function and $r (t)$ be a differentiable vector function. Prove that $\frac{d}{d t} (f (t) r (t)) = f' (t) r (t) + f (t) r' (t)$ . (This is Theorem 11.11 (b).)

Prove that the cross product of two orthogonal unit vectors is a unit vector.

Given a differentiable vector-valued function r(t), what is the definition of the unit tangent vector T(t)?

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