Chapter 11: Q. 27 (page 880)

For each of the vector-valued functions, find the unit tangent vector.
$r (t) = (3 \sin t, 5 \cos t, 4 \sin t)$

Short Answer

Expert verified

$\frac{1}{5} ⟨ 3 \cos t, - 5 \sin t, 4 \cos t ⟩$

Step by step solution

Step1. Given Information

$Consider r (t) = ⟨ 3 \sin t, 5 \cos t, 4 \sin t ⟩ The objective is to find the unit tangent vector to r (t) The unit tangent vector The unit tangent vector of r (t) denoted by T (t) is defined as T (t) = \frac{r^{'} (t)}{∥r^{'} (t)∥} Consider r (t) = ⟨ 3 \sin t, 5 \cos t, 4 \sin t ⟩ First we compute r^{'} (t) : \begin{array}{l} r^{'} (t) = \frac{d}{d t} ⟨ 3 \sin t, 5 \cos t, 4 \sin t ⟩ \\ = <\frac{d}{d t} (3 \sin t), \frac{d}{d t} (5 \cos t), \frac{d}{d t} (4 \sin t)> \\ = ⟨ 3 \cos t, - 5 \sin t, 4 \cos t ⟩ \end{array}$

Step2. Continue

$\begin{array}{l} ∥r^{'} (t)∥ = ∥ ⟨ 3 \cos t, - 5 \sin t, 4 \cos t ⟩ ∥ \\ = \sqrt{(3 \cos t)^{2} + (- 5 \sin t)^{2} + (4 \cos t)^{2}} \\ = \sqrt{9 \cos^{2} t + 25 \sin^{2} t + 16 \cos^{2} t} \\ = \sqrt{25 \cos^{2} t + 25 \sin^{2} t} \\ = \sqrt{25 (\cos^{2} t + \sin^{2} t)} \\ = 5 \sin c e \sin^{2} t + \cos^{2} t = 1 \\ The unit tangent vector to r (t) is \\ \begin{array}{l} T (t) = \frac{r^{'} (t)}{∥r^{''} (t)∥} \\ = \frac{⟨ 3 \cos t, - 5 \sin t, 4 \cos t ⟩}{5} \\ = \frac{1}{5} ⟨ 3 \cos t, - 5 \sin t, 4 \cos t ⟩ \\ Thus the unit tangent vector to the given vector function is \\ \frac{1}{5} ⟨ 3 \cos t, - 5 \sin t, 4 \cos t ⟩ \end{array} \end{array}$

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For each of the vector-valued functions, find the unit tangent vector.
$r (t) = (3 \sin t, 5 \cos t, 4 \sin t)$

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)=(3sint,5cost,4sint)

Short Answer

Step by step solution

Step1. Given Information

Step2. Continue

One App. One Place for Learning.

Most popular questions from this chapter

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For each of the vector-valued functions, find the unit tangent vector.
$r (t) = (3 \sin t, 5 \cos t, 4 \sin t)$