Chapter 11: Q. 39 (page 872)

In Exercises 35-39 a vector function $r (t)$ and scalar function $t = f (τ)$ are given. Find $\frac{d r}{d τ}$ .
$39. r (t) = \cos 3 t i + \sin 4 t j + t k, t = 5 τ - 2$

Short Answer

Expert verified

$\frac{d r}{d τ} = ⟨ - 15 \sin (15 τ - 6), 20 \cos (20 τ - 8), 5 ⟩$

Step by step solution

Step 1. Given data

The given vector function is $r (t) = \cos 3 t i + \sin 4 t j + t k, t = 5 τ - 2$

We have to find $\frac{d r}{d τ}$

Step 2. Use chain rule

If $r (t)$ is a vector function and $t = f (τ)$ , a scalar function. Then the chain rule states that $\frac{d r}{d τ} = \frac{d r}{d t} \cdot \frac{d t}{d τ}$

We have,

$\begin{aligned} r (t) = \cos 3 t i + \sin 4 t j + t k \\ r (t) = ⟨ \cos 3 t, \sin 4 t, t ⟩ \\ \frac{d r}{d t} = ⟨ - 3 \sin 3 t, 4 \cos 3 t, 1 ⟩ \end{aligned}$

It is given, $t = 5 τ - 2$

$\frac{d t}{d τ} = 5$

By the chain rule,

$\frac{d r}{d τ} = ⟨ - 3 \sin 3 t, 4 \cos 4 t, 1 ⟩ 5 = ⟨ - 15 \sin 3 t, 20 \cos 4 t, 5 ⟩ = ⟨ - 15 \sin 3 (5 τ - 2), 20 \cos 4 (5 τ - 2), 5 ⟩ = ⟨ - 15 \sin (15 τ - 6), 20 \cos (20 τ - 8), 5 ⟩$

Thus $\frac{d r}{d τ} = ⟨ - 15 \sin (15 τ - 6), 20 \cos (20 τ - 8), 5 ⟩$

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In Exercises 35-39 a vector function $r (t)$ and scalar function $t = f (τ)$ are given. Find $\frac{d r}{d τ}$ .
$39. r (t) = \cos 3 t i + \sin 4 t j + t k, t = 5 τ - 2$

Short Answer

Step by step solution

Step 1. Given data

Step 2. Use chain rule

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In Exercises 35-39 a vector function r(t)and scalar function t=f(τ)are given. Find drdτ.39.r(t)=cos⁡3ti+sin⁡4tj+tk,t=5τ−2

Short Answer

Step by step solution

Step 1. Given data

Step 2. Use chain rule

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Theoretical and Mathematical Physics

Probability and Statistics

Mechanics Maths

Geometry

Calculus

Study anywhere. Anytime. Across all devices.

In Exercises 35-39 a vector function $r (t)$ and scalar function $t = f (τ)$ are given. Find $\frac{d r}{d τ}$ .
$39. r (t) = \cos 3 t i + \sin 4 t j + t k, t = 5 τ - 2$