Chapter 11: Q. 37 (page 872)

In Exercises 35-39 a vector function $r (t)$ and scalar function $t = f (τ)$ are given. Find $\frac{d r}{d τ}$ .
$37. r (t) = ⟨ \sec t, 1 / t, e^{t} \ln t ⟩, t = τ^{- 1}$

Short Answer

Expert verified

$\frac{d r}{d τ} = ⟨ \frac{- \sec \frac{1}{τ} \tan \frac{1}{τ}}{τ^{2}}, 1, \frac{e^{\frac{1}{τ}}}{τ^{2}} (\ln τ - τ) ⟩$

Step by step solution

Step 1. Given data

The given vector function is $r (t) = ⟨ \sec t, 1 / t, e^{t} \ln t ⟩, t = τ^{- 1}$

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,

$\begin{aligned} r (t) = ⟨ \sec t, \frac{1}{t}, e^{t} \ln t ⟩ \\ \frac{d r}{d t} = ⟨ \sec t \tan t, \frac{- 1}{t^{2}}, \frac{e^{t}}{t} + e^{t} \ln t ⟩ \end{aligned}$

By the chain rule,

$\frac{d r}{d τ} \begin{aligned} = ⟨ \sec t \tan t, \frac{- 1}{t^{2}}, \frac{e^{t}}{t} + e^{t} \ln t ⟩ ⟨ \frac{- 1}{τ^{2}} ⟩ \\ = ⟨ \sec \frac{1}{τ} \tan \frac{1}{τ}, \frac{- 1}{{(\frac{1}{τ})}^{2}}, \frac{e^{\frac{1}{τ}}}{\frac{1}{τ}} + e^{\frac{1}{τ}} \ln (\frac{1}{τ}) ⟩ (\frac{- 1}{τ^{2}}) \\ = ⟨ \frac{- \sec \frac{1}{τ} \tan \frac{1}{τ}}{τ^{2}}, τ^{2} \cdot \frac{1}{τ^{2}}, \frac{- e^{\frac{1}{τ}}}{τ^{2}} (τ + \ln e^{- 1}) ⟩ \\ = ⟨ \frac{- \sec \frac{1}{τ} \tan \frac{1}{τ}}{τ^{2}}, 1, \frac{- e^{\frac{1}{τ}}}{τ^{2}} (τ - \ln τ) ⟩ \\ = ⟨ \frac{- \sec \frac{1}{τ} \tan \frac{1}{τ}}{τ^{2}}, 1, \frac{e^{\frac{1}{τ}}}{τ^{2}} (\ln τ - τ) ⟩ \end{aligned}$

Thus $\frac{d r}{d τ} = ⟨ \frac{- \sec \frac{1}{τ} \tan \frac{1}{τ}}{τ^{2}}, 1, \frac{e^{\frac{1}{τ}}}{τ^{2}} (\ln τ - τ) ⟩$

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In Exercises 35-39 a vector function $r (t)$ and scalar function $t = f (τ)$ are given. Find $\frac{d r}{d τ}$ .
$37. r (t) = ⟨ \sec t, 1 / t, e^{t} \ln t ⟩, t = τ^{- 1}$

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.37.r(t)=⟨sect,1/t,etlnt⟩,t=τ−1

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

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Statistics

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Study anywhere. Anytime. Across all devices.

In Exercises 35-39 a vector function $r (t)$ and scalar function $t = f (τ)$ are given. Find $\frac{d r}{d τ}$ .
$37. r (t) = ⟨ \sec t, 1 / t, e^{t} \ln t ⟩, t = τ^{- 1}$