/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Differentiate the following func... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 9

Differentiate the following functions. $$\mathbf{r}(t)=\left\langle 2 t^{3}, 6 \sqrt{t}, 3 / t\right\rangle$$

Short Answer

Expert verified
Answer: $$\mathbf{r'}(t)=\left\langle 6t^2, 3t^{-\frac{1}{2}}, -3t^{-2}\right\rangle$$.

Step by step solution

01

Identify the components of the vector function

The given vector function $$\mathbf{r}(t)=\left\langle 2t^3, 6\sqrt{t}, \frac{3}{t}\right\rangle$$. The components are \(2t^3\), \(6\sqrt{t}\), and \(\frac{3}{t}\).
02

Differentiate the first component

The first component, \(2t^3\), can be differentiated with respect to \(t\) using the power rule: $$\frac{d}{dt}(2t^3) = 6t^2$$.
03

Differentiate the second component

The second component, \(6\sqrt{t}\), can be rewritten as \(6t^{\frac{1}{2}}\). Differentiating with respect to \(t\), using the power rule: $$\frac{d}{dt}(6t^{\frac{1}{2}}) = \frac{6}{2}t^{-\frac{1}{2}}=3t^{-\frac{1}{2}}$$.
04

Differentiate the third component

The third component, \(\frac{3}{t}\), can be rewritten as \(3t^{-1}\). Differentiating with respect to \(t\), using the power rule: $$\frac{d}{dt}(3t^{-1}) = -3t^{-2}$$.
05

Combine the derivatives into a new vector

We now combine the derivatives of each component into a new vector, which is the first derivative of the given vector function: $$\mathbf{r'}(t)=\left\langle 6t^2, 3t^{-\frac{1}{2}}, -3t^{-2}\right\rangle$$. The first derivative of the provided vector function is $$\mathbf{r'}(t)=\left\langle 6t^2, 3t^{-\frac{1}{2}}, -3t^{-2}\right\rangle$$.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Show that for real numbers \(u_{1}, u_{2},\) and \(u_{3},\) it is true that \(\left(u_{1}+u_{2}+u_{3}\right)^{2} \leq 3\left(u_{1}^{2}+u_{2}^{2}+u_{3}^{2}\right)\). (Hint: Use the Cauchy-Schwarz Inequality in three dimensions with \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) and choose \(\mathbf{v}\) in the right way.)

Determine the equation of the line that is perpendicular to the lines \(\mathbf{r}(t)=\langle 4 t, 1+2 t, 3 t\rangle\) and \(\mathbf{R}(s)=\langle-1+s,-7+2 s,-12+3 s\rangle\) and passes through the point of intersection of the lines \(\mathbf{r}\) and \(\mathbf{R}\).

Given a fixed vector \(\mathbf{v},\) there is an infinite set of vectors \(\mathbf{u}\) with the same value of proj\(_{\mathbf{v}} \mathbf{u}\). Let \(\mathbf{v}=\langle 0,0,1\rangle .\) Give a description of all position vectors \(\mathbf{u}\) such that \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}=\operatorname{proj}_{\mathbf{v}}\langle 1,2,3\rangle\).

Consider the motion of an object given by the position function $$\mathbf{r}(t)=f(t)\langle a, b, c\rangle+\left(x_{0}, y_{0}, z_{0}\right\rangle, \text { for } t \geq 0$$ where \(a, b, c, x_{0}, y_{0},\) and \(z_{0}\) are constants and \(f\) is a differentiable scalar function, for \(t \geq 0\) a. Explain why this function describes motion along a line. b. Find the velocity function. In general, is the velocity constant in magnitude or direction along the path?

Find the domains of the following vector-valued functions. $$\mathbf{r}(t)=\frac{2}{t-1} \mathbf{i}+\frac{3}{t+2} \mathbf{j}$$
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.