/*! 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 44 Find the indefinite integral. ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the indefinite integral. $$ \int\left(4 t^{3} \mathbf{i}+6 t \mathbf{j}-4 \sqrt{t} \mathbf{k}\right) d t $$

Short Answer

Expert verified
The indefinite integral \(\int\left(4 t^{3} \mathbf{i}+6 t \mathbf{j}-4 \sqrt{t} \mathbf{k}\right) d t = t^{4}\mathbf{i} + 3t^{2}\mathbf{j} - \frac{8}{3}t^{3/2}\mathbf{k} + C\)

Step by step solution

01

Analyzing The Integrals

In the problem, recognize that each component is a separate function of \(t\). The vector \(4t^{3}\mathbf{i} + 6t\mathbf{j} - 4\sqrt{t}\mathbf{k}\) can be decomposed into three independent integrals.
02

Computing Component-Wise Integrals

Evaluate each integral independently. For the \(\mathbf{i}\)-component, \( \int 4t^{3} dt = t^{4}\); for the \(\mathbf{j}\)-component, \(\int 6t dt = 3t^{2}\); and for the \(\mathbf{k}\)-component, \( \int -4\sqrt{t} dt = -\frac{8}{3}t^{3/2}\). Remember to add the constant of integration \(C\) to each term.
03

Compiling The Results

Bring the results back together. The solution of the given integral is the vector obtained by collecting the results from the second step: \(t^{4}\mathbf{i} + 3t^{2}\mathbf{j} - \frac{8}{3}t^{3/2}\mathbf{k} + C\), where \(C\) may also be a vector, since we are dealing with vector functions.

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 91Ó°ÊÓ!

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

A projectile is launched with an initial velocity of 100 feet per second at a height of 5 feet and at an angle of \(30^{\circ}\) with the horizontal. (a) Determine the vector-valued function for the path of the projectile. (b) Use a graphing utility to graph the path and approximate the maximum height and range of the projectile. (c) Find \(\mathbf{v}(t),\|\mathbf{v}(t)\|,\) and \(\mathbf{a}(t)\) (d) Use a graphing utility to complete the table. $$ \begin{array}{|l|l|l|l|l|l|l|} \hline \boldsymbol{t} & 0.5 & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 \\ \hline \text { Speed } & & & & & & \\ \hline \end{array} $$ (e) Use a graphing utility to graph the scalar functions \(a_{\mathbf{T}}\) and \(a_{\mathrm{N}} .\) How is the speed of the projectile changing when \(a_{\mathrm{T}}\) and \(a_{\mathbf{N}}\) have opposite signs?

The position vector \(r\) describes the path of an object moving in space. Find the velocity, speed, and acceleration of the object. $$ \mathbf{r}(t)=4 t \mathbf{i}+4 t \mathbf{j}+2 t \mathbf{k} $$

Find the vectors \(\mathrm{T}\) and \(\mathrm{N},\) and the unit binormal vector \(\mathbf{B}=\mathbf{T} \times \mathbf{N},\) for the vector-valued function \(\mathbf{r}(t)\) at the given value of \(t\). $$ \mathbf{r}(t)=2 e^{t} \mathbf{i}+e^{t} \cos t \mathbf{j}+e^{t} \sin t \mathbf{k}, \quad t_{0}=0 $$

Use the model for projectile motion, assuming there is no air resistance. Eliminate the parameter \(t\) from the position function for the motion of a projectile to show that the rectangular equation is \(y=-\frac{16 \sec ^{2} \theta}{v_{0}^{2}} x^{2}+(\tan \theta) x+h\)

Use the model for projectile motion, assuming there is no air resistance. Find the angle at which an object must be thrown to obtain (a) the maximum range and (b) the maximum height.

See all solutions

Recommended explanations on Math Textbooks

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.