/*! 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 43 Let \(\mathbf{r}(t)=\ln t \mathb... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 43

Let \(\mathbf{r}(t)=\ln t \mathbf{i}+2 t \mathbf{j}+t^{2} \mathbf{k} .\) Find (a) \(\left\|\mathbf{r}^{\prime}(t)\right\|\) (b) \(\frac{d s}{d t}\) (c) \(\int_{1}^{3}\left\|\mathbf{r}^{\prime}(t)\right\| d t\).

Short Answer

Expert verified
The solutions involve derivatives and magnitudes: \( (a) \left\|\mathbf{r}^{\prime}(t)\right\| = \sqrt{\frac{1}{t^2} + 4 + 4t^2} \), \( (b) \frac{d s}{d t} = \left\|\mathbf{r}^{\prime}(t)\right\| \), \( (c) \int_{1}^{3} \left\|\mathbf{r}^{\prime}(t)\right\| d t \) requires numerical methods.

Step by step solution

01

Find the derivative of \( \mathbf{r}(t) \)

The vector function is \( \mathbf{r}(t) = \ln t \mathbf{i} + 2t \mathbf{j} + t^2 \mathbf{k} \). We differentiate each component separately: - \( \frac{d}{dt}(\ln t) = \frac{1}{t} \)- \( \frac{d}{dt}(2t) = 2 \)- \( \frac{d}{dt}(t^2) = 2t \)Therefore, the derivative \( \mathbf{r}'(t) \) is: \( \mathbf{r}'(t) = \frac{1}{t} \mathbf{i} + 2 \mathbf{j} + 2t \mathbf{k} \).
02

Compute the magnitude of \( \mathbf{r}'(t) \)

The magnitude \( \left\| \mathbf{r}'(t) \right\| \) is found using the formula for the magnitude of a vector \( \| \mathbf{a} \| = \sqrt{a_x^2 + a_y^2 + a_z^2} \):\[ \left\| \mathbf{r}'(t) \right\| = \sqrt{\left(\frac{1}{t}\right)^2 + 2^2 + (2t)^2} \] Simplifying gives:\[ \left\| \mathbf{r}'(t) \right\| = \sqrt{\frac{1}{t^2} + 4 + 4t^2} \].
03

Substitute to find \( \frac{ds}{dt} \)

By definition, \( \frac{ds}{dt} = \left\| \mathbf{r}'(t) \right\| \), so it is simply the expression from Step 2:\[ \frac{ds}{dt} = \sqrt{\frac{1}{t^2} + 4 + 4t^2} \].
04

Evaluate the integral from 1 to 3

We need to compute the integral \( \int_{1}^{3} \left\| \mathbf{r}'(t) \right\| dt \):\[ \int_{1}^{3} \sqrt{\frac{1}{t^2} + 4 + 4t^2} \, dt \]This integral may require numerical methods or special techniques (such as substitution) to solve, since it cannot be expressed in elementary functions directly. However, these methods would provide an approximate numerical value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

derivative of vector functions
The derivative of a vector function is an extension of the typical derivative concept. Instead of dealing with single-variable functions, we work with vector functions which have multiple components. If you have a vector function like \( \mathbf{r}(t) = \ln t \mathbf{i} + 2t \mathbf{j} + t^2 \mathbf{k} \), you take the derivative of each component separately.

Specifically, you can find the derivative of \( \ln t \) to be \( \frac{1}{t} \), the derivative of \( 2t \) to be 2, and the derivative of \( t^2 \) to be \( 2t \). Therefore, the derivative vector function is:
  • \( \mathbf{r}'(t) = \frac{1}{t} \mathbf{i} + 2 \mathbf{j} + 2t \mathbf{k} \)
This concept is important in understanding how a vector function changes with respect to the variable \( t \), such as determining velocity when \( \mathbf{r}(t) \) represents a position vector.
magnitude of a vector
The magnitude of a vector tells us how long the vector is, regardless of its direction. It is computed using the Pythagorean theorem in three-dimensions. For a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), the magnitude \( \| \mathbf{v} \| \) is \( \sqrt{a^2 + b^2 + c^2} \).

When you calculate the magnitude of the derivative of a vector function \( \mathbf{r}'(t) = \frac{1}{t} \mathbf{i} + 2 \mathbf{j} + 2t \mathbf{k} \), you get:
  • \( \| \mathbf{r}'(t) \| = \sqrt{\left(\frac{1}{t}\right)^2 + 2^2 + (2t)^2} \)
This simplifies to \( \sqrt{\frac{1}{t^2} + 4 + 4t^2} \).

The magnitude of a vector is crucial for applications like physics, where it can represent quantities like speed or force that have no direction attached, only size.
definite integral of functions
The definite integral of a function provides the total accumulation of a quantity over an interval. For example, if you have a function that represents speed, the integral over time would give you the total distance traveled.

In our scenario, we need to evaluate the integral \( \int_{1}^{3} \sqrt{\frac{1}{t^2} + 4 + 4t^2} \, dt \). This integration involves finding the accumulation of the magnitudes of the vector function's derivative over the interval \( t = 1 \) to \( t = 3 \).
  • This integral might not be easy to solve with elementary methods. It often requires numerical approaches like Simpson's rule or trapezoidal rule for approximation.
Evaluating such integrals helps in computing overall changes or total values, which can be used in a wide range of applications, from engineering to economics.

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Most popular questions from this chapter

(a) Suppose that at time \(t=t_{0}\) an electron has a position vector of \(\mathbf{r}=3.5 \mathbf{i}-1.7 \mathbf{j}+\mathbf{k}\), and at a later time \(t=t_{1}\) it has a position vector of \(\mathbf{r}=4.2 \mathbf{i}+\mathbf{j}-2.4 \mathbf{k}\). What is the displacement of the electron during the time interval from \(t_{0}\) to \(t_{1}\) ? (b) Suppose that during a certain time interval a proton has a displacement of \(\Delta \mathbf{r}=0.7 \mathbf{i}+2.9 \mathbf{j}-1.2 \mathbf{k}\) and its fi- nal position vector is known to be \(\mathbf{r}=3.6 \mathbf{k}\). What was the initial position vector of the proton?

Suppose that the position function of a particle moving along a circle in the \(x y\) -plane is \(\mathbf{r}=5 \cos 2 \pi t \mathbf{i}+5 \sin 2 \pi t \mathbf{j}\) (a) Sketch some typical displacement vectors over the time interval from \(t=0\) to \(t=1\). (b) What is the distance traveled by the particle during the time interval?

Determine whether the statement is true or false. Explain your answer. If \(C\) is the graph of a smooth vector-valued function \(\mathbf{r}(t)\) in 2-space, then the angle measured in the counterclockwise direction from the unit tangent vector \(\mathbf{T}(t)\) to the unit normal vector \(\mathbf{N}(t)\) is \(\pi / 2\)

Use the given information to find the position and velocity vectors of the particle. $$ \mathbf{a}(t)=\sin t \mathbf{i}+\cos t \mathbf{j}+e^{t} \mathbf{k} ; \mathbf{v}(0)=\mathbf{k} ; \mathbf{r}(0)=-\mathbf{i}+\mathbf{k} $$

Use the given information to find the position and velocity vectors of the particle. $$ \mathbf{a}(t)=\mathbf{i}+e^{-t} \mathbf{j} ; \mathbf{v}(0)=2 \mathbf{i}+\mathbf{j} ; \mathbf{r}(0)=\mathbf{i}-\mathbf{j} $$
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