/*! 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 20 Find a tangent vector at the giv... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 20

Find a tangent vector at the given value of \(t\) for the following parameterized curves. $$\mathbf{r}(t)=2 e^{t} \mathbf{i}+e^{-2 t} \mathbf{j}+4 e^{2 t} \mathbf{k}, t=\ln 3$$

Short Answer

Expert verified
Question: Find the tangent vector of the curve \(\mathbf{r}(t) = 2e^t \mathbf{i} + e^{-2t} \mathbf{j} + 4e^{2t} \mathbf{k}\) at \(t=\ln 3\). Answer: The tangent vector at \(t=\ln 3\) is \(\mathbf{T}(t) = 6 \mathbf{i} - \frac{2}{9} \mathbf{j} + 72 \mathbf{k}\).

Step by step solution

01

Find the derivative of the parametrized curve

To find the derivative of \(\mathbf{r}(t)\) with respect to \(t\), we must take the derivative of each component with respect to \(t\). The derivative of the given curve with respect to \(t\) is: $$\frac{d\mathbf{r}}{dt} = \left(\frac{d}{dt}(2e^t \mathbf{i})\right) + \left(\frac{d}{dt}(e^{-2t} \mathbf{j})\right) + \left(\frac{d}{dt}(4e^{2t} \mathbf{k})\right)$$
02

Evaluate the derivatives

Now we evaluate each derivative separately. For the first component: $$\frac{d}{dt}(2e^t \mathbf{i}) = 2 \frac{d}{dt}(e^t\mathbf{i}) = 2e^t \mathbf{i}$$ For the second component: $$\frac{d}{dt}(e^{-2t} \mathbf{j}) = -2e^{-2t} \mathbf{j}$$ And for the third component: $$\frac{d}{dt}(4e^{2t} \mathbf{k}) = 8e^{2t} \mathbf{k}$$ Putting these together, we get the derivative of the curve: $$\frac{d\mathbf{r}}{dt} = 2e^t \mathbf{i} - 2e^{-2t} \mathbf{j} + 8e^{2t} \mathbf{k}$$
03

Find the tangent vector at \(t=\ln 3\)

Now that we have the derivative of the curve, we can find the tangent vector at \(t = \ln 3\) by plugging this value into the equation we just derived. $$\frac{d\mathbf{r}}{dt}\bigg|_{t=\ln 3} = 2e^{\ln 3} \mathbf{i} - 2e^{-2\ln 3} \mathbf{j} + 8e^{2\ln 3} \mathbf{k}$$ Recall that \(e^{\ln x }= x\). Using this, we can rewrite the expression: $$\frac{d\mathbf{r}}{dt}\bigg|_{t=\ln 3} = 6 \mathbf{i} - 2\left(\frac{1}{3^2}\right) \mathbf{j} + 8(3^2) \mathbf{k}$$
04

Simplify and write the final answer

Now we simplify the expression: $$\frac{d\mathbf{r}}{dt}\bigg|_{t=\ln 3} = 6 \mathbf{i} - \frac{2}{9} \mathbf{j} + 72 \mathbf{k}$$ So, the tangent vector to the curve at \(t = \ln 3\) is \(\mathbf{T}(t) = 6 \mathbf{i} - \frac{2}{9} \mathbf{j} + 72 \mathbf{k}\).

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

Compute the indefinite integral of the following functions. $$\mathbf{r}(t)=2^{t} \mathbf{i}+\frac{1}{1+2 t} \mathbf{j}+\ln t \mathbf{k}$$

Find the function \(\mathbf{r}\) that satisfies the given conditions. $$\mathbf{r}^{\prime}(t)=\left\langle e^{t}, \sin t, \sec ^{2} t\right\rangle ; \mathbf{r}(0)=\langle 2,2,2\rangle$$

Evaluate the following definite integrals. $$\int_{-\pi}^{\pi}(\sin t \mathbf{i}+\cos t \mathbf{j}+2 t \mathbf{k}) d t$$

Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\langle\cos 3 t, \sin 4 t, \cos 6 t\rangle$$

Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\left\langle e^{4 t}, 2 e^{-4 t}+1,2 e^{-t}\right\rangle$$
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

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.