/*! 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 58 Find the domain of the following... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the domain of the following vector-valued functions. $$\mathbf{r}(t)=\cos 2 t \mathbf{i}+e^{\sqrt{t}} \mathbf{j}+\frac{12}{t} \mathbf{k}$$

Short Answer

Expert verified
Answer: The domain of the given vector-valued function is \(t > 0\).

Step by step solution

01

Analyze the first component

The first component is given by \(\cos 2t\). The cosine function is defined for all real numbers, so there are no domain restrictions for this component.
02

Analyze the second component

The second component is given by \(e^{\sqrt{t}}\). The exponential function is defined for all real numbers, but since we have \(\sqrt{t}\), we need to ensure \(t \geq 0\). So the domain restriction for this component is \(t \geq 0\).
03

Analyze the third component

The third component is given by \(\frac{12}{t}\). This function is undefined when \(t = 0\), so the domain restriction for this component is \(t \neq 0\).
04

Combine domain restrictions

To find the overall domain of the vector-valued function, we need to combine the domain restrictions found from each component. From Step 2, we have \(t \geq 0\). From Step 3, we have \(t \neq 0\). Combining these restrictions, we get the overall domain as \(t > 0\).
05

Final Answer

The domain of the given vector-valued function is \(t > 0\).

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

Properties of dot products Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle,\) and \(\mathbf{w}=\left\langle w_{1}, w_{2}, w_{3}\right\rangle .\) Prove the following vector properties, where \(c\) is a scalar. $$\mathbf{u} \cdot(\mathbf{v}+\mathbf{w})=\mathbf{u} \cdot \mathbf{v}+\mathbf{u} \cdot \mathbf{w}$$

Relationship between \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) Consider the ellipse \(\mathbf{r}(t)=\langle 2 \cos t, 8 \sin t, 0\rangle,\) for \(0 \leq t \leq 2 \pi\) Find all points on the ellipse at which \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) are orthogonal.

Graph the curve \(\mathbf{r}(t)=\left\langle\frac{1}{2} \sin 2 t, \frac{1}{2}(1-\cos 2 t), \cos t\right\rangle\) and prove that it lies on the surface of a sphere centered at the origin.

Suppose the vector-valued function \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) is smooth on an interval containing the point \(t_{0} .\) The line tangent to \(\mathbf{r}(t)\) at \(t=t_{0}\) is the line parallel to the tangent vector \(\mathbf{r}^{\prime}\left(t_{0}\right)\) that passes through \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right) .\) For each of the following functions, find an equation of the line tangent to the curve at \(t=t_{0} .\) Choose an orientation for the line that is the same as the direction of \(\mathbf{r}^{\prime}\). $$\mathbf{r}(t)=\left\langle 3 t-1,7 t+2, t^{2}\right\rangle ; t_{0}=1$$

Let \(\mathbf{u}(t)=\left\langle 1, t, t^{2}\right\rangle, \mathbf{v}(t)=\left\langle t^{2},-2 t, 1\right\rangle\) and \(g(t)=2 \sqrt{t}\). Compute the derivatives of the following functions. $$\mathbf{v}(g(t))$$

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.