\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Smooth Vector Functions

Coordinate Systems

Vectors

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Triple Integrals - 3Int

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Vector Fields

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Calculus 2 Practice

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Coordinate Systems

Vectors

Using Vectors

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Vector Functions

Partial Derivatives

Partial Integrals

Double Integrals - 2Int

Triple Integrals - 3Int

Practice

Vector Fields

SV Calculus

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On this page we explain how to determine where a vector function is smooth. You need to understand what vector functions are and how to take the derivative of vector functions in order to understand this page.

For many of our calculations with vector functions, we will require that the vector function be smooth. A smooth vector function is one where the derivative is continuous and where the derivative is not equal to zero. This is comparable to what you already know from basic continuity where a graph is continuous and does not contain any sharp corners. Here is a good video clip explaining this in more detail.

MIP4U - Determining Where a Space Curve is Smooth from a Vector Valued Function [1min-35secs]

video by MIP4U

Okay, so you are ready to work some practice problems on your own.

Practice

Unless otherwise instructed, determine the values of \(t\) where the vector function is smooth.

\( \vec{r}(t) = t^3\vhat{i} - t^5\vhat{j} \)

Problem Statement

Determine the values of \(t\) where the vector function \( \vec{r}(t) = t^3\vhat{i} - t^5\vhat{j} \) is smooth.

Final Answer

\( \vec{r}(t) \) is smooth everywhere except for \( t = 0 \)

Problem Statement

Determine the values of \(t\) where the vector function \( \vec{r}(t) = t^3\vhat{i} - t^5\vhat{j} \) is smooth.

Solution

2039 video

video by PatrickJMT

Final Answer

\( \vec{r}(t) \) is smooth everywhere except for \( t = 0 \)

close solution

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\( \vec{r}(t) = (t^2e^{-t})\vhat{i} - 2(t-1)^2\vhat{j} \)

Problem Statement

Determine the values of \(t\) where the vector function \( \vec{r}(t) = (t^2e^{-t})\vhat{i} - 2(t-1)^2\vhat{j} \) is smooth.

Final Answer

\( \vec{r}(t) \) is smooth for all \(t\)

Problem Statement

Determine the values of \(t\) where the vector function \( \vec{r}(t) = (t^2e^{-t})\vhat{i} - 2(t-1)^2\vhat{j} \) is smooth.

Solution

2040 video

video by PatrickJMT

Final Answer

\( \vec{r}(t) \) is smooth for all \(t\)

close solution

Log in to rate this practice problem and to see it's current rating.

You CAN Ace Calculus

Topics You Need To Understand For This Page

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

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Practice Instructions

Unless otherwise instructed, determine the values of \(t\) where the vector function is smooth.

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