## 17Calculus - Smooth Vector Functions

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.

$$\vec{r}(t)$$ 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

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

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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.

$$\vec{r}(t)$$ 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

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

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You CAN Ace Calculus

 vectors vector functions derivatives of vector functions

### 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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