## 17Calculus - The Unit Step Function and The Laplace Transform

### Practice

Calculus 1 Practice

Calculus 2 Practice

Practice Exams

Calculus Tools

### Articles

This page discusses the unit step function, also called the heaviside function, using the Laplace Transform.

Unit Step Function

The unit step function looks exactly as the name implies. It is a discontinuous function which is piecewise continuous. It is zero to the left of zero, one to the right of zero and $$1/2$$ at zero. The graph is shown on the right ($$x=t$$) and the equation is
$$\displaystyle{ u(t) = \left\{ \begin{array}{lr} 0 & t < 0 \\ 1/2 & t = 0 \\ 1 & t > 0 \end{array} \right. }$$

There are variations of this that you may see that usually involve the value at $$t=0$$. Some books set this value at $$1$$, others at $$0$$. These variations should not affect your use of Laplace transforms. Check with your book and instructor to see which they prefer.

Looking carefully at the graph, we can see that there is only one discontinuity at $$t=0$$, which is nonremovable. The graph is a function, since it passes the vertical line test. One way we could use this function is to multiply it by another function, say $$g(t)$$ and, when we do that, this unit step function essentially cancels out everything to the left of zero in $$g(t)$$ and everything to the right of zero stays as $$g(t)$$. Using the unit step function this way is a way to filter or isolate part of a function. Now, this would be pretty limiting if everything was centered at zero. However, we can shift the unit step function to suit our needs. This is one thing shown in this first video.
Okay, let's watch a video to see how we use this function and it's Laplace transform.

### Dr Chris Tisdell - Introduction to Heaviside step function [31min-5secs]

video by Dr Chris Tisdell

Practice

Calculate $$\mathcal{L}\{ u(t-a) \}$$ where $$u(t)$$ is the unit step function.

Problem Statement

Calculate $$\mathcal{L}\{ u(t-a) \}$$ where $$u(t)$$ is the unit step function.

Solution

### 3609 video

video by blackpenredpen

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

You CAN Ace Calculus

### 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

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

### Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

### Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

free ideas to save on books

 The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free.

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.