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

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

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### 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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