Simon's copy of Laplace of trig functions

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Simon Thomas 6 years ago

Created this as a copy of Laplace of trig functions.

Name	Status	Author	Last Modified
Laplace of trig functions	draft	Julie Crowley	09/10/2016 22:11
Inverse Laplace - trig functions	draft	Julie Crowley	09/10/2016 22:24
Simon's copy of Laplace of trig functions	draft	Simon Thomas	11/03/2019 09:34

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Question statement

Note that

$L[\cos(bt)]=\frac{s}{s^2+b^2}$

$L[\sin(bt)]=\frac{b}{s^2+b^2}$

Use these results to carry out the Laplace transforms below:

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	8
b	integer	2
c	integer	2

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random(2,4,5,8,10,20)

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Generated value: `integer`

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(a)

$L[\cos(bt)]=\frac{s}{s^2+b^2}$

In this example $b=\var{a}$

Hence $L[\cos(\var{a}t)]=\frac{s}{s^2+(\var{a})^2}$

(b)

Note: $L[\sin(bt)]=\frac{b}{s^2+b^2}$

In this example $b=\var{b}$

Hence $L[\sin(\var{b}t)]=\frac{(\var{b})}{s^2+(\var{b})^2}$

(c)

Note: $L[\cos(bt)]=\frac{s}{s^2+b^2}$ and $L[\sin(bt)]=\frac{b}{s^2+b^2}$

In the first part $b=\frac{1}{\var{a}}$ and in the second part $b=\frac{1}{\var{c}}$

Hence $L[\cos(\frac{t}{\var{a}})+\sin(\frac{t}{\var{b}})]=\frac{s}{s^2+(\frac{1}{\var{a}})^2}+\frac{\frac{1}{\var{c}}}{s^2+(\frac{1}{\var{c}})^2}=\simplify{(s/(s^2+1/{a}^2)+(1/{c})/(s^2+1/{c}^2))}$

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This question is used in the following exam:

lntroduction to Laplace Transforms by Simon Thomas in Maths support.

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Simon Thomas 6 years ago

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Generated value: `integer`

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Simon's copy of Laplace of trig functions

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 6 years ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Mathematical expression

Advanced settings

Checking accuracy

Variable value generators

Pattern restriction

Variables

String restrictions

Generated value: `integer`