Next, a time shift is demonstrated. Here our function is changed from f(t) to f(t-2). Notice that subtracting 2 from t in the function results in a positive shift of the graph.

In the next two graphs t will be scaled. Scaling t is not quite as intuitive as we may have expected. When we multiply t by 2, corresponding points of the function now occur at 1/2 the time they previously had. When we divide t by 2, each corresponding time on the graph occurs at a t that is now multiplied by 2. Notice that each of these factors directly affects the duration of the signal.

Scaling the amplitude has more intuitive results. If we multiply f(t) by 2, the amplitude of 1 is changed to 2. Multiplying f(t) by 1/2 results in an amplitude of 1/2.

Finally, multiplying t by -1 mirrors our function over the y-axis. Each time now occurs at its negative.

Example:
Here we will attempt to convert f(t) into 2*f(.5t+3). The graph of f(t) is shown below.

The easiest way to handle this type of problem without error is to manipulate the function one step at a time. First, I have converted f(t) into 2*f(t). Only the peaks are changed here (by a factor of 2).

Next, I convert 2*f(t) into 2*f( t). Notice how the actually expands our graph duration by a factor of 2 (from a 6 sec duration to a 12 sec duration).

Finally, we move from 2*f( t) to 2*f( t + 3). As shown in the discussion above, this is a time shift. Time shifts can be a little confusing because adding results in a negative shift of our graph. Try to think of it as our signal occurring 3 seconds earlier than before, reading from left to right on the graph. The easiest way to do this part is shift each x-intercept by 3 seconds (to the left, of course).

System Equation:

For example purposes, I will solve a system equation with complex roots.  A system equation with complex roots as a function of will appear in the following format (if it does not, you need to manipulate your equation to be in the form):

Roots:

So we have

which also equals

so your first step is to look at your equation and determine your roots, then write out your equation with constants.

Example with initial conditions and

so now we can write our equation as follows:

In order to solve for and we need to use our initial conditions.  To evaluate the first derivative initial condition, we must first take the derivative of our that we just found.

evaluating this equation with t = 0 and the response equal to -4v, we get this:

evaluating our equation with t = 0 and the response equal to 3v, we calculate

Using these two equations, we calculate our constants:

and

Fill these into our equation to determine the final result.

Now you know how to solve this common differential equations and linear systems problem, determine characteristic roots and modes, and write system equations.

The integral of a function is represented like so: and it can be thought of as a sum of areas like so:

Here the integral is performed on the function from point a to point b, which would make it a definite integral because the bounds are defined.  Our indefinite integral is the same procedure, except missing the bounds, which makes indefinite integral operation require a little twist.

We all know that the derivative of a constant is zero.  For instance, the derivative of 5 is equal to zero.  Once performing this derivative, we should still be able to perform the anti-derivative on this new function (zero) to obtain the original equation (5).  But how will we know what number permeates from performing an anti-derivative of zero.  An indefinite integral is called indefinite because the bounds are not defined on the integration, like so:  .  When we perform our indefinite integral we represent this long-lost constant by the letter ‘C’.

Before we integrate our golf cart velocity equation, lets go ahead and look at the laws of integration:

The integral can also be split up into separate individual integrals if there is addition in the function you are integrating.

and whenever you add these integrals together, you only need to account for 1 of the constants (C).

So the integral of our velocity will go as follows:

or

We can solve for our constant, C, if we are given initial conditions, such as the golf cart was moving at y’(0) = 1 m/s when we began collecting our data.  Otherwise, we leave the integral in this form.  If you would like to learn how to perform a definite integral, refer to our article on definite integrals.

Cartesian:

Cylindrical:

Spherical:

In most cases you will only need to work from Cartesian to Cylindrical or Spherical OR back, so I will only supply those equations. If you need to work between Cylindrical and Spherical, it would be one more simple step working from one of those, to Cartesian, then on to the other.

Cartesian Cylindrical:

Cartesian Cylindrical:

Cartesian Spherical:

Cartesian Spherical:

Plug the values from any given points into the correct equation to convert to a different type of coordinate system.

You would begin by finding the vector between these two points. The direction of this vector may be important so look for key words such as . Once we have established the direction we’re going in, to in this case, we subtract the beginning point from the end point. . This will give us the vector we are looking for. The next step would be to convert this vector into a unit vector, by dividing it by it’s magnitude.

These are the two formulas we are looking at:

Note:
Here I use as my vector from to and denotes the unit vector from to .

Implementing these formulas:

Hmm.. The End

RULE 1:

This is a scalar product that is equal to the two magnitudes multiplied together and multiplied by the cosine of the angle between them.  If  the two vectors are perpendicular to each other then the angle between them is 90 degrees, which will make the dot product equal zero.  This is an equivalent equation.

RULE 2:

When you finish your dot product, you should have a number, not a directional vector.  So if you get something like this you did something wrong: .  If you ended up with (any #) then you don’t have to completely rule out your answer.

A few vectors to practice with:

Okay, now we have found our dot product by applying RULE 2 above.  We can use this value along with our individual vector magnitudes to apply RULE 1 and obtain the angle .

We can calculate the projection of the vector A onto the vector B by this relationship:

Note that this is a scalar quantity, and that we can also define the projection of B onto vector A in a similar fashion:

Perform a dot product in MATLAB like so:

>>A = [ 1 2 3];

>>B = [2 3 4];

>>dot(A,B)

Enjoy

Vector

In MATLAB: >>A = [1 2 4];

Vector

In MATLAB: >>B = [2 3 4];

When you add vectors together, you add each individual directional component ().  Subtraction works the exact same way.  Lets go ahead and do a few, C will represent the vector that results from the addition and subtraction.

These vectors are all in 3-dimensional space with a X, Y and Z component.  The number in front of each directional component is the weight or magnitude of that particular directional component of the vector.  There it is, short and sweet

Since these integrals can be tedious and certain functions tend to reoccur, a table of Laplace Transforms has been linked:

Laplace Transforms Table

This table can be a little complex to use at first so an example is provided below to get you started. In this problem we implement the Laplace Transform and Inverse Laplace Transform to solve for .

The first step is to take the Laplace Transform of both sides of the equation. Use element 1 of our table for the right side and element 18 for the left side. Note that the initial conditions are necessary to take the Laplace Transform of the left side.

Inputting our initial conditions:

Assuming you are an engineering student and can do a little alegebra, our next step is to find the terms that have in common and factor it out. Our goal is to find because, after all, :

After solving for and factoring the denominator:

Now we arrive at the trickier part of this procedure. We must take the Inverse Laplace of to find . If our function does not match anything in the table, such as this case, factoring is a good place to start. This problem can easily be factored using the function on your TI-89. Just go to and enter your function in parenthesis. Using this function:

Noting and , we can take the Laplace Transform of each term independently and also manipulate the constant terms if necessary or just pull them out. Using the 2nd property in our Laplace Transform table:

To check your work you can plug into the original, differential equation and at we find that .

rmax = 1;
V = 0;
step = 0.02;
A = 5;

B = 10;
C = 15;
for x= -5 : step : 5
for y = -10 : step : 10
for z = -15 : step : 15
if((x/A).^2 + (y/B).^2 + (z/C).^2) < rmax
V = V + step.^3;
end
end
end
end
disp(V)

it should spit out , approximately

If n is a positive integer, this is your theorem:

let’s apply this to our equation .  Our derivative is looks like this to start where n = 1 because 5x is not raised to any power.

The first derivative will be represented like so:

or

Our equation is now equal to a constant.  If you take the derivative of ANY constant, the result is Zero.  So here

Let’s make this example a little more realistic, our new equation is

each piece of the equation loses 1 order ( second order), keeping in mind that the 3 was a constant and is now zero.

we can continue all the way to zero.