Presentation is loading. Please wait.

Presentation is loading. Please wait.

Department of Mathematics University of Leicester

Similar presentations


Presentation on theme: "Department of Mathematics University of Leicester"— Presentation transcript:

1 Department of Mathematics University of Leicester
Parametric Department of Mathematics University of Leicester

2 What is it? A parametric equation is a method of defining a relation using parameters. For example, using the equation: We can use a free parameter, t, setting: and

3 What is it? We can see that this still satisfies the equation, while defining a relationship between x and y using the free parameter, t.

4 Why do we use parametric equations
Parameterisations can be used to integrate and differentiate equations term wise. You can describe the motion of a particle using a parameterisation: r being placement.

5 Why do we use parametric equations
Now we can use this to differentiate each term to find v, the velocity:

6 Why do we use parametric equations
Parameters can also be used to make differential equations simpler to differentiate. In the case of implicit differentials, we can change a function of x and y into an equation of just t.

7 Why do we use parametric equations
Some equations are far easier to describe in parametric form. Example: a circle around the origin Cartesian form: Parametric form:

8 How to get Cartesian from parametric
Getting the Cartesian equation of a parametric equation is done more by inspection that by a formula. There are a few useful methods that can be used, which are explored in the examples.

9 How to get Cartesian from parametric
Example 1: Let: So that: and

10 How to get Cartesian from parametric
Next set t in terms of y: Now we can substitute t in to the equation of x to eliminate t.

11 How to get Cartesian from parametric
Substituting in t: Which expands to:

12 How to get Cartesian from parametric
Example 2: Let: So that: and

13 How to get Cartesian from parametric
To change this we can see that: And

14 How to get Cartesian from parametric
And as we know that We can see that:

15 How to get Cartesian from parametric
Which equals: This is the Cartesian equation for an ellipse.

16 Example Example 3: let: Be the Cartesian equation of a circle at the point (a,b). Change this into parametric form.

17 Example If we set: And: Then we can solve this using the fact that:

18 Example From this we can see that: So: Therefore:

19 Example Similarly: So: Therefore:

20 Example Compiling this, we can see that:
Which is the parametric equation for a circle at the point (a,b).

21 Polar co-ordinates Parametric equations can be used to describe curves in polar co-ordinate form: For example:

22 Polar co-ordinates Here we can see, that if we set t as the angle, then we can describe x and y in terms of t: Using trigonometry: and

23 Polar co-ordinates These can be used to change Cartesian equations to parametric equations:

24 Polar co-ordinates: example
Let: Be the equation for a circle. If we set:

25 Polar co-ordinates: example
We can see that if we substitute these in, then the equation still holds: Therefore we can use: As a parameterisation for a circle.

26 Finding the gradient of a parametric curve
To find dy/dx we need to use the chain rule:

27 How to get Cartesian from parametric: example
Let: and Then:

28 How to get Cartesian from parametric: example
Then, using the chain rule:

29 Extended parametric example
Let: Be the Cartesian equation.

30 Extended parametric example
Then to change this into parametric form, we need to find values of x and y that satisfy the equation. If we set: And:

31 Extended parametric example
Then we have: Which expands to:

32 Extended parametric example
We know that: Therefore we can see that our values of x and y satisfy the equation. Therefore:

33 Extended parametric example
Now, as this is the placement of the particle, we can find the velocity of the particle by differentiating each term:

34 Extended parametric example
Next, we can find the gradient of the curve. Using the formula:

35 Extended parametric example
Using this: And:

36 Extended parametric example
Therefore the gradient is:

37 Conclusion Parametric equations are about changing equations to just 1 parameter, t. Parametric is used to define equations term wise. We can use the chain rule to find the gradient of a parametric equation.

38 Conclusion Standard parametric manipulation of polar co- ordinates is:
x=rcos(t) Y=rsin(t)

39


Download ppt "Department of Mathematics University of Leicester"

Similar presentations


Ads by Google