Parametric functions are functions that have two equations, one for x and one for y. They can be organized into of x and y. Whenever you plug in t values, you will have an x-value and y-value returned.
This is an example of a parametric function, with `color(black)(x=t^2)` and `color(black)(y=2t)`
Vector-Valued functions, like parametric functions, contain more than one equation. There are three main vectors that you will work with in Calculus BC: position vectors, velocity vectors, and acceleration vectors. The position vector will look like `color(black)(vecR= (:x(t), y(t):) )` or something similar.
Vectors have direction and magnitutde and the magnitude of a vector can be calcuated as `color(black)(norm(vecR)=sqrt(x(t)^2+y(t)^2))`
If `color(black)(vecR)` gives the position of a moving object, then `color(black)(vecR'=vecV)` where V is velocity, and `color(black)(vecR''=vecV'=vecA)`, where A is acceleration. Speed would be the magnitude of the velocity vector and would be equivalent to `color(black)(norm(vecR')=norm(vecV)=sqrt(x'(t)^2+y'(t)^2))`
If you want to graph a parametric function, make a table with three columns, one column being t-values, another being x-values, and the final being y-values. After creating the table, plot the x and y coordinates.
Here is an example of the parametric function described in the first section: `color(black)(x=t^2)` and `color(black)(y=2t)`
| t | x | y |
|---|---|---|
| 0 | 0 | 0 |
| 1 | 1 | 2 |
| 2 | 4 | 4 |
| 3 | 9 | 6 |
| 4 | 16 | 8 |
| 5 | 25 | 10 |
After that, you should graph the x and y-values to get a graph that looks like this, which is very similar to the line graphed in the first part.
If you want to convert an equation back into rectangular form (y=x), then you need to eliminate the parameter. You do this by taking the x equation and making it so it becomes t=something. then you take that t=something and plug it into the y equation to get to rectangular form.
For example, if you have the equation `color(black)(x=t^2)` and `color(black)(y=2t)`, you take `color(black)(x=t^2)` and take the square root to get `color(black)(sqrt(x)=t)`. Afterwards, plug that into the y equation to get `color(black)(y=2sqrt(x))`. If you graph it, you will see that this equation will be equivalent to the parametric function. The orange line is the parametric function and the purple line is the rectangular form of the equation, you can see that they are equal in the graph below.
To eliminate the parameter in a vector-valued function you will undergo the same steps as you would in eliminating the parameter in a parametric function.
Here's some practice.
Question: `color(black)(x=5t+4)` and `color(black)(y=sin(t+3))`
Click on the up carrot to get the answer.
Answer: `color(black)(y=sin((x+11)/5))`
The first derivative of a parametric function is `color(black)(((dy)/(dt))/((dx)/(dt)))` because that is equivalent to `color(black)((dy)/(dt)xx(dt)/(dx)=(dy)/(dx))`
The second derivative of a parametric function is `color(black)((d/(dt)[(dy)/(dx)])/((dx)/(dt)))` because that is equivalent to `color(black)((d^2y)/(dxdt)xx(dt)/(dx)=(d^2y)/(dx)^2`
In order to find tangent lines at t=c, plug in c into the first derivative of the parametric functions, then plug in c into the x and y to get the x and y-values. Then plug in everything into the equation `color(black)(y-y_1=m(x-x_1))` where `color(black)((x_1,y_2)` is the point you are finding the tangent to and m is the slope of the line.
The first derivative of the vector-valued function `color(black)(vecR=(:R_x(t),R_y(t):))` is `color(black)(vecR'=vecV=(:R_x'(t),R_y'(t):))`
The second derivative of the vector-valued function `color(black)(vecR=(:R_x(t),R_y(t):))` is `color(black)(vecR''=vecV'=vecA=(:R_x'(t),R_y'(t):))`
Here's some practice.
Question 1: What is the `color(black)((dy)/(dx))` equal if `color(black)(x=t^2+3t+1)` and `color(black)(y=cos(t^2))`?
Question 2: What is the velocity and accleration vector if the position vector is `color(black)(vecR=(:5t^3, e^(2t):))`
Click on the up carrot to get the answer.
Answer 1: `color(black)((dy)/(dx)=(-2tsin(t^2))/(t^2+3t+1))`
Answer 2: `color(black)(vecV=(:15t^2, 2e^(2t):))` and `color(black)(vecA=(:30t, 4e^(2t):))`
The arc length for a parametric function from t=a to t=b is given by `color(black)(int_a^bsqrt(((dy)/(dt))^2 + ((dx)/(dt))^2)dt)`
The arc length, or the distance traveled by a vector-valued function is given by `color(black)(int_a^bnorm(vecR')dt)` which is equivalent to `color(black)(int_a^bsqrt((R_y'(t))^2 + (R_x'(t))^2)dt)` which equals `color(black)(int_a^bsqrt((V_y(t))^2 + (V_x(t))^2)dt)`.
Here's some practice.
If `color(black)(x=ln(t^2+5t))` and `color(black)(y=tan(2t))`, write an expression for arc length from t=0 to t=10 .
Click on the up carrot to get the answer.
Answer: `color(black)(int_0^10sqrt((2sec^2(2t))^2 + ((2t+5)/(t^2+5t))^2)dt)`