9.1 Intro to Parametric & Vector Calculus

9.1—Intro to Parametric & Vector Calculus

Parametric Equations and Curves In Algebra, equations are graphed in two variables, and . There are many situations in which both, and , depend independently on a third variable, or . There are a great many curves that we can’t even write down as a single equation in terms of only and . The third variable, or is called the parameter, and the separate equation are called parametric equations.

In those cases instead of defining in terms of we define both and in terms of parameter as

Each value of defines a point

The collection of points that we get by letting t be all possible values is the graph of the parametric equations and is called the parametric curve. Parametric curves have a direction of motion. The direction of motion is given by increasing t. So, when plotting parametric curves we also include arrows that show the direction of motion. Sketching a parametric curve is not always an easy thing to do. If possible, Methods 1 and 2 could be used to get explicit relationship from parametric equations. If it is not possible to get analytically explicit relationship , we use calculator (Method 3) to find point by point for each value of .

Method 1: Elimination of the parameter from the set of the equations and substitution: example: draw the graph: x = t2 + t y = 2t – 1 – 1 ≤ t ≤ 1 t = ½ (y+1) into x to give: x = ¼ y2 + y + ¾

this is a parabola that opens to the right.

t = – 1 x = 0 y = – 3

t = 1 x = 2 y = 1 direction of motion (arrows): from t = – 1 to t = 1

Method 2: Using trigonometry to eliminate a parameter example 1: draw the parametric curve for the following set of parametric equations. Clearly indicate direction of motion. x = 5 cos t y = 2 sin t 0 ≤ t ≤ 2π Instead solving one of these equations for t and plugging this into the other (very ugly process) we do this:

t x y 0 5 0 π/2 0 2 π -5 0 3π/2 0 – 2 2π 5 0 for direction of motion we have to go back to parameter : 2

example 2: draw the parametric curve for the following set of parametric equations. Clearly indicate direction of motion. x = 5 cos 3t y = 2 sin 3t 0 ≤ t ≤ 2π we replaced the t with 3t.

difference between example 1. and 2.:

As we’re going around faster, while we have the same ellipse that we got in the previous example we’ll trace out the curve three times, each in the following ranges: 0 ≤ t ≤ 2π/3, 2π/3 ≤ t ≤ 4π/3 and 4π/3 ≤ t ≤ 2π

3. method : there are some equations that you can’t do anything else but use calculator – as I have only TI – 84 + (silver edition) , I know only this one: TI – 84 plus CLEAR RAM follow these key strokes: "2nd", "+", "7","1","2" Graphing parametric equations graph the parametric equations x = 5cos3t, y = 5sin3t 1. Hit the MODE key. 2. Arrow down to where it says Func and then use the right arrow to choose Par. 3. Hit ENTER The calculator is now in parametric equations mode. 4. Hit the Y= key.

5. In the X1T slot, type 5(cos( 6. Hit X,T,q,n key: you get 5(cos(T 7. Hit two closing parentheses, )) to get: 5(cos(T)) .

3 8. Hit MATH 3 to get cubed. You should now have X1T=5(cos(T)) .

3 9. Hit the down arrow to go to Y1T and in that space type 5(sin(T)) 10. Finally, hit GRAPH. If you did everything right, the result looks a bit like a diamond, with all four edges bowed inward. You can also specify the T values that the calculator begins and ends with; for instance, you may limit the graph to 0

SPECIAL PLANE CURVES CYCLOID A famous curve that was named by Galileo in 1599. This is a curve traced by a point P on a circle of radius a rolling along x axis. Equations in parametric form: x = a ( - sin ) y = a(1 – cos )

Area of one arch = 3πa2

Arc length of one arch = 8a

PROLATE CYCLOID 3

The path traced out by a fixed point at a radius b > a, where a is the radius of a rolling circle, also sometimes called an extended cycloid. The prolate cycloid contains loops, and has parametric equations x = a  - b sin  y = a – b cos 

HYPOCYCLOID WITH FOUR CUSPS Equation in rectangular coordinates: x2/3 + y2/3 = a2/3 Equations in parametric form: x = a cos3  y = a sin3 

Area bounded by curve = 3πa2/8 Arc length of entire curve = 6a This is a curve described by a point P on a circle of radius a/4 as it rolls on the inside of a circle of radius a.

PARAMETRIC FORM OF THE DERIVATIVE If and are differentiable functions of , then the derivatives and measure the rates of change of and with respect to : and tell how fast each variable is changing with respect to . The derivative is the slope of the line tangent to the parametric graph To calculate we need to use the Chain Rule:

If is also differentiable function of , then the second derivative is given by

Don’t be surprised if you get two tangents lines at a point. A quick graph of the parametric curve will explain what is going on here. 4

The area enclosed by function y = y(x) and x-axis in Cartesian plane in parametric form:

Example 1: The location of an object is given by the parametric equations x(t) = t3 + 1 feet and y(t) = t2 + t feet at time t seconds.

(a) Evaluate x(t) and y(t) at t = –2, –1, 0, 1, and 2, and then graph the path of the object for –2 ≤ t ≤ 2. (b) Evaluate dy/dx for t = –2, –1, 0, 1, and 2. Do your calculated values for dy/dx agree with the shape of your graph in part (a)? (c) Find the equation of the line tangent to the graph of the parametric equations when t = 3. a) t x y –2 –7 2 –1 0 0 0 1 0 1 2 2 2 9 6 b) and so .

t dy/dx –2 –1/4 –1 –1/3 0 ∞ 1 1 5

2 5/12 c) x(t) = t3 + 1 y(t) = t2 + t

⟹ when t = 3 ⟹ =

tangent: y – y1 = y’(x – x1) y – 12 = (x-28) ⟹ y = x

Particle Motion

Velocity and Acceleration Component VECTORS, Speed and Direction of Motion

Suppose a particle moves along a smooth curve in the plane so that its position at any time t is , where x and y are differentiable functions of t.

is velocity of a particle in the horizontal direction.

is velocity of a particle in the vertical direction.

is the particle’s position vector at time

is the velocity vector at time

is the acceleration vector at time

is the speed of a particle or magnitude/ length/ norm of the velocity vector.

is the acceleration of a particle or magnitude/ length/ norm of acceleration. is the direction vector representing the particle’s direction of motion

Displacement & Distance Traveled

Suppose a particle moves along a path in the plane so that its velocity at any time is

vector is the displacement from t1 to t2

is the distance traveled from t1 to t2 6