Parametric Curves:

Number 6

For 

(a) Eliminate the parameter to find a Cartesian equation of the curve.

(b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases.

Solution:

(a) With sine and cosine around, squaring is a good thing to try. However, x has a 3 in it and y has a 2. It would be handy if these two numbers were the same (otherwise you'll get 9cos^2 and 4sin^2 which don't add nicely). If we multiply x by 2 and y by 3, they'll both have a 6 for a coefficient, we can then square the results, add them and simplify using the identity [sin(x)]^2 + [cos(x)]^2 = 1. That is, (2x)^2 + (3y)^2 =  = 36, so (ignoring the middle stuff) we get 4x^2 + 9y^2 = 36.

(b) Using the parametric mode on the calculator, you should see that the motion starts at (3,0) and travels couterclockwise around the oval curve below.

> plots[animatecurve]([3*cos(theta), 2*sin(theta),theta=0..2*Pi],
color=black, labels=["x","y"],
title="motion", thickness=2,
scaling=constrained, numpoints=100,frames=50);

Number 13

Describe the motion of a particle with position (x,y) as t varies in the given interval, where x = 2sin(t) and y = 3cos(t).

Solution:

As a reminder, the graphs of x and y (both as seperate functions of t) are shown first, with x red and y blue.

> plot([2*sin(t),3*cos(t)],t=0..2*Pi,
color=[red,blue], labels=["t","x and y"],
title="x and y seperate", thickness=2);

Thus, x will start at 0 and first increase to 2, then decrease to -2, and finally increase back up to 0. In terms of motion, that means the particle will begin at x = 0, move 2 units to the right, move 4 units to the left, and then 2 units to the right. Similarly, the y coordinate of the particle will start at y = 3, move down to y = -3, then climb back up to y = 3. Here is the path of this motion (a parametric plot of x and y):

> plot([2*sin(t),3*cos(t),t=0..2*Pi],
color=red, labels=["x","y"],
title="particle path", thickness=2,
scaling=constrained);

Here is an animation illustrating the motion:

> plots[animatecurve]([2*sin(t),3*cos(t),t=0..2*Pi],
color=red, labels=["x","y"],
title="motion", thickness=2,
scaling=constrained, numpoints=100,frames=50);

Number 32

One particle is at x1 = 3sin(t), y1 = 2cos(t), and another is at x2 = -3 + cos(t), y2 = 1 + sin(t), for t between 0 and 2 Pi.

(a) Graph the paths of their positions. How many points of intersection are there?

(b) Are any of the intersection points collision points? That is, are the 2 particles ever at the same location at the same time?

(c) Describe what happens if the second particle is at x2 = 3 + cos(t), y2 = 1 + sin(t), for t between 0 and 2 Pi.

Solution:

(a) The plot below shows the first particle's path in red and the second's path in blue. The paths intersect twice.

> p1:=plot([3*sin(t), 2*cos(t), t=0..2*Pi],color=red):
p2:=plot([-3+cos(t), 1+sin(t), t=0..2*Pi],color=blue):
plots[display]([p1,p2],title="2 Paths", scaling=constrained);

(b) If either of the intersection points is a collision point, then there would be a time t where the x and y coordinates of both particles would be the same. This means that x1(t) = x2(t) and y1(t) = y2(t) (and the t in the first of these equations is the same t that appears in the second). We saw in class a method to check whether this was true or not, that mixed a graphical approach with an algebraic approach. The idea was to figure out when the red path was at (or near) the intersection point at (-3, 0), and then check when the blue graph was at (-3,0). This yielded both paths at the location (-3,0) when t = 3Pi/2. The other intersection point is close to (-2.1, 1.4). The red plot is in that vicinity near t = 5.5, while the blue path is near (-2.1,1.4) for values of t close to 0.45, so the other intersection point is not a collision point. Here is another way to verify the same thing. First, we plot all 4 of the functions x1, y1, x2, and y2 for t = 0 to 2Pi, and look for a time t, where the 2 x graphs intersect AND the two y graphs also intersect. The x functions are red, the y functions are blue. The functions x1 and y1 are solid, while x2 and y2 are dashed. For a collision, we need the two red graphs to cross at the same t value where the blue graphs cross. This happens near t = 4.75 only.

> plot([3*sin(t), 2*cos(t), -3+cos(t), 1+sin(t)],
t = 0..2*Pi,color=[red,blue,red,blue],linestyle=[1,1,4,4],
title="The x and y functions",numpoints=25,adaptive=false);

The plot above suggests a possible collision point near t = 4.75. The red graphs are close to -3 there, so the x coordinate of this potential collision point is about x = -3. The blue graphs are close to zero, so the y coordinate of this potential collision point is about y = 0. Now we pin down this point. For the x coordinates to be equal, we need the two red graphs to intersect. Using your calculator, you should be able to find the 2 intersection points of the red graphs above. The first has t between 4 and 5. Here it is (just look at the blue stuff):

> t1=fsolve(3*sin(t)=-3+cos(t), t=4..5);

Just to check, here is a decimal version (in blue) of 3Pi/2

> evalf(3*Pi/2);

The second intersection of the red plots is between t = 5 and 6:

> t2=fsolve(3*sin(t)=-3+cos(t), t=5..6);

For the y coordinates to be equal, the 2 blue graphs must intersect. This also happens twice, once for t between 0 and 1, and again for t between 4 and 5. These intersections can also be found via the calculator.

> b1=fsolve(2*cos(t)=1+sin(t), t=0..1);
b2=fsolve(2*cos(t)=1+sin(t), t=4..5);

Thus, when t = 4.7123 etc. (or 3Pi/2), we have a collision!

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