Worked Sample Problems - Math 114

CHAPTER 9 - Parametrized Curves and Polar Coordinates

Section 9.4, page 741

Problem 6

We eliminate the parameter as follows:

>	solve({x=4*sin(t),y=2*cos(t)},{x,t});

Or, of course , which is an ellipse. We'll plot the whole thing, and also the part that is covered by the parametrization as to goes from 0 to .

>	with(plots,display):

>	A:=plot([4*sin(t),2*cos(t),t=-Pi..Pi],color=red,thickness=1):

>	B:=plot([4*sin(t),2*cos(t),t=0..Pi],color=blue,thickness=3):

>	display({A,B},scaling=constrained);

The right half of the curve is traversed from top to bottom.

Problem 36

The point P rotates through the same angle as the wheel. So when the wheel has rotated clockwise through the angle , the center of the wheel has moved  to the right. If the point P starts out at [0,a+b], then after the wheel has turned  radians, it will be at [, ]. Let's plot this curve, with a=2 and b=1, a=2, b=2 (cycloid) and a=2, b=3 (spoke is longer than the radius of the wheel):

>	plot({seq([2*t+k*sin(t),2+k*cos(t),t=0..4*Pi],k=1..3)},color=blue,thickness=2,scaling=constrained);

Section 9.5, page 749

Problem 9

>	x:=2*t^2+3; y:=t^4;

Tangent line:

>	tanline:=subs(t=-1,y)+subs(t=-1,diff(y,t)/diff(x,t))*(xx-subs(t=-1,x));

(we used xx as the variable on the line because x was already taken!)

>	second_deriv:=diff(diff(y,t)/diff(x,t),t)/diff(x,t);

Problem 18

>	x:=t^3; y:=3*t^2/2;

>	arclength:=int(sqrt(diff(x,t)^2+diff(y,t)^2),t=0..sqrt(3));

Section 9.6, page 755

Problem 60

>

restart;

>	simplify(subs(x=r*cos(theta),y=r*sin(theta),(x-5)^2+y^2=25));

>	solve(%,r);

So the equation becomes .

Section 9.8, page 768

Problem 37

>	r:=1/(1+cos(theta));

>	plot([r*cos(theta),r*sin(theta),theta=0..2*Pi],view=[-4..4,-4..4],color=blue,thickness=2);

Clearly this is a parabola, vertex is when , so directrix is the line when x=1.

Section 9.9, page 775

Problem 15

First plot, then solve for intersections and go from there:

>	r1:=6; r2:=3*csc(theta);

>	plot({[r1*cos(theta),r1*sin(theta),theta=-Pi..Pi],[r2*cos(theta),r2*sin(theta),theta=-Pi/2..Pi/2]},color=blue,thickness=2,scaling=constrained,view=[-7..7,-7..7]);

We want above the line and below the circle -- where do they intersect?

>	solve(r1=r2);

Looks like  and  -- so we integrate:

>	area:=int(r1^2/2-r2^2/2,theta=Pi/6..5*Pi/6);