x=cos2t, y=sin2t, find rectangular equation for each curve?

Answer 1

x = cos[2t] and y = sin[2t]

x² = cos²[2t] and y² = sin²[2t]

Since we already know that sin²[2t] + cos²[2t] = 1,

Therefore,

x² + y² = 1

This is the equation of the unit circle, a circle centered at the origin with a radius of 1.

Answer 2

This is a parametric curve; it describes a curve in the plane, and how we traverse it through time. The advantage is that the curve we describe need not be the graph of a function. t = the parameter = the independent variable ; x and y = dependent variables


A circle of radius 1 centered at (0,0) : x2+y2 = 1

x(t) = cost , y(t) = sint 0 £ t £ 2p

Twice as fast around: x(t) = cos2t , y(t) = sin2t 0 £ t £ p

A circle of radius r centered at (a,b) : (x-a)2+(y-b)2 = r2

Think: x-a = rcost , y-b = rsint

x(t) = a+rcost , y(t) = b+rsint 0 £ t £ 2p


An ellipse: (x/a)2+(y/b)2 = 1

x(t) = acost , y(t) = bsint 0 £ t £ 2p

A line through (a,b) and (c,d)

x(t) = a+t(c-a) , y(t) = b+t(d-b)


Finding an (x,y) equation from a parametric equation: (if possible) solve for x = x(t) or y = y(t) as t = expression in x or y, then plug into the other equation.


Ex: x = t2-1 , y = t3+t-1 , then x+1 = t2 so t = ±[Ö(x+1)], so y = (±[Ö(x+1)])3+(±[Ö(x+1)])-1


Calculus of curves


Thinking of a parametric curve as a path that we are traversing, we are at each instant aware of (at least) two things: how fast we are going and what direction we are going. Each can be computed essentially as we would for a graph.


Speed = the limit of (distance)/(time interval) as the time interval shrinks to 0.


average speed = Ö{(Dx)2+(Dy)2}/Dt = Ö{(Dx/Dt)2+(Dy/Dt)2}


instantaneous speed = [Ö((dx/dt)2+(dy/dt)2)] = Ö{(x¢(t))2 + (y¢(t))2}


direction = slope of tangent line - limit of slopes of secant lines


secant lines: slope = Dy/Dx = (Dy/Dt)/(Dx/Dt)

tangent lines: slope = (dy/dt)/(dx/dt) = y¢(t)/x¢(t)


We can encode both of these in the velocity vector (x¢(t),y¢(t))


A parametric curve x = x(t) , y = y(t) , a £ t £ b with x(a) = x(b) , y(a) = y(b) ends where it begins; it is a closed curve. Such a curve surrounds and encloses a region R in the plane.


If the curve goes around the region counterclockwise, then the area of the region can be computed as



Area = òab x(t)y¢(t) dt = -òab y(t)x¢(t) dt



We will see why this formula is true later in this class....


Arclength and surface area


Just as with graphs of functions, we can compute the length of a paramentric curve and the surface area when a curve is rotated around an axis:


Length: we approximate it the same way, as a sum of lengths of line seqments that approximate the curve. Each segment has length



Ö{(Dx)2+(Dy/)2} = Ö{(Dx/Dt)2+(Dy/Dt)2}Dt » Ö{(x¢(t))2 + (y¢(t))2} dt



so the length of the curve is òab Ö{(x¢(t))2 + (y¢(t))2} dt


Surface area: if we spin the curve x = x(t) , y = y(t) , a £ t £ b around the line y = c then, just like before, we can approximate the surface by frustra of cones, each having area approximately


2p|y(t)-c|Ö{(x¢(t))2 + (y¢(t))2} dt = (2p)(radius)(length)


and so the area of the surface of revolution is



2pòab |y(t)-c|Ö{(x¢(t))2 + (y¢(t))2} dt



Ex: for the ellipse x = 3cost , y = 5sint , 0 £ t £ 2p, spun around y = 7, we have


Area = 2pò02p (7-3sint)Ö{9sin2 t + 25cos2 t} dt = 2pò02p (7-3sint)Ö{9 + 16cos2 t} dt


Polar coordinates


Idea: describe points in the plane in terms of (distance,direction).

r = (x2+y2)1/2 = distance , q = arctan(y/x) = angle with the positive x-axis.

x = rcosq , y = rsinq

The same point in the plane can have many representations in polar coordinates:

(1,0)rect = (1,0)pol = (1,2p)pol = (1,16p)pol = ¼

A negative distance is interpreted as a positive distance in the opposite direction (add p to the angle):

(-2,p/2)pol = (2,p/2+p)pol = (0,-2)rect

An equation in polar coordinates can (in principal) be converted to rectangular coords, and vice versa:

E.g., r = sin(2q) = 2sinqcosq can be expressed as

r3 = (x2+y2)3/2 = 2(rsinq)(rcosq) = 2yx, i.e., (x2+y2)3 = 4x2y2



Given an equation in polar coordinates

r = f(q) , i.e., the curve (f(q),q)pol , q1 £ q £ q2

we can compute the slope of its tangent line, by thinking in rectangular coords:

x = f(q)cosq, y = f(q)sinq , so

[dy/dx] = [(dy/dq)/(dx/dq)] = [(f¢(q)sinq+ f(q)cosq)/(f¢(q)cosq- f(q)sinq)]


Arclength: the polar curve r = f(q) is really the (rectangular) parametrized curve

x = f(q)cosq, y = f(q)sinq, and (x¢(q))2+(y¢(q))2)1/2 = (f¢(q))2+(f(q))2)1/2,

so the arclength for a £ q £ b is displaystyle òab (f¢(q))2+(f(q))2)1/2 dq


Area: if r = f(q) , a £ q £ b describes a closed curve (f(a) = f(b) = 0), then we can compute the area inside the curve as a sum of areas of sectors of a circle, each with area approximately

pr2 (Dq/2p = [((f(q))2)/2]Dq

so the area can be computed by the integral òab [1/2](f(q))2 dq

Answer 3

x^2 + y^2 = cos^2 2t + sin^2 2t = 1, so this curve is the unit circle.

Answer 4

cos^2(2t)+sin^2(2t)=1
so x^2+y^2=1