Question

Match each of the parametric equations 1-5 with the curve it represents from the list A through E below. 1. r=3 2. r=2sinθ 3. θ=π4 4. r=4θ 5. r=2cosθ A. A circle of radius 1 with center on the y-axis. B. None of these C. A circle of radius 3 centered at the origin. D. The straight line y=x. E. A circle of radius 1 with center on the x-axis.

Answer 1

The parametric equations are matched with their corresponding curves A through E.

Step-by-step explanation:

The parametric equation r=3 represents a circle of radius 3 centered at the origin, so it matches option C.

The parametric equation r=2sinθ represents a circle of radius 2 with center on the y-axis, so it matches option E.

The parametric equation θ=π/4 represents a single point on the circle, so it matches option B.

The parametric equation r=4θ represents a spiral approaching the origin, so it does not match any of the given options.

The parametric equation r=2cosθ represents a circle of radius 2 with center on the x-axis, so it matches option A.

Answer 2

Step-by-step explanation:

1. r=3

To convert from polar coordinates (r,θ) to rectangular coordinates (x,y), we use the following equations

x=rCosθ

Cosθ=x/r

y=rSinθ

Sinθ=y/r

Also, r²=x²+y²

When r =3,

Then we have

r=3,

x²+y²=3

All points with r = 3 are at

distance 3 from the origin, so r = 3 describes the circle of radius 3, with center at the origin (0,0).

Option C

2. r=2sinθ

When r=2sinθ

To convert from polar coordinates (r,θ) to rectangular coordinates (x,y), we use the following equations

x=rCosθ

Cosθ=x/r

y=rSinθ

Sinθ=y/r

Also, r²=x²+y²

Now, apply the given information

r=2sinθ

Since Sinθ=y/r

r=2y/r

Cross multiply

r²=2y

x²+y²=2y

x²+y²-2y =0

x²+ (y-1)² -1 =0

x²+(y-1)²=1²

Then,

It is a circle with center (0,1) and radius 1.

Because the sine is periodic, we know that we will get the entire curve for values of θ in [0, 2π). As θ runs from 0 to π/2, r increases

from 0 to 2. Then as θ continues to π, r decreases again to 0. When θ runs from π to

2π, r is negative, and it is not hard to see that the first part of the curve is simply traced

out again, so in fact we get the whole curve for values of θ in [0, π). Thus, check attachment for curve, Now, this suggests that the curve could possibly be a circle,

and if it is, it would have to be the circle x² + (y − 1)² = 1. Having made this guess, we can easily check it. First we substitute for x and y to get (r cos θ)² + (r sin θ − 1)² = 1;

expanding and simplifying does indeed turn this into r = 2 sin θ.

Option E

A circle of radius 1 with center on the x-axis

3. θ=π/4

This is a point on the circle

p=(r, θ)

So, r=0

x = rCosπ/4=rCos45

x=r√2 /2

and

y = rsin(π/4) = rSin45

y=r√2 /2

This makes it very

easy to convert equations from rectangular to polar coordinates.

x²+y²= (r√2/2)²+(r√2/2)²

x²+y²= 2r²/4 + 2r²/4

x²+y²= r²

Since r=0

x²+y²= 0

x²=y²

Therefore, x=y

Option D

The straight line y=x

4. r=4θ

Here the distance from the origin

exactly matches the angle, so a bit of thought makes it clear that when θ ≥ 0 we get the spiral of Archimedes ( check attachment ) When θ < 0, r is also negative, and so the full graph is the right hand picture in the figure. Check attachment

The correct option is B

None of the above

5. r=2Cosθ

To convert from polar coordinates (r,θ) to rectangular coordinates (x,y), we use the following equations

x=rCosθ

Cosθ=x/r

y=rSinθ

Sinθ=y/r

Also, r²=x²+y²

Now, apply the given information

r=2cosθ

Since cosθ=x/r

r=2x/r

Cross multiply

r²=2x

x²+y²=2x

x²+y²-2x=0

(x-1)² + y² -1 =0

(x-1)² + y² = 1²

Then,

It is a circle with center (1,0) and radius 1.

A circle of radius 1 with center on the y-axis

Option A is correct