# polar/cartesian curves

• November 22nd 2007, 08:44 AM
eigenvector11
polar/cartesian curves
For each of the described curves, decide if the curve would be more easily given by a polar or a Cartesian equation. Then write an equation for the curve:

1. A circle with radius 5 and center (2,3).

2. A circle centered at the origin with radius 4.

A length of curve problem:

Find the length of the curve x = 3cos(t) - cos(3t), y = 3sin(t) - sin(3t) for
0 <= t <= pi.

Thanks.
• November 22nd 2007, 03:41 PM
Soroban
Hello, eigenvector11!

Quote:

For each of the described curves, decide if the curve would
be more easily given by a polar or a Cartesian equation.
Then write an equation for the curve:

1. A circle with radius 5 and center (2,3).

Cartesian: . $(x-2)^2 + (y-3)^2 \:=\:25$

Quote:

2. A circle centered at the origin with radius 4.
Polar: . $r \:=\:4$

Quote:

A length of curve problem:
Find the length of the curve: . $\begin{Bmatrix}x &= & 3\cos(t) - \cos(3t) \\ y & = & 3\sin(t) - \sin(3t)\end{Bmatrix}\quad\text{ for }0 \leq t \leq \pi$

Formula: . $L \;=\;\int^b_a\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$

We have: . $\frac{dx}{dt} \:= \: -3\sin(t) + 3\sin(3t) \:=\:-3[\sin(t) - \sin(3t)]$

. . . . . . . . $\frac{dy}{dt} \: = \: 3\cos(t) - 3\cos(3t) \: = \: 3[\cos(t) - \cos(3t)]$

Then: . $\left(\frac{dx}{dt}\right)^2\;=\;9\left[\sin^2(t) - 2\sin(t)\sin(3t) + \sin^2(3t)\right]$

. . . . . $\left(\frac{dy}{dt}\right)^2 \;=\;9\left[\cos^2(t) - 2\cos(t)\cos(3t) + \cos^2(3t)\right]$

$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \;=$ . $9\left\{[\underbrace{\sin^2(t) + \cos^2t}_{\text{This is 1}} - 2\underbrace{[\cos(t)\cos(3t) + \sin(t)\sin(3t)]}_{\text{This is }\cos(2t)} + \underbrace{\sin^2(3t) + \cos^2(3t)}_{\text{This is 1}} ]\right\}$

. . . . . . $= \;9\left[2 - 2\cos(2t)\right] \;=\;9\left[2 - 2(1 - 2\sin^2t)\right] \;=\;9\cdot4\sin^2t$

Hence: . $\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \;=\;\sqrt{36\sin^2t} \;=\;6\sin(t)$

Therefore: . $L \;=\;6\int^{2\pi}_0 \sin(t)\,dt$

Since this definite integral comes out to zero,
. . I suspect the graph has symmetry about the origin.
I'll let you work out the details . . .
.