Curves #1

In this two part series I though I would discuss some famous curves with interesting shapes.

Cochleoid

Polar Form: r = a sin(θ)/θ

The name means the ‘snail-form’ curve. It was discussed by J Peck in 1700, however the form given above was discovered by Joseph Neuberg.

Fermat’s Spiral

Polar Equation: r2 = a2θ

This spiral was studied by Fermat in 1636. For any given positive value of θ there are two corresponding values of r: one negative and the other positive. Due to this, the spiral is symmetrical about the line y  = -x.

Lituus

Polar Equation: r2 = a2/θ

The lituus curve, originated with Cotes in 1722, is the locus of the point P moving in such a way that the area of a circular sector remains constant.

Watt’s Curve

Polar Equation: r2 = b2 – [a sin(θ) ± √(c2a2cos2(θ))]2

This curve was named after James Watt, a Scottish engineer who developed the steam engine, due to the fact that the curve comes from the linkages of rods connecting two wheels of equal diameter.

Newton’s Diverging Parabolas

Cartesian Equation: ay2 = x(x2 – 2bx + c), a > 0

The above equation represents the third class of Newton’s classification of cubic curves. These cases can be divided as follows:

All the roots are real and unequal. The figure is a diverging parabola with the shape of a bell, with an oval at its vertex. This is the case depicted in the image.

Two of the roots are equal. A parabola will be formed, either Nodated (by touching an oval) or Punctate (by having the oval infinitely small).

The three roots are equal. This is the semicubical parabola.

Only one real root. If two of the roots are not real, there will be a pure parabola with a bell-like form.

Stay tuned for part 2 on Friday! M x

Advertisements

One comment

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s