The curve that is parametrized by x(t) = \cos^3 t and y(t) = \sin^3 t, with 0 \leq t \leq 2\pi,...


The curve that is parametrized by {eq}x(t) = \cos^3 t\text{ and }y(t) = \sin^3 t {/eq}, with {eq}0 \leq t \leq 2\pi {/eq}, is called a hypocycloid.

a. Show that the curvature is {eq}k(t) = \frac 1{|3\sin t \cos t|} {/eq} .

b. What is the minimum curvature and where on the curve does it occur?

c. For what t is the curvature undefined? At what points on the hypocycloid does that occur? Explain what is happening on the path at those point.

Curvature of parametric functions

This example illustrates how to find the curvature of a function whose equation is specified parametrically.

The example goes on to investigate anomalous points on the graph of one such curve.

Answer and Explanation:

Part a

We will use the formula below for curvature:

{eq}\kappa = \dfrac{|x'y'' - y'x''|}{(x'^2 + y'^2)^{\frac{3}{2}}} {/eq}

and where the dashes are used to denote differentiation with respect to the parameter t.

Differentiating using the chain rule:

{eq}x(t) = \cos^3t \\ \Rightarrow x' = 3(\cos^2t)(-\sin t) = -3\cos^2t\sin t {/eq}

Differentiating again using a mixture of the product rule and chain rule:

{eq}x'' = (-3\cos^2t)(\cos t) + (-6\cos t(-\sin t))(\sin t) = -3\cos^3t + 6\sin^2t\cos t {/eq}

In a similar way:

{eq}y(t) = \sin^3 t \\ \Rightarrow y' = 3\sin^2t \cos t \\ \Rightarrow y'' = 3\sin^2t(-\sin t) + (6\sin t(\cos t))(\cos t) = -3\sin^3t + 6\sin t\cos^2t {/eq}

Then we have the curvature as:

{eq}\kappa = \dfrac{|x'y'' - y'x''|}{(x'^2 + y'^2)^{\frac{3}{2}}} \\ \Rightarrow \kappa = \dfrac{|(-3\cos^2t\sin t)(-3\sin^3t + 6\sin t\cos^2t) - (3\sin^2t \cos t)(-3\cos^3t + 6\sin^2t\cos t)|}{((-3\cos^2t\sin t)^2 + (3\sin^2t \cos t)^2)^{\frac{3}{2}}} \\ = \dfrac{|9\sin^4t\cos^2t - 18\sin^2t\cos^4t + 9\sin^2t\cos^4t - 18\sin^4t\cos^2t|}{(9\sin^2t\cos^4t + 9\sin^4t\cos^2t)^\frac{3}{2}} \\ = \dfrac{|-9\sin^4t\cos^2t - 9\sin^2t\cos^4t|}{(9\sin^2t\cos^2t(\cos^2t + \sin^2t))^\frac{3}{2}} \\ = \dfrac{9\sin^2t\cos^2t(\sin^2t + \cos^2t)}{(9\sin^2t\cos^2t(\cos^2t + \sin^2t))^\frac{3}{2}} \\ = \dfrac{9\sin^2t\cos^2t(1)}{(9\sin^2t\cos^2t(1))^\frac{3}{2}} \\ = \dfrac{9\sin^2t\cos^2t}{|27\sin^3t\cos^3t|} \\ = \dfrac{1}{|3\sin t\cos t|} {/eq}

as required.

Note the introduction of the modulus sign to ensure that the denominator maintains its positive state following the cancellation of terms.

Part b

Using the identity {eq}\sin t\cos t = \dfrac{1}{2}\sin 2t, {/eq} we have

{eq}\kappa = \dfrac{1}{|\dfrac{3}{2}\sin 2t|} = \dfrac{2}{3|\sin 2t|}. {/eq}

The minimum value of {eq}\kappa {/eq} will occur where the denominator is a maximum.

This occurs when {eq}\sin 2t = \pm1, {/eq} giving a minimum curvature {eq}\kappa = \dfrac{2}{3}. {/eq}

The corresponding value of t comes from

{eq}\sin 2t = \pm 1 \\ \Rightarrow 2t = \sin^{-1}(1) = \dfrac{1}{2}\pi \text { or } 2t = \sin^{-1}(-1) = \dfrac{3}{2}\pi \\ \Rightarrow t = \dfrac{1}{4}\pi \text { or } t = \dfrac{3}{4}\pi {/eq}

These values of t give

{eq}x = \cos^3(\dfrac{1}{4}\pi) = 0.35, \: y = \sin^3(\dfrac{1}{4}\pi) = 0.35 \\ x = \cos^3(\dfrac{3}{4}\pi) = -0.35, \: y = \sin^3(\dfrac{3}{4}\pi) = 0.35 {/eq}

i.e. the points {eq}(0.35, 0.35), \: (-0.35, 0.35). {/eq}

Part c

The curvature is undefined when the denominator of {eq}\kappa {/eq} is zero. Hence:

{eq}\sin 2t = 0 \\ \Rightarrow 2t = \sin^{-1}(0) = 0, \: \pi, \: \2\pi, \: 3\pi \\ \Rightarrow t = 0, \: \dfrac{1}{2}\pi, \: \pi, \dfrac{3}{2}\pi {/eq}

The corresponding x and y coordinates are easily found to be {eq}(1, 0), \: (0,1), (-1, 0), (0, -1). {/eq}

Below is a sketch of the curve.

It can be seen that the 4 points found above are at the "cusps" of the curve, and as such the curve is not differentiable at hese points, hence the undefined curvature.

Learn more about this topic:

Graphs of Parametric Equations

from Precalculus: High School

Chapter 24 / Lesson 5

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