MAT37x

Franklin College

Spring2004

Erich Prisner,

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x= cos^{3}(t), y= sin^{3}(t)

x^{2/3}+y^{2/3}=1

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The slope of the trajectory is ...

The length of the trajectory is ...

The area enclosed by the trajectory ...

The astroid has the property that for every tangent the sum of both
intercept values is constant. To derive this, we concentrate on the first quadrant. Here we can solve the equation for y to get
y = (1-x^{2/3})^{3/2}.
We differentiate and get the slope
dy/dx = 3/2(1-x^{2/3})^{1/2}*(-2/3)x^{-1/3}
= - (1-x^{2/3})^{1/2}x^{-1/3}.
We plug this into the point slope formula to get the equation of the tangent at the point
(a,(1-a^{2/3})^{3/2}, namely

- (1-a^{2/3})^{1/2}a^{-1/3} =
(y-(1-a^{2/3})^{3/2})/(x-a)

To find the x-intercept, we plug in y=0
and get

a^{-1/3} =
(1-a^{2/3})/(x-a)

or
x = (1-a^{2/3})a^{1/3}+a = a^{1/3}-a+a = a^{1/3}.

To find the y-intercept, we plug in x=0 and obtain

(1-a^{2/3})^{1/2}a^{2/3} =
y-(1-a^{2/3})^{3/2} or

y = (1-a^{2/3})^{1/2} [a^{2/3}+(1-a^{2/3})] =
(1-a^{2/3})^{1/2}.

The distance between these x-and y-intercepts is
(a^{2/3}+(1-a^{2/3}))^{1/2} = 1

(Another way uses implicit differentiation to get
(2/3)x^{-1/3}+(2/3)y^{-1/3}dy/dx = 0, or
dy/dx = -(y/x)^{1/3}.
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