Question

Derivation of the equation of a Hypocycloid?

https://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/assignments/ps3.pdf (question 6)

b) The equation involving x2 and y2 that expresses the property that ladder L2 has length one x2^2+y2^2=1 y2=(1-x2^2)^0.5 The equation of the line made by the ladder with its foot at (x2,0) is y=-y2/x2(x-x2)=-(1-x2^2)^0.5/x2(x-x2) After computing the derivative with respect to x2, how can I use the information about the derivative at x2 to know more about the derivative at x1?

dy/dx2=x-x2^3/x2^2(1-x2^2)^0.5

Answer 1

See below.

Explanation:

Calling

`p_1=(0,y_1)`
`q_1=(x_1,0)`

and

`p_2=(0,y_2)`
`q_2=(x_2,0)`

and defining

`s_1 = p_1+lambda_1(q_1-p_1)`
`s_2 = p_2+lambda_2(q_2-p_2)`

we have `s_1 nn s_2` at

`s_1=p_1+lambda_1(q_1-p_1)=p_2+lambda_2(q_2-p_2)=s_2`

for

`{(lambda_1 =(x_2 (y_1 - y_2))/(x_2 y_1 - x_1 y_2) ),(lambda_2 = (x_1 ( y_2-y_1))/( x_1 y_2-x_2 y_1)):}`

and the intersection point is

`P = ((x_1 x_2 (y_1 - y_2))/(x_2 y_1 - x_1 y_2), ((x_1 - x_2) y_1 y_2)/( x_1 y_2-x_2 y_1))`

now considering that

`{(x_1^2+y_1^2=L^2->y_1=sqrt(L^2-x_1^2)),(x_2^2+y_2^2=L^2->y_2=sqrt(L^2-x_2^2)):}`

now substituting into `P(x_1,y_1,x_2,y_2)` we have

#P = {(p_x=(x_1 x_2 (L^2 + x_1 x_2 - sqrt(L^2 - x_1^2) sqrt(L^2 - x_2^2) ))/(L^2 (x1 + x2))),(p_y=(sqrt(L^2 - x_1^2) (x_2-x_1) sqrt(L^2 - x_2^2))/(x_2sqrt(L^2 - x_1^2) + x_1 sqrt(L^2 - x_2^2))):}#

now making `lim_({(x_1->x),(x_2->x):})P(x_1,x_2)` we get

`P(x) = 1/L^2(x^3, (L^2-x^2)^(3/2))` which is the intersection point equation

Attached the plot of `P` for `0 le x le L`

NOTE: This last result can be understood as a parametric description for `P` as `P(lambda) = (x,y)` and then

`{(bar x=xL^2=lambda^3),(bar y=yL^2= (L^2-lambda^2)^(3/2)):}` and solving we arrive at

`bar y = (L^2-(bar x)^(2/3))^(3/2)` and finally

`x^(2/3)+y^(2/3)=L^(2/3)` as the cartesian hypocycloid equation.