Question

Graph the epitrochoid with the following equations. Use
Mathematica to find the approximate length of this curve. You will
have the figure out the interval that will generate exactly one
cycle of this curve. Do this in mathmatica

x = 11cost t - 4 cos((11t)/2) y =
11sin t - 4sin((11t)/2)

Answer #1

WE have

x = 11*cos(t) - 4*cos((11*t)/2)

y = 11*sin(t) - 4*sin((11*t)/2)

to find the time period of the function

we can see that sin(t) and cos(t) have a period of 2*pi

while sin(11t/2) and cos(11t/2) have a period of 2pi*(2/11)= 4pi/11

since 4pi/11<2pi hence, time period of function is 4pi/11.

from t=0 to t=4pi/11, one cycle of function will be completed and correspondingly we'll use it to determine length of the curve

1. The figure below shows the curve plotted for 1 cycle and 2 cycle respectively

To determine length of curve, we run a code which integrates over 1 cycle

integrate from t=0 to t=4pi/11, sqrt(1+(dy/dx)^2) *(dx/dt) (script written in mathematica (wolframalpha) to integrate)

Finally length of curve we get answer as 7.47159544 by numerical integration.

Mathematica Question:
Using Mathematica's Plot and FindRoot functions, approximate the
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