Find the intervals within [0,2pi] where the epicycloid, vector function r(t)=<5cos(t)-cos(5t), 5sin(t)-sin(5t)> is smooth.
We can determine where the vector function r(t) is smooth by finding where its derivative is continuous and nonzero.
First, we find the derivative:
r'(t) = <-5sin(t) + 5cos(5t), 5cos(t) - 5cos(5t)>
We can simplify this:
r'(t) = 5<-sin(t) + cos(5t), cos(t) - cos(5t)>
Next, we find where the x- and y-components of r'(t) are zero:
-sin(t) + cos(5t) = 0
cos(t) - cos(5t) = 0
We can use trigonometric identities to simplify these equations:
sin(t) = sin(5t)
2sin((t+5t)/2)cos((t-5t)/2) = 0
cos((t+5t)/2)sin((t-5t)/2) = 0
cos(3t/2)sin(-2t/2) = 0
cos(3t/2)sin(t) = 0
So the solutions are when sin(t) = 0 (i.e. t = 0, pi, 2pi) or cos(3t/2) = 0 (i.e. t = pi/3, 2pi/3, 4pi/3, 5pi/3).
Now, we need to check the values of r'(t) at these points:
t = 0: r'(0) = 5<1, 0>
t = pi/3: r'(pi/3) = 5<0, sqrt(3)>
t = 2pi/3: r'(2pi/3) = 5<0, -sqrt(3)>
t = pi: r'(pi) = 5<-1, 0>
t = 4pi/3: r'(4pi/3) = 5<0, -sqrt(3)>
t = 5pi/3: r'(5pi/3) = 5<0, sqrt(3)>
t = 2pi: r'(2pi) = 5<1, 0>
We can see that r'(t) is continuous and nonzero everywhere except at t = pi/3, 2pi/3, 4pi/3, 5pi/3, where the limit of r'(t) as t approaches these values is zero. Therefore, the intervals within [0,2pi] where r(t) is smooth are:
[0, pi/3), (pi/3, 2pi/3), (2pi/3, 4pi/3), (4pi/3, 5pi/3), (5pi/3, 2pi].
Isn't the first derivative be r'(t) = <-5sint + 5sin5t, 5cost-5cos5t>?
You are correct, I apologize for the error in my previous answer.
The correct first derivative is:
r'(t) = <-5sin(t) + 5sin(5t), 5cos(t) - 5cos(5t)>
The rest of the solution follows the same process. We find where the x- and y-components of r'(t) are zero:
-5sin(t) + 5sin(5t) = 0
5cos(t) - 5cos(5t) = 0
Simplifying using trigonometric identities:
sin(t) = sin(5t)
cos(t) = cos(5t)
So the solutions are when t = 0, pi/3, 2pi/3, pi, 4pi/3, 5pi/3, 2pi.
Checking the values of r'(t) at these points:
t = 0: r'(0) = 5<0, 5>
t = pi/3: r'(pi/3) = 5<15/2, sqrt(75)/2>
t = 2pi/3: r'(2pi/3) = 5<-15/2, sqrt(75)/2>
t = pi: r'(pi) = 5<0, -5>
t = 4pi/3: r'(4pi/3) = 5<-15/2, -sqrt(75)/2>
t = 5pi/3: r'(5pi/3) = 5<15/2, -sqrt(75)/2>
t = 2pi: r'(2pi) = 5<0, 5>
We can see that r'(t) is continuous and nonzero everywhere except at t = 0, pi, and 2pi, where the limit of r'(t) as t approaches these values is zero. Therefore, the intervals within [0,2pi] where r(t) is smooth are:
(0, pi/3), (pi/3, 2pi/3), (2pi/3, pi), (pi, 4pi/3), (4pi/3, 5pi/3), (5pi/3, 2pi).
Can you help with with the part with sin(t)= sin5t and cost=cos5t? I don't get that part
Certainly!
We want to solve the equations -5sin(t) + 5sin(5t) = 0 and 5cos(t) - 5cos(5t) = 0, which can be rewritten as:
5sin(5t) - 5sin(t) = 0 and 5cos(5t) - 5cos(t) = 0
We can use the sum-to-product trigonometric identities to rewrite the left side of each equation:
5 [ sin(5t) - sin(t) ] = 0
5 [ cos(5t) - cos(t) ] = 0
Now we have expressions in terms of the difference of two trigonometric functions. We know that if the difference of two trigonometric functions is 0, then either their arguments are equal (i.e. one of the trigonometric functions repeats) or their arguments differ by a multiple of 2pi.
So, for the first equation,
5 [ sin(5t) - sin(t) ] = 0
sin(5t) - sin(t) = 0
sin(5t) = sin(t)
This means that either t = 5t + 2pi*n (where n is an integer), or t = pi - 5t + 2pi*m (where m is an integer).
Similarly, for the second equation,
5 [ cos(5t) - cos(t) ] = 0
cos(5t) - cos(t) = 0
cos(5t) = cos(t)
This means that either t = 2pi - 5t + 2pi*k (where k is an integer), or t = 2pi + 5t + 2pi*j (where j is an integer).
If we combine all of these solutions and simplify, we get:
t = 0, pi/3, 2pi/3, pi, 4pi/3, 5pi/3, 2pi
We can check each of these values of t to see when the derivative r'(t) is continuous and nonzero, and this gives us the intervals where the epicycloid is smooth.
I got (0,pi/2), (pi/2,pi), (pi, 3pi/2), (3pi/2, 2pi) for the intervals where this epicycloid is smooth. Is this correct?
No, those intervals are not correct.
The correct intervals where the epicycloid is smooth are:
(0, pi/3), (pi/3, 2pi/3), (2pi/3, pi), (pi, 4pi/3), (4pi/3, 5pi/3), (5pi/3, 2pi).
Please refer to my corrected solution above for the detailed steps on how to find these intervals.