Find the intervals [0,2pi] where the epicycloid, vector function r(t)=<5cost-cos5t, 5sint-sin5t> is smooth.

Answer 1

We need to check the following conditions for the vector function r(t) to be smooth:

1. The function should be continuous on the interval [0,2π].
2. The first derivative should exist and be continuous on the interval [0,2π].
3. The first derivative should not be equal to the zero vector on the interval [0,2π].

Let's first find the first derivative of r(t):

r'(t) = <-5sint + 5sin5t, 5cost - 5cos5t>

Now, let's check the first condition:

1. r(t) is made up of continuous trigonometric functions, so it is continuous on the interval [0,2π].

Next, let's check the second condition:

2. The first derivative of r(t) exists and is continuous on the interval [0,2π], since it is also made up of continuous trigonometric functions.

Finally, let's check the third condition:

3. We need to find where r'(t) is equal to the zero vector. This happens when:

-5sint + 5sin5t = 0
5cost - 5cos5t = 0

Dividing these two equations, we get:

tan(t) = tan(5t)

This equation has solutions at t = 0, π/6, π/2, 5π/6, π, 7π/6, 3π/2, 11π/6, and 2π. We need to check whether r'(t) is the zero vector at these points:

- At t = 0, r'(t) = <-5, 5>, which is not the zero vector.
- At t = π/6, r'(t) = <0, 5-5√3>, which is not the zero vector.
- At t = π/2, r'(t) = <5, 0>, which is not the zero vector.
- At t = 5π/6, r'(t) = <0, -5+5√3>, which is not the zero vector.
- At t = π, r'(t) = <5, 0>, which is not the zero vector.
- At t = 7π/6, r'(t) = <0, 5+5√3>, which is not the zero vector.
- At t = 3π/2, r'(t) = <-5, 0>, which is not the zero vector.
- At t = 11π/6, r'(t) = <0, -5-5√3>, which is not the zero vector.
- At t = 2π, r'(t) = <-5, 0>, which is not the zero vector.

Therefore, r(t) is smooth on the interval [0,2π] except at the points t = π/6 and t = 5π/6, where it has cusps.

Answer 2

Hey how did you get tan(t)=tan(5t) when you divide

Answer 3

To find where r'(t) is equal to the zero vector, we need to solve the following system of equations:

-5sint + 5sin5t = 0
5cost - 5cos5t = 0

We can simplify these by factoring out the common factor of 5:

5(-sint + sin5t) = 0
5(cost - cos5t) = 0

Dividing both sides by 5 and rearranging, we get:

sint = sin5t
cost = cos5t

Using the trigonometric identity tanθ = sinθ/cosθ, we can divide the second equation by the first to get:

tan(t) = tan(5t)