Find the intervals [0,2pi] where the epicycloid, vector function r(t)=<5cost-cos5t, 5sint-sin5t> is smooth.
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.
Hey how did you get tan(t)=tan(5t) when you divide
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)
To determine where the given vector function is smooth on the interval [0, 2π], we need to find where its derivatives exist and are continuous.
1. First, let's find the first derivative of r(t). The first derivative will help us check the continuity of r(t) and identify any points where it might not be smooth.
dr(t)/dt = <d/dt (5cos(t) - cos(5t)), d/dt (5sin(t) - sin(5t))>
= <-5sin(t) + 5sin(5t), 5cos(t) - 5cos(5t)>
= <-5sin(t) + 5sin(5t), 5cos(t) - 5cos(5t)>
2. Now, let's find the second derivative of r(t). The second derivative will help us check the continuity of the first derivative and identify any points where it might not be smooth.
d^2r(t)/dt^2 = <d/dt (-5sin(t) + 5sin(5t)), d/dt (5cos(t) - 5cos(5t))>
= <-5cos(t) + 25cos(5t), -5sin(t) + 25sin(5t)>
= <-5cos(t) + 25cos(5t), -5sin(t) + 25sin(5t)>
3. To determine where these derivatives exist and are continuous, we need to find the points where sin(t) and cos(t) have discontinuities or where the denominators become zero. In this case, there are no denominators involved.
4. Since sin(t) and cos(t) are continuous functions for all t, the derivatives dr(t)/dt and d^2r(t)/dt^2 are also continuous for all t.
Therefore, the vector function r(t) = <5cos(t) - cos(5t), 5sin(t) - sin(5t)> is smooth for all values of t in the interval [0, 2π].