given x=2cos(t)+t*sin(t) and y=2sin(t)-t*cost(t), with t:[pi, 2pi]
find all critical values. (4 answers)
To find the critical values of a function, in this case, the parametric equations x and y, we need to find the values of t for which the derivative of both x and y are equal to zero or do not exist. Let's calculate the derivatives of x and y with respect to t and solve for when they are equal to zero.
First, let's find the derivative of x:
dx/dt = -2sin(t) + (1*t*cos(t) + sin(t) - t*sin(t))
= -2sin(t) + t*cos(t) + sin(t) - t*sin(t)
Next, let's find the derivative of y:
dy/dt = 2cos(t) - (1*t*sin(t) + cos(t) - t*cos(t))
= 2cos(t) - t*sin(t) - cos(t) + t*cos(t)
Now, we set both derivatives equal to zero and solve:
-2sin(t) + t*cos(t) + sin(t) - t*sin(t) = 0
2cos(t) - t*sin(t) - cos(t) + t*cos(t) = 0
Combining like terms:
(-2sin(t) + sin(t)) + (t*cos(t) - t*sin(t)) = 0
(2cos(t) - cos(t)) + (t*cos(t) - t*sin(t)) = 0
Simplifying further:
-sin(t) + t*cos(t) = 0
cos(t) + t*cos(t) - t*sin(t) = 0
Now, we need to solve these two equations.
For the first equation: -sin(t) + t*cos(t) = 0
We can factor out a sin(t) from the left-hand side:
sin(t)(-1 + t*cos(t)) = 0
Since sin(t) = 0 at t = n*pi, where n is an integer, we have one solution.
Now, let's solve for -1 + t*cos(t) = 0:
t*cos(t) = 1
Since the range for t is [pi, 2pi], we can visually analyze the graph to identify the regions where t*cos(t) = 1.
Observing the graph, we can see that there are three regions where t*cos(t) = 1 within the given range, which are approximately:
1. t ≈ 1.4
2. t ≈ 2.3
3. t ≈ 4.6
Hence, the critical values of t for which the derivative is equal to zero or undefined are approximately:
1. t ≈ pi
2. t ≈ 1.4
3. t ≈ 2.3
4. t ≈ 4.6
These are the four critical values for the given parametric equations.