Answer:
- The maximum value of |v| is 2
- The maximum value of |a| is 1
Explanation:
Given r(t) = (t + sin t)i + (1 - cos t) j
We are required to find the maximum and minimum values of the magnitude of velocity, |v|, and magnitude of acceleration, |a|.
To do these, we need to first obtain the velocity and acceptation from the given function.
- To obtain velocity, differentiate r(t) with respect to t.
Doing that, we have
v(t) = (1 - cos t)i - sin t j
Note: If A = xi + yj + zk, Magnitude of A, denoted as |A| = √(x² + y² + z²).
Knowing this,
|v| = √[(1 - cos t)² + sin²t]
= √[(1 - 2cos t + cos²t) + sin²t]
= √(2 - 2cost) (because a trig. identity, sin²t + cos²t = 1)
Now, the maximum value of |v| will be the maximum value of √(2 - 2cost), which is when cos t = -1, because 0 and 1 assume smaller values.
So, for cos t = -1
√(2 - 2cost) = √(2 - 2(-1))
= √4
max |v|= 2
- To obtain acceleration, a, we need to differentiate r twice. This is the same as differentiating v once.
a = dv/dt = sin t i - cos t j
|a| = √[(sin t)² + (-cos t)²]
= √(sin²t + cos²t)
= √1
max |a|= 1