A particle moves in the xy-plane in such a way that its position at time t is Bold r (t )= (t- Bold sin t ) i+ (1- Bold cos t ) j. Find the maximum and minimum values of |v| and |a|.

- The maximum value of |v| is 2

- The maximum value of |a| is 1

Step-by-step 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

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

step-by-step explanation: