Could you simplify calculation? position vector is r=(a cos(wt))i + (a sin(wt))j

[Indranil]

Could you simplify the calculation below?
1. Where it comes from '(-aw sin wt)i + (aw cos wt)j because there is (a cos wt)i + (a sin wt)j?
2. why the slop of velocity vector = -a cos wt/sin wt ( why - a here it should be cos wt/sin wt)

 

[Dr.Peterson]

Could you simplify the calculation below?
1. Where it comes from '(-aw sin wt)i + (aw cos wt)j because there is (a cos wt)i + (a sin wt)j?
2. why the slop of velocity vector = -a cos wt/sin wt ( why - a here it should be cos wt/sin wt)

1. They differentiated the position vector with respect to t.

2. You've copied wrong. There is a in both numerator and denominator, which cancel out, resulting in -cos wt/sin wt) = -1/tan wt, as they say.

 

[Indranil]

1. They differentiated the position vector with respect to t.

2. You've copied wrong. There is a in both numerator and denominator, which cancel out, resulting in -cos wt/sin wt) = -1/tan wt, as they say.

1. could you please show me how to differentiate '(a cos wt)i '?
2. My question is why '-cos wt/sin wt', it should be 'cos wt/sin wt = 1/ tan wt'?

 

[Dr.Peterson]

1. could you please show me how to differentiate '(a cos wt)i '?
2. My question is why '-cos wt/sin wt', it should be 'cos wt/sin wt = 1/ tan wt'?

1. Because a, w, and the vector i are constants, d/dt[(a cos wt)i] = a d/dt[cos wt]i = a [w * -sin wt]i = -(aw sin wt) i.
The key is the chain rule; we are taking u = wt, so d/dt(cos u) = d/du(cos u) * du/dt = -sin u * w = -sin wt * w.

2. The negative sign is present in the derivative. I don't understand why you think it shouldn't be there. They did "move" it; the ratio of the j-component to the i-component of v is, literally, [a cos wt]/[-asin wt], but the a's cancel out (because a/a = 1) and the - can be moved to the top or the outside (because 1/-1 = -1/1 = -1). It doesn't go away; it just moves. To put it another way, a/-a = -1.