Vector Calculus Question

[generalkorn12]

Just a question that's been baffling me,

'Position vector is given by, r(t)= (t-3)i-(t)j, the particle is closest to origin when t is....'

I figured that it would be closest to the origin at t=0 or a really minute value, such as, t=0.000000001.
Anything higher than that, would be pretty far from origin. 

[Hancock]

Just a question that's been baffling me,

'Position vector is given by, r(t)= (t-3)i-(t)j, the particle is closest to origin when t is....'

I figured that it would be closest to the origin at t=0 or a really minute value, such as, t=0.000000001.
Anything higher than that, would be pretty far from origin. 

You need to find when it's distance is at a minimum from the origin.

So it's distance is the magnitude or r(t) or ((t-3)^2+(-t)^2)^0.5 = ((t-3)^2+t^2)^0.5 = D = (2t^2 - 6t + 9)^0.5
To find the minimum value of D, you differentiate and make that equal to zero.

D = (2t^2 - 6t + 9)^0.5
dD/dt = (4t - 6) / 2(2t^2 - 6t + 9)^0.5 = 0

Therefore, dD/dt = 0 when 4t - 6 = 0 so t = 3/2

r(3/2) = (-3/2)i - (3/2)j
d(3/2) = ((3/2)^2 + (3/2)^2)^0.5 = 3 / (2^0.5) = 3(2^0.5)/2

[Hancock]

If you think about it quanlitatively, the particle starts at -3i and moves in the positive i direction 1 distance per unit time, while it does the same in the negative j direction (moving at an angle of -45 degrees with respect to the positive i direction). When it is at -3/2 i - 3/2j, it is the closest it will ever be to the origin before moving away further South-East