Hello, Nerd!

The particle's position vector is: .A particle moves along a path given by:

. . x(t) .= .3·cos(πt)

. . y(t) .= .5·sin(πt)

where 0 ≤ t ≤ 6.

1) Velocity vector fot the particle at any time t?

I just took the derivative of x(t) and y(t). Is that correct?

What did you do with thetwoexpressions?s.= .[3·cos(πt)]i+ [5·sin(πt)]j

Its velocity vector is: .v.= .[-3π·sin(πt)]i+ [5π·cos(πt)]j

Do they mean "integral" or "integer"? .Either way, it doesn't make sense.2) Write an integral expression that gives the distance

between t = 1.25 and t = 1.75

. . . . . . . . . . . . . . . . . . . . . . . . . . . ._ . . . . ._

When t = 5π/4, the particle is at: P(-3√2/2, -5√2/2)

. . . . . . - . . - . . . . . . . . . . . . . . . . . _ . . . . ._

When t = 7π/4, the particle is at: Q(3√2/2, -5√2/2)

. . . - . . - . . . . . . . . . _

The distance PQ is: .3√2

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I think I get it . . . They want theArc Length.

. . . - . . - . . _______________

. . L . = . ∫ √(dx/dt)² + (dy/dt)² dt

. . . . . . . . . . . . . . . .________________________

So we have: .L .= .∫ √9π²·sin²(πt) + 25π²·cos²(πt) dt

. . evaluated from t = 5/4 to t = 7/4.