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Given x(t)=3sin(t) - 3, y(t)=t-1 for 0 is less than or equal to t is less than or equal to 2pi How do you find the velocity of the particle at t=3?

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mason m Share
Jun 23, 2018

Answer:

The particle's velocity is #sqrt(9cos^2(3)+1)approx3.1338#

Explanation:

We can find the #x#- and #y#-velocities by taking the derivatives of each position function.

#{(x'(t)=3cos(t)),(y'(t)=1):}#

So at #t=3#, we have

#{(x'(3)=3cos(3)),(y'(3)=1):}#

The magnitude of the velocity is then the hypotenuse of the right triangle whose legs are the #x# and #y# components of the velocity:

#v(3)=sqrt((x'(3))^2+(y'(3))^2)=sqrt(9cos^2(3)+1)approx3.1338#

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