What is the instantaneous velocity of an object moving in accordance to f(t)= (t^2,tcos(t-(5pi)/4)) at t=(pi)/3 ?

1 Answer
Jun 2, 2016

v(pi/3)=1/3sqrt(4pi^2+9cos^2(pi/12)+pisin^2(pi/12)+6picos(pi/12)sin(pi/12))

Explanation:

The equation f(t)=(t^2;tcos(t-(5pi)/4)) gives you the object's coordinates with respect to time:

x(t)=t^2
y(t)=tcos(t-(5pi)/4)

To find v(t) you need to find v_x(t) and v_y(t)

v_x(t)=(dx(t))/dt=(dt^2)/dt=2t

v_y(t)=(d(tcos(t-(5pi)/4)))/dt=cos(t-(5pi)/4)-tsin(t-(5pi)/4)

Now you need to replace t with pi/3

v_x(pi/3)=(2pi)/3

v_y(pi/3)=cos(pi/3-(5pi)/4)-pi/3 cdot sin(pi/3-(5pi)/4)
=cos((4pi-15pi)/12)-pi/3 cdot sin((4pi-15pi)/12)
=cos((-11pi)/12)-pi/3 cdot sin((-11pi)/12)
=cos(pi/12)+pi/3 cdot sin(pi/12)

Knowing that v^2=v_x^2+v_y^2 you find:

v(pi/3)=sqrt(((2pi)/3)^2+(cos(pi/12)+pi/3 cdot sin(pi/12))^2)
=sqrt((4pi^2)/9+cos^2(pi/12)+pi^2/9 cdot sin^2(pi/12)+(2pi)/3 cdot cos(pi/12)sin(pi/12))
=1/3sqrt(4pi^2+9cos^2(pi/12)+pisin^2(pi/12)+6picos(pi/12)sin(pi/12))