I have the answer ready in paint but i am not able to upload. 1) F.pdf

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Ankitagarwaleleraipu

I have the answer ready in paint but i am not able to upload. 1) F(0)=(1,0), F(pi/4)=(0,1), F(pi/2) =(-1,0), F(3pi/4)=(0,-1) & F(pi/2)=(1,0) Hence the and moves in a circle of radius 1 unit. 2) F\'(t)= (-2sin2t, 2cos2t) F\'(pi/2) = (0, -2) This represents the Speed at t= pi/2 3) Integral 0 to 1 F(t) dt = Area of circle = pir^2 = pi (as r = 1) Solution I have the answer ready in paint but i am not able to upload. 1) F(0)=(1,0), F(pi/4)=(0,1), F(pi/2) =(-1,0), F(3pi/4)=(0,-1) & F(pi/2)=(1,0) Hence the and moves in a circle of radius 1 unit. 2) F\'(t)= (-2sin2t, 2cos2t) F\'(pi/2) = (0, -2) This represents the Speed at t= pi/2 3) Integral 0 to 1 F(t) dt = Area of circle = pir^2 = pi (as r = 1).

I have the answer ready in paint but i am not able to upload.
1) F(0)=(1,0), F(pi/4)=(0,1), F(pi/2) =(-1,0), F(3pi/4)=(0,-1) & F(pi/2)=(1,0)
Hence the and moves in a circle of radius 1 unit.
2) F'(t)= (-2sin2t, 2cos2t)
F'(pi/2) = (0, -2)
This represents the Speed at t= pi/2
3) Integral 0 to 1 F(t) dt = Area of circle = pir^2 = pi (as r = 1)
Solution
I have the answer ready in paint but i am not able to upload.
1) F(0)=(1,0), F(pi/4)=(0,1), F(pi/2) =(-1,0), F(3pi/4)=(0,-1) & F(pi/2)=(1,0)
Hence the and moves in a circle of radius 1 unit.
2) F'(t)= (-2sin2t, 2cos2t)
F'(pi/2) = (0, -2)
This represents the Speed at t= pi/2
3) Integral 0 to 1 F(t) dt = Area of circle = pir^2 = pi (as r = 1)

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