the velocity of the particle noving in the x-y plane is given by dx/dt=8asin2at,dy/dt=5acos2at.when t=0,x=8andy=0,the path of the particle is a.................here a=3.14sir plz tell some 2 way to find path of particle

Dear jauneet

to find the path means you have to find the ralation ship between x and y with the given information .

here we have to eleminate the t .

dx/dt=8asin2at

intigrate

 x= -(8acos2at )/2a   + c1

 x = -4cos2at + c1

now use the given condition

at t=0 ,x=8

so 8=-4 +c1

 c1=12

 so equation become

x =-4cos2at + 12 

or  (x-12)/4 = -cos2at ............1

now similerly for y

dy/dt=5acos2at

intigrate

y= (5asin2at)/2a  +c2

y =5/2 sin2at + c2

now apply condition

y=0 at t=0

0 =0+c2

c2=0

so equation become

y =5/2 sin2at

or 2y/5 = sin2at ............2

now square equation 1 and 2 and add

((x-12)/4)² + (2y/5)² = 1

this the path of the article

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