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A particle moves along the curve x 2 /9 + y 2 /4 = 1 With constant speed v . Express its velocity vectorially as a function of (x,y).
So, effectively, the displacement of the particle with respect to the reference co-ordinate axes is:-s= (x^2)/9 + (y^2)/4 =1.Now, velocity is ds/dt.So, we partially differentiate the equation, first with respect to X, treating you as a constant. This gives the velocity of the particle. (Vx) in X direction.Vx= (2/9)x (i)^Similarly, to get Vy, partially differentiate with respect to y.Vy= y/2 (j)^So, the velocity vector isV= (2/9)X (i)^ + y/2 (j)^.
So, effectively, the displacement of the particle with respect to the reference co-ordinate axes is:-
s= (x^2)/9 + (y^2)/4 =1.
Now, velocity is ds/dt.
So, we partially differentiate the equation, first with respect to X, treating you as a constant. This gives the velocity of the particle. (Vx) in X direction.
Vx= (2/9)x (i)^
Similarly, to get Vy, partially differentiate with respect to y.
Vy= y/2 (j)^
So, the velocity vector is
V= (2/9)X (i)^ + y/2 (j)^.
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