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A particle moves along the curve x^2/9+x^2/4=1 with constant speed v.express its velocity vectorially as a function of(x,y)

A particle moves along the curve x^2/9+x^2/4=1 with constant speed v.express its velocity vectorially as a function of(x,y)

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1 Answers

Khimraj
3007 Points
2 years ago

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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