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A particle moves along the curve x square upon 9 + Y square by 4 is equals to 1 with the constant speed V Express the velocity Vectorially as a function of ( x , y )

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one year 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 isV= (2/9)X (i)^ + y/2 (j)^.
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one year ago
```							Since speed is tangential to the ellipse, we write 𝑑𝑦𝑑𝑥=𝑣𝑦𝑣𝑥 Differentiating the equation of the curve wrt x, you’d get 𝑦2𝑑𝑦𝑑𝑥=−2𝑥9 𝑑𝑦𝑑𝑥=−4𝑥9𝑦 So 𝑣𝑦𝑣𝑥=−4𝑥9𝑦 Now, the vector 9𝑦𝑖̂ −4𝑥𝑗̂ is parallel to the velocity vector. We can now go on and write the velocity vector 𝑣⃗ as 𝑣⃗ =𝑣9𝑦𝑖̂ −4𝑥𝑗̂ 16𝑥2+81𝑦2‾‾‾‾‾‾‾‾‾‾‾‾√
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one year ago
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