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Grade: 11
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 )
8 months ago

Answers : (2)

Vikas TU
10470 Points

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)^.

8 months ago
3008 Points
Since speed is tangential to the ellipse, we write ๐‘‘๐‘ฆ๐‘‘๐‘ฅ=๐‘ฃ๐‘ฆ๐‘ฃ๐‘ฅ Differentiating the equation of the curve wrt x, youโ€™d get ๐‘ฆ2๐‘‘๐‘ฆ๐‘‘๐‘ฅ=โˆ’2๐‘ฅ9 ๐‘‘๐‘ฆ๐‘‘๐‘ฅ=โˆ’4๐‘ฅ9๐‘ฆ So ๐‘ฃ๐‘ฆ๐‘ฃ๐‘ฅ=โˆ’4๐‘ฅ9๐‘ฆ Now, the vector 9๐‘ฆ๐‘–ฬ‚ โˆ’4๐‘ฅ๐‘—ฬ‚ is parallel to the velocity vector. We can now go on and write the velocity vector ๐‘ฃโƒ— as ๐‘ฃโƒ— =๐‘ฃ9๐‘ฆ๐‘–ฬ‚ โˆ’4๐‘ฅ๐‘—ฬ‚ 16๐‘ฅ2+81๐‘ฆ2โ€พโ€พโ€พโ€พโ€พโ€พโ€พโ€พโ€พโ€พโ€พโ€พโˆš
7 months ago
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