1.

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`.

Answer» Given `(x^(2))/(9)+(v^(2))/(4)=1`
`rArr (2x)/(9) [(dx)/(dt)]+(2y)/(4)[(dy)/(dt)]=1 [(dx)/(dt)=v_(x),(dy)/(dt)=v_(y)]`
`rArr v_(x) =-(9)/(4)(y)/(x)v_(y)`
Also `v_(x)^(2)+v_(y)^(2) =v^(2) rArr (-(9)/(4)(y)/(x)v_(y))^(2) +v_(y)^(2)=v^(2)`
`rArr v_(y)^(2) =(16x^(2)v^(2))/(16x^(2)+81y^(2)) rArr v_(y)= (+-4xv)/(sqrt(16x^(2)+81y^(2))`
From (i) and (ii), we get ,`v_(x)=(+-yv)/(sqrt(16x^(2)+81y^(2))`
So, velocity is given by `v_(x) jat i+ v_(y) hat j = ((+- 9 y hat i+- 4 x hat j)v)/(sqrt(16x^(2)+81y^(2)))`.


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