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Question

# sin3 tcos t7.cos 2tcos 2t

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Solution

## It is given that x and y are parametrically connected by the equations, x= sin 3 t cos2t (1) And, y= cos 3 t cos2t (2) Differentiate equation (2) with respect to t. dy dt = d dt ( cos 3 t cos2t ) dy dt = d dt ( cos 3 t ). cos2t − d( cos2t ) dt . cos 3 t ( cos2t ) 2 dy dt = 3 cos 2 t d( cost ) dt cos2t − 1 2 cos2t d( cos2t ) dt . cos 3 t cos2t dy dt = −3 cos 2 tsint cos2t + 1 cos2t sin2t. cos 3 t cos2t Further simplify. dy dt = −3 cos 2 tsintcos2t+sin2t. cos 3 t cos2t cos2t Differentiate equation (1) with respect to t. dx dt = d dt ( sin 3 t cos2t ) dx dt = d( sin 3 t ) dt . cos2t − d( cos2t ) dt . sin 3 t ( cos2t ) 2 dx dt = 3 sin 2 t. d( sint ) dt cos2t − 1 2 cos2t d( cos2t ) dt . sin 3 t cos2t dx dt = 3 sin 2 t.cost cos2t − 1 2 cos2t ( −sin2t ).2. sin 3 t cos2t Further simplify. dx dt = 3 sin 2 t.costcos2t+( sin2t ). sin 3 t ( cos2t ) cos2t We know that, dy dx = dy dt dx dt Substitute the value of dy dt and dx dt . dy dx = −3 cos 2 tsintcos2t+sin2t. cos 3 t cos2t cos2t 3 sin 2 t.costcos2t+( sin2t ). sin 3 t ( cos2t ) cos2t dy dx = −3 cos 2 tsint( 2 cos 2 t−1 )+( 2sintcost ). cos 3 t 3 sin 2 t.cost( 1−2 sin 2 t )+( 2sintcost ). sin 3 t dy dx = sintcost( −6 cos 3 t+3cost+2 cos 3 t ) sintcost( 3sint−3 sin 3 t+2 sin 3 t ) dy dx = −4 cos 3 t+3cost 3sint− sin 3 t Further simplify. dy dx = −cos3t sin3t dy dx =−cot3t Thus, the solution is dy dx =−cot3t.

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