More practice questions

5. Q5. Find dy/dx if x = sin³ t/√(cos 2t), y = cos³ t/√(cos 2t) (NCERT Exercise 5.6, Q7, page 137).

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   The real parametrisation requires cos 2t > 0. Let D = (cos 2t)^3/2. The quotient rule, with sin 2t = 2 sin t cos t and cos 2t = 1 − 2 sin² t = 2 cos² t − 1, gives dx/dt = sin t cos t × (3 sin t − 4 sin³ t)/D and dy/dt = sin t cos t × (3 cos t − 4 cos³ t)/D. Since 3 sin t − 4 sin³ t = sin 3t and 3 cos t − 4 cos³ t = −cos 3t, dy/dx = −cos 3t/sin 3t = −cot 3t.

6. Q6. Find the slope of the tangent to the parabola x = at², y = 2at at the point where t = 2, and the point itself.

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   dy/dx = 1/t = 1/2 at the point (4a, 4a).

7. Q7. For the curve x = t² − 1, y = t³ − t, find dy/dx, and the values of t where the tangent is horizontal.

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   dy/dx = (3t² − 1)/(2t), for t ≠ 0. The tangent is horizontal where 3t² − 1 = 0, that is t = ±1/√3.

8. Q8. Find dy/dx for x = e^t cos t, y = e^t sin t.

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   dx/dt = e^t(cos t − sin t) and dy/dt = e^t(sin t + cos t), so dy/dx = (sin t + cos t)/(cos t − sin t), which equals tan (t + π/4).