Exercise 5.6, Question 1
\(x = 2at^2,\ y = at^4\)
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- \(\dfrac{dx}{dt} = 4at\), \(\dfrac{dy}{dt} = 4at^3\).
Answer: \(\dfrac{dy}{dx} = t^2\)
NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6
Exercise 5.6: Parametric differentiation. If \(x = f(t)\) and \(y = g(t)\), then \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\) wherever \(\tfrac{dx}{dt} \ne 0\); simplify the ratio with identities (half angles often help).
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Exercise 5.6 questions and solutions
Exercise 5.6, Question 2
\(x = a\cos\theta,\ y = b\cos\theta\)
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- \[\dfrac{dy/d\theta}{dx/d\theta} = \dfrac{-b\sin\theta}{-a\sin\theta}\]
Answer: \(\dfrac{dy}{dx} = \dfrac{b}{a}\)
Exercise 5.6, Question 3
\(x = \sin t,\ y = \cos 2t\)
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- \[\dfrac{-2\sin 2t}{\cos t} = \dfrac{-4\sin t\cos t}{\cos t}\]
Answer: \(\dfrac{dy}{dx} = -4\sin t\)
Exercise 5.6, Question 4
\(x = 4t,\ y = \dfrac4t\)
Show solution
- \(\dfrac{-4/t^2}{4}\).
Answer: \(\dfrac{dy}{dx} = -\dfrac{1}{t^2}\)
Exercise 5.6, Question 5
\(x = \cos\theta - \cos 2\theta,\ y = \sin\theta - \sin 2\theta\)
Show solution
- \[\dfrac{dx}{d\theta} = -\sin\theta + 2\sin 2\theta\], \[\dfrac{dy}{d\theta} = \cos\theta - 2\cos 2\theta\]
Answer: \[\dfrac{dy}{dx} = \dfrac{\cos\theta - 2\cos 2\theta}{2\sin 2\theta - \sin\theta}\]
Exercise 5.6, Question 6
\(x = a(\theta - \sin\theta),\ y = a(1 + \cos\theta)\)
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- \[\dfrac{-a\sin\theta}{a(1 - \cos\theta)} = -\dfrac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}\]
Answer: \(\dfrac{dy}{dx} = -\cot\dfrac{\theta}{2}\)
Exercise 5.6, Question 7
\(x = \dfrac{\sin^3 t}{\sqrt{\cos 2t}},\ y = \dfrac{\cos^3 t}{\sqrt{\cos 2t}}\)
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- \[\begin{aligned}\dfrac{dx}{dt} &= \dfrac{3\sin^2 t\cos t\cos 2t + \sin^3 t\sin 2t}{(\cos 2t)^{3/2}} \\ &= \dfrac{\sin t\cos t\,(3\sin t\cos 2t + 2\sin^3 t)}{(\cos 2t)^{3/2}}\end{aligned}\]; note \(3\sin t - 4\sin^3 t = \sin 3t\) gives \[\begin{aligned}3\sin t\cos 2t + 2\sin^3 t &= 3\sin t - 4\sin^3 t \\ &= \sin 3t\end{aligned}\]
- Similarly \[\dfrac{dy}{dt} = \dfrac{\sin t\cos t\,(-3\cos t\cos 2t + 2\cos^3 t)}{(\cos 2t)^{3/2}}\] and \[\begin{aligned}-3\cos t\cos 2t + 2\cos^3 t &= 3\cos t - 4\cos^3 t \\ &= -\cos 3t\end{aligned}\]
- Ratio: \(\dfrac{-\cos 3t}{\sin 3t}\).
Answer: \(\dfrac{dy}{dx} = -\cot 3t\)
Exercise 5.6, Question 8
\(x = a\left(\cos t + \log\tan\tfrac{t}{2}\right),\ y = a\sin t\)
Show solution
- \[\begin{aligned}\dfrac{d}{dt}\log\tan\tfrac{t}{2} &= \dfrac{\frac12\sec^2\frac{t}{2}}{\tan\frac{t}{2}} \\ &= \dfrac{1}{\sin t}\end{aligned}\], so \[\begin{aligned}\dfrac{dx}{dt} &= a\left(\dfrac{1}{\sin t} - \sin t\right) \\ &= \dfrac{a\cos^2 t}{\sin t}\end{aligned}\]
- \(\dfrac{dy}{dt} = a\cos t\).
Answer: \(\dfrac{dy}{dx} = \tan t\)
Exercise 5.6, Question 9
\(x = a\sec\theta,\ y = b\tan\theta\)
Show solution
- \[\dfrac{b\sec^2\theta}{a\sec\theta\tan\theta} = \dfrac{b\sec\theta}{a\tan\theta}\]
Answer: \(\dfrac{dy}{dx} = \dfrac{b}{a}\csc\theta\)
Exercise 5.6, Question 10
\(x = a(\cos\theta + \theta\sin\theta),\ y = a(\sin\theta - \theta\cos\theta)\)
Show solution
- \[\begin{aligned}\dfrac{dx}{d\theta} &= a(-\sin\theta + \sin\theta + \theta\cos\theta) \\ &= a\theta\cos\theta\end{aligned}\]; \(\dfrac{dy}{d\theta} = a\theta\sin\theta\).
Answer: \(\dfrac{dy}{dx} = \tan\theta\)
Exercise 5.6, Question 11
\(x = \sqrt{a^{\sin^{-1}t}},\ y = \sqrt{a^{\cos^{-1}t}}\). Show \(\dfrac{dy}{dx} = -\dfrac{y}{x}\).
Show solution
- \[\begin{aligned}xy &= \sqrt{a^{\sin^{-1}t + \cos^{-1}t}} \\ &= \sqrt{a^{\pi/2}}\end{aligned}\], a constant.
- Differentiate \(xy = \text{const}\) with respect to x: \(y + x\dfrac{dy}{dx} = 0\).
Answer: Shown
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