Class 12 Maths NCERT Solutions Chapter 5 Ex 5.6 – Derivatives in Parametric Form

We're given x and y both as functions of the parameter t. First differentiate each with respect to t separately.

Differentiate x=2at^2 with respect to t: \dfrac{dx}{dt}=2a\cdot2t=4at.

Differentiate y=at^4 with respect to t: \dfrac{dy}{dt}=a\cdot4t^3=4at^3.

Now use the parametric-form rule \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}: \dfrac{dy}{dx}=\dfrac{4at^3}{4at}.

Cancel the common factor 4at from numerator and denominator (valid since t\neq0).

Differentiate x=a\cos\theta with respect to \theta: \dfrac{dx}{d\theta}=-a\sin\theta.

Differentiate y=b\cos\theta with respect to \theta: \dfrac{dy}{d\theta}=-b\sin\theta.

Divide: \dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}=\dfrac{-b\sin\theta}{-a\sin\theta}.

The -\sin\theta factor cancels from top and bottom.

Differentiate x=\sin t with respect to t: \dfrac{dx}{dt}=\cos t.

Differentiate y=\cos2t with respect to t, using the chain rule (derivative of \cos u is -\sin u\cdot u' with u=2t): \dfrac{dy}{dt}=-\sin2t\cdot2=-2\sin2t.

Use the double-angle identity \sin2t=2\sin t\cos t to rewrite: \dfrac{dy}{dt}=-2(2\sin t\cos t)=-4\sin t\cos t.

Divide: \dfrac{dy}{dx}=\dfrac{-4\sin t\cos t}{\cos t}.

Cancel the common factor \cos t (valid where \cos t\neq0).

Differentiate x=4t with respect to t: \dfrac{dx}{dt}=4.

Rewrite y=\dfrac4t=4t^{-1} and differentiate: \dfrac{dy}{dt}=4\cdot(-1)t^{-2}=-\dfrac{4}{t^2}.

Divide: \dfrac{dy}{dx}=\dfrac{-4/t^2}{4}.

Differentiate x=\cos\theta-\cos2\theta term by term with respect to \theta.

For the first term, \dfrac{d}{d\theta}\cos\theta=-\sin\theta.

For the second, using the chain rule with u=2\theta: \dfrac{d}{d\theta}\cos2\theta=-2\sin2\theta, so -\dfrac{d}{d\theta}\cos2\theta=2\sin2\theta.

So \dfrac{dx}{d\theta}=-\sin\theta+2\sin2\theta.

Differentiate y=\sin\theta-\sin2\theta term by term.

\dfrac{d}{d\theta}\sin\theta=\cos\theta, and \dfrac{d}{d\theta}\sin2\theta=2\cos2\theta (chain rule), so -\dfrac{d}{d\theta}\sin2\theta=-2\cos2\theta.

So \dfrac{dy}{d\theta}=\cos\theta-2\cos2\theta.

Divide the two results: \dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}=\dfrac{\cos\theta-2\cos2\theta}{2\sin2\theta-\sin\theta}.

Differentiate x=a(\theta-\sin\theta) with respect to \theta: \dfrac{dx}{d\theta}=a(1-\cos\theta).

Differentiate y=a(1+\cos\theta) with respect to \theta: the constant 1 vanishes, leaving \dfrac{dy}{d\theta}=a(-\sin\theta)=-a\sin\theta.

Divide: \dfrac{dy}{dx}=\dfrac{-a\sin\theta}{a(1-\cos\theta)}=\dfrac{-\sin\theta}{1-\cos\theta} (the constant a cancels).

Now simplify using the half-angle identities \sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2} and 1-\cos\theta=2\sin^2\frac{\theta}{2}.

Substitute: \dfrac{dy}{dx}=\dfrac{-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}.

Cancel one factor of 2\sin\frac{\theta}{2} from numerator and denominator: \dfrac{dy}{dx}=\dfrac{-\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}.

Write x=\sin^3t\cdot(\cos2t)^{-1/2} so we can use the product rule with u=\sin^3t and v=(\cos2t)^{-1/2}.

Differentiate u: u'=3\sin^2t\cos t (chain rule, since u=(\sin t)^3).

Differentiate v: v'=-\tfrac12(\cos2t)^{-3/2}\cdot(-2\sin2t)=\dfrac{\sin2t}{(\cos2t)^{3/2}} (chain rule, differentiating \cos2t gives -2\sin2t).

By the product rule, \dfrac{dx}{dt}=u'v+uv'=\dfrac{3\sin^2t\cos t}{\sqrt{\cos2t}}+\dfrac{\sin^3t\cdot\sin2t}{(\cos2t)^{3/2}}.

Substitute the double-angle identity \sin2t=2\sin t\cos t into the second term: it becomes \dfrac{2\sin^4t\cos t}{(\cos2t)^{3/2}}.

Both terms now share the factor \dfrac{\sin^2t\cos t}{(\cos2t)^{3/2}}. Factor it out: \dfrac{dx}{dt}=\dfrac{\sin^2t\cos t}{(\cos2t)^{3/2}}\left[3\cos2t+2\sin^2t\right].

Simplify the bracket using \cos2t=1-2\sin^2t: 3(1-2\sin^2t)+2\sin^2t=3-6\sin^2t+2\sin^2t=3-4\sin^2t.

Recognise the triple-angle identity \sin3t=3\sin t-4\sin^3t=\sin t(3-4\sin^2t), so 3-4\sin^2t=\dfrac{\sin3t}{\sin t}.

Substitute this in: \dfrac{dx}{dt}=\dfrac{\sin^2t\cos t}{(\cos2t)^{3/2}}\cdot\dfrac{\sin3t}{\sin t}=\dfrac{\sin t\cos t\sin3t}{(\cos2t)^{3/2}}.

Now repeat the same process for y=\cos^3t\cdot(\cos2t)^{-1/2}, with u=\cos^3t so u'=-3\cos^2t\sin t, and the same v' as before.

By the product rule: \dfrac{dy}{dt}=\dfrac{-3\cos^2t\sin t}{\sqrt{\cos2t}}+\dfrac{\cos^3t\cdot\sin2t}{(\cos2t)^{3/2}}.

Substitute \sin2t=2\sin t\cos t in the second term to get \dfrac{2\cos^4t\sin t}{(\cos2t)^{3/2}}.

Factor out \dfrac{\cos^2t\sin t}{(\cos2t)^{3/2}}: \dfrac{dy}{dt}=\dfrac{\cos^2t\sin t}{(\cos2t)^{3/2}}\left[-3\cos2t+2\cos^2t\right].

Simplify the bracket using \cos2t=2\cos^2t-1: -3(2\cos^2t-1)+2\cos^2t=-6\cos^2t+3+2\cos^2t=3-4\cos^2t.

Use the triple-angle identity \cos3t=4\cos^3t-3\cos t=\cos t(4\cos^2t-3), so 3-4\cos^2t=-\dfrac{\cos3t}{\cos t}.

Substitute this in: \dfrac{dy}{dt}=\dfrac{\cos^2t\sin t}{(\cos2t)^{3/2}}\cdot\left(-\dfrac{\cos3t}{\cos t}\right)=\dfrac{-\cos t\sin t\cos3t}{(\cos2t)^{3/2}}.

Finally divide: \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{-\cos t\sin t\cos3t}{\sin t\cos t\sin3t}. The \sin t\cos t and the (\cos2t)^{3/2} factors both cancel.

Differentiate x term by term. For \cos t: \dfrac{d}{dt}\cos t=-\sin t.

For \log\tan\frac{t}{2}: by the chain rule, \dfrac{d}{dt}\log\tan\dfrac{t}{2}=\dfrac{1}{\tan\frac{t}{2}}\cdot\sec^2\dfrac{t}{2}\cdot\dfrac{1}{2} (the extra factor \frac12 comes from differentiating the inner t/2).

Simplify: \dfrac{1}{\tan\frac{t}{2}}\cdot\sec^2\dfrac{t}{2}=\dfrac{\cos\frac{t}{2}}{\sin\frac{t}{2}}\cdot\dfrac{1}{\cos^2\frac{t}{2}}=\dfrac{1}{\sin\frac{t}{2}\cos\frac{t}{2}}.

Using the double-angle identity \sin t=2\sin\frac{t}{2}\cos\frac{t}{2}, we get \dfrac{1}{\sin\frac{t}{2}\cos\frac{t}{2}}=\dfrac{2}{\sin t}, so with the \frac12 factor: \dfrac{d}{dt}\log\tan\dfrac{t}{2}=\dfrac{1}{2}\cdot\dfrac{2}{\sin t}=\dfrac{1}{\sin t}.

So \dfrac{dx}{dt}=a\left(-\sin t+\dfrac{1}{\sin t}\right)=a\cdot\dfrac{-\sin^2t+1}{\sin t}=a\cdot\dfrac{\cos^2t}{\sin t} (using 1-\sin^2t=\cos^2t).

Differentiate y=a\sin t: \dfrac{dy}{dt}=a\cos t.

Divide: \dfrac{dy}{dx}=\dfrac{a\cos t}{a\cos^2t/\sin t}=\cos t\cdot\dfrac{\sin t}{\cos^2t}.

Cancel one factor of \cos t: \dfrac{dy}{dx}=\dfrac{\sin t}{\cos t}.

Differentiate x=a\sec\theta with respect to \theta, using the standard result \dfrac{d}{d\theta}\sec\theta=\sec\theta\tan\theta: \dfrac{dx}{d\theta}=a\sec\theta\tan\theta.

Differentiate y=b\tan\theta with respect to \theta, using \dfrac{d}{d\theta}\tan\theta=\sec^2\theta: \dfrac{dy}{d\theta}=b\sec^2\theta.

Divide: \dfrac{dy}{dx}=\dfrac{b\sec^2\theta}{a\sec\theta\tan\theta}.

Cancel one factor of \sec\theta: \dfrac{dy}{dx}=\dfrac{b\sec\theta}{a\tan\theta}.

Write \sec\theta=\dfrac{1}{\cos\theta} and \tan\theta=\dfrac{\sin\theta}{\cos\theta}, so \dfrac{\sec\theta}{\tan\theta}=\dfrac{1/\cos\theta}{\sin\theta/\cos\theta}=\dfrac{1}{\sin\theta}=\csc\theta.

Differentiate x=a(\cos\theta+\theta\sin\theta) term by term. For \cos\theta: derivative is -\sin\theta.

For \theta\sin\theta, use the product rule (with u=\theta, v=\sin\theta): \dfrac{d}{d\theta}(\theta\sin\theta)=1\cdot\sin\theta+\theta\cdot\cos\theta=\sin\theta+\theta\cos\theta.

Add the two pieces: \dfrac{dx}{d\theta}=a\left[-\sin\theta+\sin\theta+\theta\cos\theta\right]. The -\sin\theta and +\sin\theta cancel, leaving \dfrac{dx}{d\theta}=a\theta\cos\theta.

Differentiate y=a(\sin\theta-\theta\cos\theta) term by term. For \sin\theta: derivative is \cos\theta.

For \theta\cos\theta, use the product rule (with u=\theta, v=\cos\theta): \dfrac{d}{d\theta}(\theta\cos\theta)=1\cdot\cos\theta+\theta\cdot(-\sin\theta)=\cos\theta-\theta\sin\theta.

So \dfrac{dy}{d\theta}=a\left[\cos\theta-(\cos\theta-\theta\sin\theta)\right]=a\left[\cos\theta-\cos\theta+\theta\sin\theta\right]. The \cos\theta terms cancel, leaving \dfrac{dy}{d\theta}=a\theta\sin\theta.

Divide: \dfrac{dy}{dx}=\dfrac{a\theta\sin\theta}{a\theta\cos\theta}. Cancel a\theta from top and bottom.

Square both sides of each equation to remove the square roots: x^2=a^{\sin^{-1}t} and y^2=a^{\cos^{-1}t}.

Take the natural logarithm of both sides of each equation. For the first: \log(x^2)=\sin^{-1}t\cdot\log a, i.e. 2\log x=(\log a)\sin^{-1}t.

For the second: 2\log y=(\log a)\cos^{-1}t.

Differentiate the first equation with respect to t, treating x as a function of t (chain rule on the left, standard inverse-sine derivative on the right): \dfrac{2}{x}\dfrac{dx}{dt}=\dfrac{\log a}{\sqrt{1-t^2}}.

Solve for \dfrac{dx}{dt}: \dfrac{dx}{dt}=\dfrac{x\log a}{2\sqrt{1-t^2}}.

Differentiate the second equation the same way, using \dfrac{d}{dt}\cos^{-1}t=-\dfrac{1}{\sqrt{1-t^2}}: \dfrac{2}{y}\dfrac{dy}{dt}=\dfrac{-\log a}{\sqrt{1-t^2}}.

Solve for \dfrac{dy}{dt}: \dfrac{dy}{dt}=\dfrac{-y\log a}{2\sqrt{1-t^2}}.

Divide the two results: \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{-y\log a/\left(2\sqrt{1-t^2}\right)}{x\log a/\left(2\sqrt{1-t^2}\right)}.

Every factor in the denominator of each fraction — \log a and 2\sqrt{1-t^2} — is identical top and bottom, so they all cancel, leaving only the ratio of -y to x.