Ch 5

Class 12 Mathematics • Chapter 5 • NCERT Part I

5.6 Derivatives of Functions in Parametric Forms

A curve may be given by parametric equations \(x=f(t),\;y=g(t)\) where \(t\) is a parameter. Examples: a circle \(x=a\cos t,\; y=a\sin t\); a projectile \(x=vt,\; y=vt-\tfrac12 gt^2\).

Reasoning. By chain rule \(\dfrac{dy}{dt}=\dfrac{dy}{dx}\cdot\dfrac{dx}{dt}\); dividing by \(dx/dt\) gives the stated formula.

Example 29 — Circle

\(x=a\cos t,\;y=a\sin t\): \(\dfrac{dx}{dt}=-a\sin t,\;\dfrac{dy}{dt}=a\cos t\). So \(\dfrac{dy}{dx}=\dfrac{a\cos t}{-a\sin t}=-\cot t.\)

Example 30 — Ellipse

\(x=a\cos\theta,\;y=b\sin\theta\). \(\dfrac{dy}{dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac{b}{a}\cot\theta.\)

Example 31 — Cycloid-like

\(x=a(t+\sin t),\;y=a(1-\cos t)\). \(\dfrac{dx}{dt}=a(1+\cos t),\;\dfrac{dy}{dt}=a\sin t\). So \(\dfrac{dy}{dx}=\dfrac{\sin t}{1+\cos t}=\tan(t/2)\) (using \(\sin t=2\sin(t/2)\cos(t/2),\;1+\cos t=2\cos^2(t/2)\)).

Example 32 — \(x=a\sec\theta,\;y=b\tan\theta\)

\(\dfrac{dx}{d\theta}=a\sec\theta\tan\theta,\;\dfrac{dy}{d\theta}=b\sec^2\theta\). Hence \(\dfrac{dy}{dx}=\dfrac{b\sec^2\theta}{a\sec\theta\tan\theta}=\dfrac{b}{a\sin\theta}.\)

5.7 Second Order Derivative

The second derivative is the derivative of \(f'\): \(f''(x)=\dfrac{d}{dx}\!\left(\dfrac{dy}{dx}\right)=\dfrac{d^2y}{dx^2}.\) Higher orders: \(y''',y^{(4)}\), etc.

Example 33 — \(y=x^3\)

\(y'=3x^2,\;y''=6x,\;y'''=6,\;y^{(4)}=0.\)

Example 34 — \(y=\sin x\)

\(y'=\cos x,\;y''=-\sin x=-y.\) So \(y'' +y=0\) — the simple-harmonic differential equation.

Example 35 — Parametric second derivative

\(x=a\cos t,\;y=a\sin t\). We found \(y'=-\cot t\). Now \(\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(-\cot t)=\dfrac{d(-\cot t)/dt}{dx/dt}=\dfrac{\csc^2 t}{-a\sin t}=-\dfrac{1}{a\sin^3 t}.\)

Example 36 — Verify a differential equation

If \(y=e^{3x}\sin 4x\), show \(y''-6y'+25y=0\).

Solution. \(y'=3e^{3x}\sin 4x+4e^{3x}\cos 4x=e^{3x}(3\sin 4x+4\cos 4x).\) \(y''=e^{3x}[3(3\sin 4x+4\cos 4x)+(12\cos 4x-16\sin 4x)]=e^{3x}(-7\sin 4x+24\cos 4x).\) Then \(y''-6y'+25y = e^{3x}[(-7\sin 4x+24\cos 4x)-6(3\sin 4x+4\cos 4x)+25\sin 4x]=e^{3x}[(−7−18+25)\sin 4x+(24−24)\cos 4x]=0.\;\blacksquare\)

Worked Examples

Summary — Chapter 5

• Continuity: \(\lim_{x\to c} f(x)=f(c)\); algebra of continuous functions.
• Differentiability implies continuity; converse is false (\(|x|\)).
• Chain rule: \((f\circ g)'=f'(g)\cdot g'\).
• Implicit differentiation for relations \(F(x,y)=0\).
• Standard derivatives: \((\sin^{-1}x)'=1/\sqrt{1-x^2}\), \((\tan^{-1}x)'=1/(1+x^2)\), \((e^x)'=e^x\), \((\log x)'=1/x\), \((a^x)'=a^x\log a\).
• Logarithmic differentiation for \([f(x)]^{g(x)}\).
• Parametric: \(dy/dx=(dy/dt)/(dx/dt)\).
• Second derivative: \(d^2y/dx^2=d(y')/dx\).