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Examples of calculus

August 16, 2020
Category: Uncategorized

Examples :

1 If xe^xy – y = sin²x. Differentiating both the sides with respect to x.

Soln. $\frac{dy}{dx}$ ( xe^xy – $\frac{dy}{dx}$ = ( sin²x ) ⟹ x $\frac{dy}{dx}$ e^xy + e^xyx – $\frac{dy}{dx}$ = 2 sin x cos x

⟹ xe^xy $\left\{x\frac{d}{\mathrm{dx}}+y\right\}$ + e^xy – $\frac{dy}{dx}$ = 2 sin x cos x

⟹ (x²e^xy – 1 ) $\frac{dy}{dx}$ = sin 2x – xye^xy – e^xy ⟹ $\frac{dy}{dx}$ = $\frac{sin{2}x-(\mathrm{xy\ } +\mathrm{\ 1\ } )e^{\mathrm{xy}}}{x^2e^{\mathrm{xy}}-1}$

2. If y = b tan^-1, find $\frac{dy}{dx}$ .

Soln. We have, tan = + tan $\frac{y}{x}$

Differentiating both the sides with respect to x, we get

Sec² $\frac{y}{b}.\frac{1}{b}$ $\frac{dy}{dx}$ = $\frac{1}{a} + \frac{1}{1+\frac{y^2}{x^2}}\left\{\frac{x\frac{\mathrm{dx}}{\mathrm{dy}}-y}{x^2}\right\}$

= $\frac{1}{a} + \frac{x\frac{\mathrm{dx}}{\mathrm{dy}}-y}{x^2+y^2}$

$\therefore$ $\frac{dy}{dx}$ $\left[\frac{1}{b}{sec}^2{\frac{y}{b}}-\frac{x}{x^2+y^2}\right]$ = $\frac{1}{a} - \frac{y}{x^2+y^2} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$ = $\frac{\frac{1}{a}\frac{y}{x^2+y^2}}{\frac{1}{b}{sec}^2{\frac{y}{b}}-\frac{x}{x^2+y^2}}$

3 If x = a cos²t, y = b sin³ t then, find \frac{\mathrm{dy}}{\mathrm{dx}}.

Soln. x = a cos²t & y = b sin³ t

∴ $\frac{\mathrm{dx}}{\mathrm{dt}}$ = – 3a cos²t sin t

∴ $\frac{\mathrm{dy}}{\mathrm{dt}}$ = 3b sin²t cos t

∴ $\frac{\mathrm{dy}}{\mathrm{dx}}$ = $\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$ = $\frac{3\mathrm{b\ si}\mathrm{n}^2t}{\mathrm{-3a\ co}\mathrm{s}^2\mathrm{t\ sin\ t} }$ = – $\frac{b} {a}\ \frac{sin{t}}{\mathrm{cos\ t}} = - \frac{b}{a}tan t$

4 If x = a sin 2θ ( 1 + cos 2θ ) and y = b cos 2θ ( 1 – cos 2θ ). Then,

find $\frac{\mathrm{dy}}{\mathrm{dx}}$ , when θ = $\frac{\pi}{6}$

Soln.

x = a sin 2θ ( 1 + cos 2θ )

∴ $\frac{\mathrm{dx}}{d\theta}$ = $\frac{d}{d\theta}{a sin 2\theta ( 1 + cos 2\theta )}$

= a [sin 2θ ( – sin 2θ ) + 2 cos 2θ ( 1 + cos 2θ)]

= 2a { cos 4θ + cos 2θ }

y = b cos 2θ ( 1 – cos 2θ )

$\frac{\mathrm{dy}}{d\theta}$ = \frac{d}{d\theta}{ b cos 2θ ( 1 – cos 2θ )}

= b [cos 2θ ( 2 sin 2θ) + ( 1 – cos 2θ )(– 2 sin 2θ)]

= 2b { sin 4θ – sin 2θ }

$\frac{\mathrm{dy}}{\mathrm{dx}}$ = $\frac{b}{a}\left[\frac{sin{4}\theta-sin{2}\theta}{cos{4}\theta+cos{2}\theta}\right]$ = $\frac{b}{a}$ tanθ

∴ $\frac{\mathrm{dy}}{\mathrm{dx}}$ = $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{6}}$ = $\frac{b}{a}\times tan \frac{\pi}{6} = \frac{b}{a\sqrt3}$

5 If y = sin x, then find $\frac{d^\mathbf{3}y}{\mathrm{d}\mathrm{x}^\mathbf{3}}$ .

Soln. y = sin x $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$ = cos x $\Rightarrow \frac{d^2y}{\mathrm{d}\mathrm{x}^2}$ = $\frac{d}{\mathrm{dx}}$ ( cos x ) = – sin x

$\Rightarrow \frac{d^3y}{\mathrm{d}\mathrm{x}^3}$ = $\frac{d}{\mathrm{dx}}$ ( – sin x ) = – cos x

6 If y = e then find $\frac{d^\mathbf{2}y}{\mathrm{d}\mathrm{x}^\mathbf{2}}.$

Soln. y = e^{tan x} $\large \Rightarrow$ log y = tan x $\Rightarrow \frac{1}{y}\frac{\mathrm{dy}}{\mathrm{dx}}$ = sec² x $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$ = y sec²x

$\therefore\frac{d^2y}{\mathrm{d}\mathrm{x}^2}$ = $\frac{d}{\mathrm{dx}}$ ( y sec²x ) = y ( 2 sec²x tan x ) + $\frac{\mathrm{dy}}{\mathrm{dx}}$ sec²x

= y ( 2 sec²x tan x ) + y sec²x ^.sec²x [ $\because \frac{\mathrm{dy}}{\mathrm{dx}}$ = y sec²x ]

= y ( 2 tan x + sec²x ) sec²x = e^{tan x} . sec²x ( 2 tan x + secx )

7 If x = a sec²θ, y = a tan³θ, then find $\frac{d^\mathbf{3}y}{\mathrm{d}\mathrm{x}^\mathbf{3}}.$ .

Soln. $\frac{\mathrm{dx}}{d\theta}$ = 2a sec²θ tan & = 3a tan²θ^. sec²θ

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}$ = $\frac{\frac{\mathrm{dy}}{d\theta}}{\frac{\mathrm{dx}}{d\theta}}$ = $\frac{3\mathrm{a\ ta}\mathrm{n}^2\theta{sec}^2{\theta}}{2\mathrm{a\ se}\mathrm{c}^2\theta t a n{\theta}}$ = $\frac{3}{2}tan\theta$

$\therefore \frac{d^2y}{\mathrm{d}\mathrm{x}^2}$ = $\frac{d}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$ = $\frac{d}{\mathrm{dx}}\left(\frac{3}{2}tan{\theta}\right)$ = $\frac{3}{2}sec^{2} \theta. \frac{d\theta}{\mathrm{dx}}$ [ Apply chain rule ]

= $\frac{3}{2}sec^{2} \theta. \frac{1}{2\mathrm{a\ se}\mathrm{c}^2\theta t a n{\theta}}$ = $\frac{3}{4a}cot\theta \ \ \ [\because\frac{\mathrm{dx}}{d\theta} = 2a sec^{2} \theta tan\theta ]$

Further,

$\frac{d^3y}{\mathrm{d}\mathrm{x}^3}$ = $\frac{d}{\mathrm{dx}}\left(\frac{d^2y}{\mathrm{d}\mathrm{x}^2}\right)$ = $\frac{d}{\mathrm{dx}}\left(\frac{3}{4a}cot{\theta}\right)$ = $\frac{3}{4a}. (- cosec^{2} \theta ) \frac{d\theta}{\mathrm{dx}}$

= – $\frac{3}{4a}cosec^{2} \theta. \frac{1}{2\mathrm{a\ se}\mathrm{c}^2\theta t a n{\theta}}$ = $\frac{3}{8a^2}cot^{3} \theta$

8 At which point on the curve $\sqrt x + \sqrt y$ = 4, the tangent is equally inclined with the axes ?

Soln. We have $\sqrt x + \sqrt y$ = 4, Differentiating with respect to x,

$\frac{1}{2\sqrt x} + \frac{1}{2\sqrt y}\frac{\mathrm{dy}}{\mathrm{dx}}$ = 0 ⇒ $\frac{\mathrm{dy}}{\mathrm{dx}}$ = – $\frac{\sqrt y}{\sqrt x}$

Since, the tangent is equally inclined to the axes, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ = ±1

Here, $\frac{\mathrm{dy}}{\mathrm{dx}}$ = ±1 ⇒ – $\frac{\mathrm{dy}}{\mathrm{dx}}$ = ±1 ⇒ y = x

∴We get $\sqrt x + \sqrt y$ = 4 ⇒ $\sqrt x + \sqrt x$ = 4 ⇒ x = 4 and y = 4

∴The point is ( 4, 4 ).

9 Find the locus of all the points on the curve y² = 4a $\left(\mathrm{x\ }+\mathrm{\ a\ sin\ }\frac{x}{a}\right)$ at which the tangent is parallel to x – axis.

Soln. We have y² = 4a $\left(\mathrm{x\ }+\mathrm{\ a\ sin\ }\frac{x}{a}\right)$ ……..(i)

Differentiating w.r. to x, 2y $\frac{\mathrm{dy}}{\mathrm{dx}}$ = 4a $\left(\mathrm{1\ }+\mathrm{\ cos\ }\frac{x}{a}\right)$

Since, the tangent is parallel to x–axis, $\frac{\mathrm{dy}}{\mathrm{dx}}$ = 0

This gives 4a $\left(\mathrm{1\ }+\mathrm{\ cos\ }\frac{x}{a}\right)$ = 0 ⟹ $cos\frac{x}{a}$ = – 1 ⟹ $sin \frac{x}{a}$ = 0

∴ From (i) y² = 4a ( x + 0 ) ⟹ y² = 4ax, which is the required locus.

10 Find the Equation of normal to the curve x + y = x^y, where it cuts x – axis.

Soln. The given curve x + y = x^y …… (i)

It cuts the x–axis, when y = 0 ⟹x = 1

∴ The required point is ( 1, 0 ).

Taking log of (i), we get log ( x + y ) = y log x

Differentiating w.r.t x, ${\frac{1}{x+y}}_{\left[1+\frac{dy}{dx}\right]}$ = $\frac{dy}{dx} log x + \frac{y}{x}$

When, x = 1, y = 0, we get ${\frac{1}{1+0}}_{\left[1+\frac{dy}{dx}\right]}$ = $\frac{dy}{dx} \times 0 +$ $\frac{0}{1}$

∴ $\frac{dy}{dx}$ = – 1. The slop of normal = 1.

Equation of the normal is y – 0 = 1 ( x – 1 ) ⟹ y – x + 1 = 0.

11 Whether the curves x³ – 3xy² = a² and 3x²y – y³ = b³…..are orthogonal or not.

Soln. Differentiating (i), 3x² – 3y² – 6xy $\frac{\mathrm{dy}}{\mathrm{dx}}$ = 0 ⇒ m₁ = $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x^2-y^2}{\mathrm{2xy}}$

(ii), 6xy + 3x² $\frac{\mathrm{dy}}{\mathrm{dx}} - 3y^2\frac{\mathrm{dy}}{\mathrm{dx}}$ = 0 ⇒ m₂ = $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2\mathrm{xy}}{y^2-x^2}$

We note that m₁ m₂ = – 1

∴ The curves (i) and (ii) are orthogonal curves.

12 Find the angle between the curves y = 2 sin² x and y = cos 2x

Soln. Solving (i) and (ii), we get 2sin² x = cos 2x = 1 – 2sin²x

⇒ sin²x = $\frac{1}{4}$ ⇒ sin x = ± $\frac{1}{2}$ ⇒ x = ± $\frac{\pi}{6}$ + nπ, n ∈ I

When x = ± $\frac{\pi}{6}, y = \frac{1}{2}$ . ∴Point of intersection are $\left(\frac{\pi}{6},\frac{1}{2}\right) \left(-\frac{\pi}{6},\frac{1}{2}\right)$

Differentiating (i), $\frac{\mathrm{dy}}{\mathrm{dx}} = 4 sin x \ cos x = 2 sin 2x$

Differentiating (ii), $\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 sin 2x$

At the point $\left(\frac{\pi}{6},1\right)$ , m₁ = 2 sin $\frac{2\pi}{6}$ = $\sqrt3$ and m₂ = – 2 sin $\frac{2\pi}{6} = - \sqrt3$

$\therefore tan \theta = \frac{m_1-m_2}{1+m_1m_2}$ = $\frac{2\sqrt3}{1-3} = \frac{2\sqrt3}{-2}$ = – $\sqrt3 \Rightarrow \theta = \frac{\pi}{3} or \frac{2\pi}{3}$

Similarly,

At the point $\left(-\frac{\pi}{6},1\right)$ , m₁ = – $\sqrt3$ and m₂ = $\sqrt3$

$\therefore \ \theta = \frac{\pi}{3} or \frac{2\pi}{3}.$

13. The time period T of oscillation of a simple pendulum of length l is given by T = $2\pi \sqrt{\frac{l}{g}}$ . Suppose an error in measurement of l is found to be 1 %. Calculate the corresponding error is measurement of T.

Soln. We have, T = $2\pi \sqrt{\frac{l}{g}} \Rightarrow$ log T = $log 2\pi + \frac{1}{2}log l - \frac{1}{2}log\ g$

Differentiating we get, $\frac{1}{T}\frac{\mathrm{dT}}{dl}$ = 0 + $\frac{1}{2l} + 0 \Rightarrow \frac{\mathrm{dT}}{dl} = \frac{T}{2l}$

$\therefore\frac{\mathrm{dT}}{T}$ = $\frac{dl}{2l} \Rightarrow \left(\frac{\mathrm{dT}}{T}\right) \times 100$ = $\frac{1}{2} \left(\frac{dl}{l}\times100\right)$