{"id":626434,"date":"2023-09-11T12:38:05","date_gmt":"2023-09-11T12:38:05","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=626434"},"modified":"2023-09-11T12:39:28","modified_gmt":"2023-09-11T12:39:28","slug":"kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/","title":{"rendered":"KC Sinha: Exercise 11.5- Mathematics Solution Class 12 Chapter 11 \u0905\u0935\u0915\u0932\u0928"},"content":{"rendered":"\n

<\/span><\/p>\n\n\n\n

Question 1<\/h4>\n\n\n\n

$\\frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c
[Find $\\frac{d y}{d x}$ when]<\/strong><\/p>\n\n\n\n

(i)<\/strong> x=acos\u03b8 , y=asin\u03b8
Sol :
Differentiate w.r.t \u03b8<\/p>\n\n\n\n

$\\frac{d x}{d \\theta}=-a \\sin \\theta$..(i)<\/p>\n\n\n\n

$\\frac{d y}{d \\theta}=a \\cos \\theta$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 (ii) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d \\theta}}{\\frac{d x}{d \\theta}}=\\frac{a \\cos \\theta}{- a \\sin \\theta}$<\/p>\n\n\n\n

=-cot\u03b8<\/p>\n\n\n\n

(ii)<\/strong> x=acos\u03b8 , y=bcos\u03b8
Sol :
Differentiate w.r.t \u03b8<\/p>\n\n\n\n

$\\frac{d x}{d \\theta}=-a \\sin \\theta$..(i)<\/p>\n\n\n\n

$\\frac{d y}{d \\theta}=-b \\sin \\theta$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 (ii) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d \\theta}}{\\frac{d x}{d \\theta}}=\\frac{-b \\sin \\theta}{- a \\sin \\theta}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\frac{b}{a}$<\/p>\n\n\n\n

Question 2<\/h4>\n\n\n\n

$\\frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c
[Find $\\frac{d y}{d x}$ when]
<\/strong><\/p>\n\n\n\n

(i)<\/strong> x=at2<\/sup>,y=2at
Sol :
Differentiate w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d x}{d t}=2 a t$..(i)<\/p>\n\n\n\n

$\\frac{d y}{d t}=2 a$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d t}}{\\frac{d x}{d t}}=\\frac{2a}{2at}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\frac{1}{t}$<\/p>\n\n\n\n

(ii)<\/strong> x=4t , $y=\\frac{4}{t}$
Sol :
Differentiate w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d x}{d t}=4$..(i)<\/p>\n\n\n\n

$\\frac{d y}{d t}=-\\frac{4}{t^{2}}$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\dfrac{\\frac{d y}{d t}}{\\frac{dx}{dt}}=\\dfrac{\\frac{-4}{t^2}}{4}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\frac{-1}{t^{2}}$<\/p>\n\n\n\n

Question 3<\/h4>\n\n\n\n

$\\frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c
[Find $\\frac{d y}{d x}$ when]
<\/strong><\/p>\n\n\n\n

(i)<\/strong> x=sint,y=cos2t
Sol :
Differentiate w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d x}{d t}=\\cos \\theta$..(i)<\/p>\n\n\n\n

$\\frac{d y}{d t}=-\\sin 2 t \\times 2$<\/p>\n\n\n\n

$\\frac{dy}{d t}=-2 .\\sin 2t$<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d t}}{\\frac{d x}{d t}}=\\frac{-2 \\cdot \\cos t}{\\cos t}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\frac{-2 \\times 2\\sin t \\cos t}{\\cos t}$<\/p>\n\n\n\n

\u21d2-4sint<\/p>\n\n\n\n

(ii)<\/strong> x=asec\u03b8,y=btan\u03b8
Sol :<\/p>\n\n\n\n

Differentiating w.r.t \u03b8<\/p>\n\n\n\n

$\\frac{d x}{d \\theta}=a \\sec \\theta \\tan \\theta$..(i)<\/p>\n\n\n\n

$\\frac{d y}{d \\theta}=b \\sec ^{2} \\theta$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d \\theta}}{\\frac{d x}{d \\theta}}=\\frac{b \\sec ^{2} \\theta}{a \\sec \\theta \\tan \\theta}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\frac{b}{a}\\times \\frac{1}{\\cos \\theta}$<\/p>\n\n\n\n

$=\\frac{b}{a} \\operatorname{cosec} \\theta$<\/p>\n\n\n\n

Question 4<\/h4>\n\n\n\n

$\\frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c
[Find $\\frac{d y}{d x}$ when]
<\/strong><\/p>\n\n\n\n

(i)<\/strong> x=a(\u03b8-sin\u03b8), y=a(1-cos\u03b8)
Sol :
Differentiating w.r.t \u03b8<\/p>\n\n\n\n

$\\frac{d x}{d \\theta}=a(1-\\cos \\theta)$<\/p>\n\n\n\n

$\\frac{d x}{d \\theta}=a \\cdot 2 \\sin ^{2} \\frac{\\theta}{2}$<\/p>\n\n\n\n

$\\frac{d x}{d \\theta}=2 a \\sin ^{2} \\frac{\\theta}{2}$..(i)<\/p>\n\n\n\n

\u0905\u092c , y=(1-cos\u03b8)<\/p>\n\n\n\n

Differentiating w.r.t \u03b8<\/p>\n\n\n\n

$\\frac{d y}{d \\theta}=a \\sin \\theta=a \\cdot 2 \\sin \\frac{\\theta}{2} \\cos \\frac{\\theta}{2}$<\/p>\n\n\n\n

$\\frac{d y}{d \\theta}=2 a \\sin \\frac{\\theta}{2} \\cos \\frac{\\theta}{2}$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d \\theta}}{\\frac{d x}{d \\theta}}=\\frac{2 a \\sin \\frac{\\theta}{2} \\cos \\frac{\\theta}{2}}{2 a \\sin^2 \\frac{\\theta}{2}}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\cot \\frac{\\theta}{2}$<\/p>\n\n\n\n

(ii)<\/strong> x=a(\u03b8-sin\u03b8),y=a(1+cos\u03b8)
Sol :<\/p>\n\n\n\n

Question 5<\/h4>\n\n\n\n

$\\frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c
[Find $\\frac{d y}{d x}$ when]
<\/strong><\/p>\n\n\n\n

(i)<\/strong> x=sin\u03b8,y=\u03b8+cos\u03b8
Sol :
Differentiating w.r.t \u03b8<\/p>\n\n\n\n

$\\frac{d x}{d \\theta}=\\cos \\theta$..(i)<\/p>\n\n\n\n

$\\frac{d y}{d \\theta}=1-\\sin \\theta$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d \\theta}}{\\frac{dx}{d\\theta}}=\\frac{1-\\sin \\theta}{\\cos \\theta}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\frac{1-\\sin \\theta}{\\cos \\theta}$<\/p>\n\n\n\n

(ii)<\/strong> x=10(t-sint),y=12(1-cost),$-\\frac{\\pi}{2}<t \\leq \\frac{\\pi}{2}$
Sol :
Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d x}{d t}=10(1-\\cos t)$<\/p>\n\n\n\n

$\\frac{d x}{d t}=10 \\times 2 \\sin ^{2} \\frac{t}{2}$<\/p>\n\n\n\n

$\\frac{d x}{d t}=20 \\sin^2 \\frac{t}{2}$<\/p>\n\n\n\n

\u0905\u092c , y=12(1-cost)<\/p>\n\n\n\n

Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d y}{d t}=12 \\sin t=12 \\times 2 \\sin \\frac{t}{2} \\cos \\frac{t}{2}$<\/p>\n\n\n\n

$\\frac{d y}{d t}=24 \\sin \\frac{t}{2} \\cos \\frac{t}{2}$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d t}}{\\frac{d x}{d t}}=\\frac{24 \\sin \\frac{t}{2} \\cos \\frac{t}{2}}{20 \\sin ^{2} \\frac{t}{2}}$<\/p>\n\n\n\n

$\\frac{dy}{d t}=\\frac{6}{5} \\cos \\frac{t}{2}$<\/p>\n\n\n\n

Question 6<\/h4>\n\n\n\n

$\\frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c
[Find $\\frac{d y}{d x}$ when]
<\/strong><\/p>\n\n\n\n

(i)<\/strong> x=3cos\u03b8-cos3<\/sup>\u03b8,y=3sin\u03b8-sin3<\/sup>\u03b8
Sol :
Differentiating w.r.t \u03b8<\/p>\n\n\n\n

$\\frac{d x}{d \\theta}=-3 \\sin \\theta-3 \\cos ^{2} \\theta(-\\sin \\theta)$<\/p>\n\n\n\n

$\\frac{d x}{d \\theta}=-3 \\sin \\theta+3 \\cos ^{2} \\theta \\sin \\theta$<\/p>\n\n\n\n

$\\frac{d x}{d \\theta}=-3 \\sin \\theta\\left(1-\\cos^{2} \\theta\\right)$<\/p>\n\n\n\n

$\\frac{d x}{d t}=-3 \\sin^{3} \\theta$..(i)<\/p>\n\n\n\n

\u0905\u092c, y=3sin\u03b8-sin3<\/sup>\u03b8<\/p>\n\n\n\n

Differentiating w.r.t \u03b8<\/p>\n\n\n\n

$\\frac{d y}{d \\theta}=3 \\cos \\theta-3 \\sin ^{2} \\theta \\cdot \\cos \\theta$<\/p>\n\n\n\n

=3cos\u03b8(1-sin2<\/sup>\u03b8)<\/p>\n\n\n\n

$\\frac{d y}{d \\theta}=3 \\cos ^{3} \\theta$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d \\theta}}{\\frac{d x}{d \\theta}}=\\frac{3 \\cos ^{3} \\theta}{-3 \\sin^3 \\theta}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=-\\cot ^{3} \\theta$<\/p>\n\n\n\n

(ii)<\/strong> x=cos\u03b8-cos2\u03b8,y=sin\u03b8-sin3<\/sup>\u03b8
Sol :
Differentiating w.r.t \u03b8<\/p>\n\n\n\n

$\\frac{d x}{d \\theta}=-\\sin \\theta+\\sin 2 \\theta \\cdot 2$<\/p>\n\n\n\n

$\\frac{d x}{d \\theta}=2 \\sin 2 \\theta-\\sin \\theta$..(i)<\/p>\n\n\n\n

\u0905\u092c , y=sin\u03b8–sin3<\/sup>\u03b8<\/p>\n\n\n\n

Differentiating w.r.t \u03b8<\/p>\n\n\n\n

$\\frac{d y}{d t}=\\cos \\theta-3 \\sin ^{2} \\theta \\cdot \\cos \\theta$<\/p>\n\n\n\n

$\\frac{d y}{d \\theta}=\\cos \\theta\\left(1-3 \\sin ^{2} \\theta\\right)$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d \\theta}}{\\frac{d x}{d \\theta}}=\\frac{\\cos \\theta\\left(1-3 \\sin ^{2} \\theta\\right)}{2\\sin 2\\theta-\\sin \\theta}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\frac{\\cos \\theta\\left(1-3 \\sin ^{2} \\theta\\right)}{2 \\sin 2 \\theta-\\sin \\theta}$<\/p>\n\n\n\n

Question 7<\/h4>\n\n\n\n

$\\frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c
[Find $\\frac{d y}{d x}$ when]
<\/strong><\/p>\n\n\n\n

(i)<\/strong> $x=\\sqrt{1+t^{2}}, y=\\sqrt{1-t^{2}}$ , $x=t+\\frac{1}{t}, y=t-\\frac{1}{t}$
Sol :
$x=\\sqrt{1+t^{2}}$<\/p>\n\n\n\n

Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d x}{d t}=\\frac{1}{2 \\sqrt{1+t^2}} \\times 2t$<\/p>\n\n\n\n

$\\frac{d x}{d t}=\\frac{t}{\\sqrt{1+t^{2}}}$..(i)<\/p>\n\n\n\n

\u0905\u092c, $y=\\sqrt{1-t^{2}}$<\/p>\n\n\n\n

Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d y}{d t}=\\frac{1}{2 \\sqrt{1-t^{2}}} \\times(-2 t)$<\/p>\n\n\n\n

$\\frac{d y}{d t}=\\frac{-t}{\\sqrt{1-t^{2}}}$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d t}}{\\frac{d x}{d t}}=\\frac{\\frac{-t}{\\sqrt{1-t^{2}}}}{\\frac{t}{\\sqrt{1+t^{2}}}}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=-\\frac{\\sqrt{1+t^{2}}}{\\sqrt{1-t^{2}}}$<\/p>\n\n\n\n

(ii)<\/strong> $x=t+\\frac{1}{t}, y=t-\\frac{1}{t}$
Sol :
$x=t+\\frac{1}{t}$<\/p>\n\n\n\n

Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d x}{d t}=1-\\frac{1}{t^{2}}$<\/p>\n\n\n\n

$\\frac{d x}{d t}=\\frac{t^{2}-1}{t^{2}}$..(i)<\/p>\n\n\n\n

\u0905\u092c ,$y=t-\\frac{1}{t}$<\/p>\n\n\n\n

Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{dy}{d t}=1+\\frac{1}{t^{2}}$<\/p>\n\n\n\n

$\\frac{d y}{d t}=\\frac{t^{2}+1}{t^{2}}$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d t}}{\\frac{d x}{d t}}=\\frac{\\frac{t^{2}+1}{t^{2}}}{\\frac{t^{2}-1}{t^{2}}}$<\/p>\n\n\n\n

$\\frac{d y}{d x}-\\frac{t^{2}+1}{t^{2}-1}$<\/p>\n\n\n\n

Question 8<\/h4>\n\n\n\n

$\\frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c
[Find $\\frac{d y}{d x}$ when]
<\/strong><\/p>\n\n\n\n

(i)<\/strong> x=et<\/sup>+sint,t=logt
Sol :
Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d x}{d t}=e^{t}+\\cos t$..(i)<\/p>\n\n\n\n

$\\frac{d y}{d t}=\\frac{1}{t}$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d t}}{\\frac{d x}{d t}}=\\frac{\\frac{1}{t}}{e^{t}+\\cos t}$<\/p>\n\n\n\n

$\\frac{d_{y}}{d x}=\\frac{1}{t\\left(e^{t}+\\cos t\\right)}$<\/p>\n\n\n\n

(ii)<\/strong> $x=a\\left(\\cos t+\\log \\tan \\frac{t}{2}\\right), y=a \\sin t$
Sol :
Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d x}{d t}=a\\left(-\\sin t+\\frac{1}{\\tan \\frac{t}{2}} \\times \\sec ^{2} \\frac{t}{2} \\times \\frac{1}{2}\\right)$<\/p>\n\n\n\n

$\\frac{d x}{d t}=a\\left(-\\sin t+\\frac{1}{\\frac{\\sin t \/ 2}{\\cos t \/ 2}} \\times \\frac{1}{\\cos ^{2} \\frac{t}{2}} \\times \\frac{1}{2}\\right)$<\/p>\n\n\n\n

$\\frac{d x}{d t}=a\\left(-\\sin t+\\frac{1}{2 \\sin \\frac{t}{2} \\cos \\frac{t}{2}}\\right)$<\/p>\n\n\n\n

$\\frac{d x}{d t}=a\\left(-\\sin t+\\frac{1}{\\sin t}\\right)$<\/p>\n\n\n\n

$\\frac{d x}{d t}=a\\left(\\frac{-\\sin ^{2} t+1}{\\sin t}\\right)$<\/p>\n\n\n\n

$\\frac{d x}{d t}=a \\frac{\\cos ^{2} t}{\\sin t}$..(i)<\/p>\n\n\n\n

\u0905\u092c,y=asint<\/p>\n\n\n\n

Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d y}{d t}=a \\cos t$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d t}}{\\frac{d x}{d t}}=\\frac{a \\cos t}{\\frac{a \\cos^{2} t}{sin t}}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\tan t$<\/p>\n\n\n\n

Question 9<\/h4>\n\n\n\n

$\\frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c
[Find $\\frac{d y}{d x}$ when]
<\/strong><\/p>\n\n\n\n

$x=\\frac{2 t}{1+t^{2}}, y=\\frac{1-t^{2}}{1+t^{2}}$<\/p>\n\n\n\n

Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d x}{d t}=\\frac{2 \\cdot\\left(1+t^{2}\\right)-2 t(2 t)}{\\left(1+t^{2}\\right)^{2}}$<\/p>\n\n\n\n

$\\frac{d x}{d t}=\\frac{2+2 t^{2}-4 t^{2}}{\\left(1+t^{2}\\right)^{2}}$<\/p>\n\n\n\n

$\\frac{d x}{d t}=\\frac{2-2 t^{2}}{\\left(1+t^{2}\\right)^{2}}$<\/p>\n\n\n\n

$\\frac{d x}{d t}=\\frac{2\\left(1-t^{2}\\right)}{\\left(1+t^{2}\\right)^{2}}$..(i)<\/p>\n\n\n\n

\u0905\u092c, $y=\\frac{1-t^{2}}{1+t^{2}}$<\/p>\n\n\n\n

Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d y}{d t}=\\frac{-2 t\\left(1+t^{2}\\right)-\\left(1-t^{2}\\right) \\cdot 2 t}{\\left(1+t^{2}\\right)^{2}}$<\/p>\n\n\n\n

$=\\frac{-2 t-2 t^{3}-2 t+2 t^{3}}{\\left(1+t^{2}\\right)^{2}}$<\/p>\n\n\n\n

$\\frac{d y}{d t}=-\\frac{4 t}{\\left(1+t^{2}\\right)^{2}}$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d t}}{\\frac{d x}{d t}}=\\frac{\\frac{-4 t}{\\left(1+t^{2}\\right)^{2}}}{\\frac{2\\left(1-t^{2}\\right)}{\\left(1+t^{2}\\right)^{2}}}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\frac{-2 t}{1-t^{2}}$<\/p>\n\n\n\n

Question 10<\/h4>\n\n\n\n

\u092f\u0926\u093f(If) $x=\\frac{1-t^{2}}{1+t^{2}}, y=\\frac{2 t}{1+t^{2}}$ , \u0938\u093f\u0926\u094d\u0927 \u0915\u0930\u0947 \u0915\u093f (prove that) $\\frac{d y}{d x}+\\frac{x}{y}=0$
<\/strong> Sol :
$x=\\frac{1-t^{2}}{1+t^{2}}$<\/p>\n\n\n\n

Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d x}{d t}=\\frac{-2 t\\left(1+t^{2}\\right)-\\left(1-t^{2}\\right) \\cdot 2t}{\\left(1+t^{2}\\right)^{2}}$<\/p>\n\n\n\n

$=\\frac{-2 t-2 t^{3}-2 t+2t^{3}}{\\left(1+t^{2}\\right)^{2}}$<\/p>\n\n\n\n

$\\frac{d x}{d t}=\\frac{-4 t}{\\left(1+t^{2}\\right)^{2}}$..(i)<\/p>\n\n\n\n

\u0905\u092c , $y=\\frac{2 t}{1+t^{2}}$<\/p>\n\n\n\n

Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d y}{d t}=\\frac{2\\left(1+t^{2}\\right)-2 t \\cdot 2 t}{\\left(1+t^{2}\\right)^{2}}$<\/p>\n\n\n\n

$=\\frac{2+2 t^{2}-4 t^{2}}{\\left(1+t^{2}\\right)^{2}}$<\/p>\n\n\n\n

$\\frac{d y}{d t}=\\frac{2-2 t^{2}}{\\left(1+t^{2}\\right)^{2}}$<\/p>\n\n\n\n

$\\frac{d y}{d t}=\\frac{2\\left(1-t^{2}\\right)}{\\left(1+t^{2}\\right)^{2}}$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d{y}}{d t}}{\\frac{dx}{d t}}=\\dfrac{\\frac{2(1-t^2)}{(1+t^2)^2}}{\\frac{-4t}{(1+t^2)^2}}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=-\\frac{1-t^{2}}{2 t}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=-\\frac{x}{y}$<\/p>\n\n\n\n

$\\frac{dy}{d x}+\\frac{x}{y}=0$<\/p>\n\n\n\n

Question 11<\/h4>\n\n\n\n

\u092f\u0926\u093f(If) y=acos2<\/sup>\u03b8,x=bsin2<\/sup>\u03b8 , \u0938\u093f\u0926\u094d\u0927 \u0915\u0930\u0947 \u0915\u093f (prove that) $\\frac{d y}{d x}+\\frac{a}{b}=0$
<\/strong> Sol :
y=acos2<\/sup>\u03b8<\/p>\n\n\n\n

Differentiating w.r.t \u03b8<\/p>\n\n\n\n

$\\frac{d y}{d \\theta}=2 a \\cos \\theta \\cdot(-\\sin \\theta)$<\/p>\n\n\n\n

$\\frac{d y}{d \\theta}=-a \\sin 2 \\theta$..(i)<\/p>\n\n\n\n

\u0905\u092c , x=bsin2<\/sup>\u03b8<\/p>\n\n\n\n

Differentiating w.r.t \u03b8<\/p>\n\n\n\n

$\\frac{d x}{d \\theta}=2b \\sin \\theta \\cdot \\cos \\theta$<\/p>\n\n\n\n

$\\frac{d x}{d \\theta}=b\\sin2 \\theta$<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 (ii) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d\\theta}}{\\frac{d x}{d\\theta}}=\\frac{-a \\sin 2\\theta}{b \\sin 2 \\theta}$<\/p>\n\n\n\n

$\\frac{dy}{dx}+\\frac{a}{b}=0$<\/p>\n\n\n\n

Question 12<\/h4>\n\n\n\n

$t=\\frac{\\pi}{3}$ \u092a\u0930 $\\frac{d y}{d x}$ \u0928\u093f\u0915\u0932\u0947 \u091c\u092c (Find $\\frac{d y}{d x}$ at $t=\\frac{\\pi}{3}$ when) x=a(cost+tsint) and y=a(sint-tcost)
<\/strong> Sol :
x=a(cost+tsint)<\/p>\n\n\n\n

Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d x}{d t}=a(-\\sin t+1.\\sin t+t \\cos t)$<\/p>\n\n\n\n

$\\frac{d x}{d t}=a t \\cos t$..(i)<\/p>\n\n\n\n

\u0905\u092c,<\/p>\n\n\n\n

y=a(sint-tcost)<\/p>\n\n\n\n

Differentiating w.r.t t<\/em><\/p>\n\n\n\n

$\\frac{d y}{d t}=a[\\cos t-1 \\cdot \\cos t-t(-\\sin t)]$<\/p>\n\n\n\n

$\\frac{d y}{d t}=a t \\sin t$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{dy}{d t}}{\\frac{d x}{dt}}=\\frac{at\\sin t}{a t \\cos t}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\tan t$<\/p>\n\n\n\n

At $t=\\frac{\\pi}{3}$<\/p>\n\n\n\n

$\\left(\\frac{d y}{d x}\\right)_{t=\\frac{\\pi}{3}}=\\tan \\frac{\\pi}{3}=\\sqrt{3}$<\/p>\n\n\n\n

Question 13<\/h4>\n\n\n\n

sin2<\/sup>x \u0915\u093e ecosx<\/sup> \u0915\u0947 \u0938\u093e\u092a\u0947\u0915\u094d\u0937 \u0905\u0935\u0915\u0932\u091c \u0928\u093f\u0915\u093e\u0932\u0947 \u0964
[Find the derivative of sin2<\/sup>x w.r.t ecosx<\/sup> ]
<\/strong> Sol :
\u092e\u093e\u0928\u093e u=sin2<\/sup>x , v=ecosx<\/sup><\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{d u}{d x}=2 \\sin x \\cos x$..(i)<\/p>\n\n\n\n

$\\frac{d v}{d x}=e^{\\cos x}(-\\sin x)$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 (ii) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d u}{d x}}{\\frac{d v}{d x}}=\\frac{2 \\sin x \\cos x}{e^{\\cos x}(-\\sin x)}$<\/p>\n\n\n\n

$\\frac{d u}{d v}=-\\frac{2 \\cos x}{e^{\\cos x}}$<\/p>\n\n\n\n

Question 14<\/h4>\n\n\n\n

tan x \u0915\u093e cot x \u0915\u0947 \u0938\u093e\u092a\u0947\u0915\u094d\u0937 \u0905\u0935\u0915\u0932\u091c \u0928\u093f\u0915\u093e\u0932\u0947 \u0964
[Find the derivative of  tan x w.r.t cot x]
<\/strong> Sol :
\u092e\u093e\u0928\u093e u=tanc, v=cotx<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{du}{d{x}}=\\sec ^{2} x$..(i)<\/p>\n\n\n\n

$\\frac{d v}{d x}=-\\text{cosec}^2x$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 (ii) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{du}{dx}}{\\frac{d v}{d x}}=\\frac{\\sec ^{2} x}{-cosec^2x}$<\/p>\n\n\n\n

$\\frac{d u}{d v}=-\\tan ^{2} x$<\/p>\n\n\n\n

Question 15<\/h4>\n\n\n\n

cosx \u0915\u093e x3<\/sup> \u0915\u0947 \u0938\u093e\u092a\u0947\u0915\u094d\u0937 \u0905\u0935\u0915\u0932\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0902 \u0964
[Find the derivative of cosx w.r.t x3<\/sup>]
<\/strong> Sol :
\u092e\u093e\u0928\u093e u=cosx , v=x3<\/sup><\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{d u}{d x}=-\\sin x$..(i)<\/p>\n\n\n\n

$\\frac{d v}{d x}=3 x^{2}$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d u}{d x}}{\\frac{d v}{d x}}=\\frac{-\\sin x}{3 x^{2}}$<\/p>\n\n\n\n

$\\frac{d u}{d v}=-\\frac{\\sin x}{3 x^{2}}$<\/p>\n\n\n\n

Question 16<\/h4>\n\n\n\n

$\\frac{x}{\\sin x}$ \u0915\u093e sinx \u0915\u0947 \u0938\u093e\u092a\u0947\u0915\u094d\u0937 \u0905\u0935\u0915\u0932\u0928 \u0917\u0941\u0923\u093e\u0902\u0915 \u0928\u093f\u0915\u093e\u0932\u0947 \u0964<\/strong><\/p>\n\n\n\n

[Find the derivative of <\/strong>$\\frac{x}{\\sin x}$ w.r.t sin x<\/strong>]<\/strong><\/p>\n\n\n\n

Sol :
\u092e\u093e\u0928\u093e $u=\\frac{x}{\\sin x}$ , v=sinx<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{d u}{d x}=\\frac{1 \\sin x-x \\cdot \\cos x}{\\sin ^{2} x}$<\/p>\n\n\n\n

$\\frac{d u}{d x}=\\frac{\\sin x-x \\cos x}{\\sin ^{2} x}$..(i)<\/p>\n\n\n\n

v=sinx<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{d v}{d x}=\\cos x$..(i)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 (ii) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d y}{d x}}{\\frac{dv}{dx}}=\\frac{\\frac{\\sin x-x \\cos x}{\\sin ^{2} x}}{\\cos x}$<\/p>\n\n\n\n

$\\frac{du}{d v}=\\frac{\\sin x-x \\cos x}{\\sin ^{2} x \\cos x}$<\/p>\n\n\n\n

Question 17<\/h4>\n\n\n\n

$\\sqrt{\\sin \\left(1+x^{2}\\right)^{2}}$ \u0915\u093e 1+x2<\/sup> \u0915\u0947 \u0938\u093e\u092a\u0947\u0915\u094d\u0937 \u0905\u0935\u0915\u0932\u0928 \u0917\u0941\u0923\u093e\u0902\u0915 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964<\/strong><\/p>\n\n\n\n

[Find the d.c. of $<\/strong>\\sqrt{\\sin \\left(1+x^{2}\\right)^{2}}<\/strong>$ w.r.t $1+x^2$]<\/strong><\/p>\n\n\n\n

Sol :
\u092e\u093e\u0928\u093e $u=\\sqrt{\\sin \\left(1+x^{2}\\right)^{2}}$ ,v=1+x2<\/sup><\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{d u}{d x}=\\frac{1}{2 \\sqrt{\\sin \\left(1+x^{2}\\right)^{2}}} \\times \\cos \\left(1+x^{2}\\right)^{2} \\times 2\\left(1+x^{2}\\right) \\cdot 2 x$<\/p>\n\n\n\n

$\\frac{d u}{d x}=\\frac{2 x\\left(1+x^{2}\\right) \\cdot \\cos \\left(1+x^{2}\\right)^{2}}{\\sqrt{\\sin \\left(1+x^{2}\\right)^{2}}}$..(i)<\/p>\n\n\n\n

\u0905\u092c,<\/p>\n\n\n\n

v=1+x2<\/sup><\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{d v}{d x}=2 x$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 (ii) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d u}{d v}}{\\frac{dv}{dx}}=\\frac{\\frac{2x\\left(1+x^{2}\\right) \\cos \\left(1+x^{2}\\right)^{2}}{\\sqrt{\\sin \\left(1+x^{2}\\right)^{2}}}}{2x}$<\/p>\n\n\n\n

$\\frac{d u}{d v}=\\frac{\\left(1+t^{2}\\right) \\cdot \\cos \\left(1+x^{2}\\right)^{2}}{\\sqrt{\\sin \\left(1+x^{2}\\right)^{2}}}$<\/p>\n\n\n\n

Question 18<\/h4>\n\n\n\n

$\\tan ^{-1} \\frac{2 x}{1-x^{2}}$ \u0915\u093e $\\cos ^{-1} \\frac{1-x^{2}}{1+x^{2}}$ \u0915\u0947 \u0938\u093e\u092a\u0947\u0915\u094d\u0937 \u0905\u0935\u0915\u0932\u0928 \u0917\u0941\u0923\u093e\u0902\u0915 \u0928\u093f\u0915\u093e\u0932\u0947\u0902 \u091c\u0939\u093e\u0901 |x|<1<\/strong><\/p>\n\n\n\n

[Find the d.c. of $\\tan ^{-1} \\frac{2 x}{1-x^{2}}$ w.r.t $\\cos ^{-1} \\frac{1-x^{2}}{1+x^{2}}$ where |x|<1]<\/strong><\/p>\n\n\n\n

Sol :
\u092e\u093e\u0928\u093e $u=\\tan ^{-1} \\frac{2 x}{1-x^{2}}$ $v=\\cos ^{-1} \\frac{1-x^{2}}{1+x^{2}}$<\/p>\n\n\n\n

u=2tan-1<\/sup>x,v=2tan-1<\/sup>x<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{d u}{d x}=\\frac{2}{1+x^{2}}$..(i)<\/p>\n\n\n\n

$\\frac{d v}{dx}=\\frac{2}{1+t^{2}}$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 (ii) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d u}{d x}}{\\frac{d v}{dx}}=\\frac{\\frac{2}{1+x^{2}}}{\\frac{2}{1+x^{2}}}$<\/p>\n\n\n\n

$\\frac{d u}{d v}=1$<\/p>\n\n\n\n

Question 19<\/h4>\n\n\n\n

$\\tan ^{-1} \\sqrt{\\frac{1-x^{2}}{1+x^{2}}}$ \u0915\u093e cos-1<\/sup>x2<\/sup> \u0915\u0947 \u0938\u093e\u092a\u0947\u0915\u094d\u0937 \u0905\u0935\u0915\u0932\u0928 \u0917\u0941\u0923\u093e\u0902\u0915 \u0928\u093f\u0915\u093e\u0932\u0947\u0902 \u0964<\/strong><\/p>\n\n\n\n

[Find the d.c. of $\\tan ^{-1} \\sqrt{\\frac{1-x^{2}}{1+x^{2}}}$ w.r.t cos-1<\/sup>x2<\/sup>]<\/strong><\/p>\n\n\n\n

Sol :
\u092e\u093e\u0928\u093e $u=\\tan ^{-1} \\sqrt{\\frac{1-x^{2}}{1+x^{2}}}$ , v=cos-1<\/sup>x2<\/sup><\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\frac{1}{1+\\left(\\sqrt{\\frac{1-x^{2}}{1+x^2}}\\right)^{2}} \\times \\frac{1}{2\\sqrt{\\frac{1-x^2}{1+x^2}}}\\times \\frac{-2x(1+x^2)-(1-x^2)2x}{(1+x^2)^2}$<\/p>\n\n\n\n

$=\\frac{1}{\\frac{1+x^{2}+1-x^{2}}{1+x^{2}}} \\times \\frac{1}{2} \\sqrt{\\frac{1+x^{2}}{1-x^{2}}} \\times \\frac{-2 x-2 x^{3}-2 x+2 x^{3}}{\\left(1+x^{2}\\right)^{2}}$<\/p>\n\n\n\n

$=\\frac{1+x^{2}}{2} \\times \\frac{1}{2} \\sqrt{\\frac{1+x^{2}}{1-x^{2}}} \\times \\frac{-4 x}{\\left(1+x^{2}\\right)^{2}}$<\/p>\n\n\n\n

$\\frac{d u}{d x}=\\frac{-x}{{1+x^{2}}{}} \\times \\sqrt{\\frac{1+x^{2}}{1-x^{2}}}$<\/p>\n\n\n\n

$=\\frac{-x}{\\sqrt{1^{2}-\\left(x^{2}\\right)^{2}}}$<\/p>\n\n\n\n

$\\frac{d u}{d x}=\\frac{-x}{\\sqrt{1-x^{4}}}$..(i)<\/p>\n\n\n\n

\u0905\u092c, v=cos-1<\/sup>x2<\/sup><\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{d u}{d x}=\\frac{-1}{\\sqrt{1-\\left(x^{2}\\right)^{2}}} \\times 2 x$<\/p>\n\n\n\n

$\\frac{d v}{d x}=\\frac{-2 x}{\\sqrt{1-x^{4}}}$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 (ii) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d u}{d x}}{\\frac{d v}{d x}}=\\frac{\\frac{-x}{\\sqrt{1-x^4}}}{\\frac{-2 x}{\\sqrt{1-x^{4}}}}$<\/p>\n\n\n\n

$\\frac{d y}{d v}=\\frac{1}{2}$<\/p>\n\n\n\n

Question 20<\/h4>\n\n\n\n

$\\tan ^{-1} \\frac{2 x}{1-x^{2}}$ \u0915\u093e $\\sin ^{-1} \\frac{2 x}{1+x^{2}}$ \u0915\u0947 \u0938\u093e\u092a\u0947\u0915\u094d\u0937 \u0905\u0935\u0915\u0932\u0928 \u0917\u0941\u0923\u093e\u0902\u0915 \u0928\u093f\u0915\u093e\u0932\u0947\u0902 \u0964<\/strong><\/p>\n\n\n\n

[Find the d.c. of $\\tan ^{-1} \\frac{2 x}{1-x^{2}}$ w.r.t $\\sin ^{-1} \\frac{2 x}{1+x^{2}}$ ]
<\/strong> Sol :
\u092e\u093e\u0928\u093e $\\tan ^{-1} \\frac{2 x}{1-x^{2}}$ , $\\sin ^{-1} \\frac{2 x}{1+x^{2}}$<\/p>\n\n\n\n

u=2tan-1<\/sup>x,v=2tan-1<\/sup>x<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{d u}{d x}=\\frac{2}{1+x^{2}}$..(i)<\/p>\n\n\n\n

$\\frac{d u}{d x}=\\frac{2}{1+x^{2}}$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 (ii) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{d u}{d x}}{\\frac{d v}{d x}}=\\frac{\\frac{2}{1+x^{2}}}{\\frac{2}{1+x^{2}}}$<\/p>\n\n\n\n

$\\frac{d u}{d v}=1$<\/p>\n\n\n\n

Question 21<\/h4>\n\n\n\n

\u092f\u0926\u093f $x=\\frac{\\sin ^{3} t}{\\sqrt{\\cos 2 t}}, y=\\frac{\\cos ^{3} t}{\\sqrt{\\cos 2 t}}$ , \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f $t=\\frac{\\pi}{6}$ \u092a\u0930 $\\frac{d y}{d x}=0$<\/strong><\/p>\n\n\n\n

[If <\/strong>$x=\\frac{\\sin ^{3} t}{\\sqrt{\\cos 2 t}}, y=\\frac{\\cos ^{3} t}{\\sqrt{\\cos 2 t}}$ show that $\\frac{dy}{dx}=0$ at $t=\\frac{\\pi}{6}$<\/strong>]<\/strong><\/p>\n\n\n\n

Sol :
Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{d x}{d t}=\\frac{\\frac{3 \\sin ^{2} t \\cdot \\cos t \\sqrt{\\cos 2 t}}{1}-\\sin ^{3} t \\cdot \\frac{1}{2 \\sqrt{\\cos ^{2} t}} \\times(-\\sin 2 t) \\cdot 2}{(\\sqrt{\\cos ^{2} t})^{2}}$<\/p>\n\n\n\n

$=\\frac{\\frac{3 \\sin^{2}+\\cos t-\\cos^2 t+\\sin^{3}+\\sin 2 t}{\\sqrt{\\cos^ 2 t}}}{\\frac{\\cos 2 t}{1}}$<\/p>\n\n\n\n

$\\frac{d x}{d t}=\\frac{3 \\sin ^{2} t \\cos t\\left(1-2 \\sin ^{2} t\\right)+\\sin ^{3} t \\cdot 2 \\sin t \\cos \\theta}{\\cos ^{2} t \\sqrt{\\cos 2 t}}$<\/p>\n\n\n\n

$\\frac{d x}{d t}=\\frac{3 \\sin ^{2}t\\cos t-6 \\sin ^{2} t \\cos t+2 \\sin ^{4} t \\cos t}{\\cos 2 t \\sqrt{\\cos 2 t}}$<\/p>\n\n\n\n

$\\frac{d x}{d t}=\\frac{3 \\sin ^{2} t \\cos t-4 \\sin ^{4} t \\cos t}{\\cos 2 t \\sqrt{\\cos 2 t}}$<\/p>\n\n\n\n

$\\frac{d{x}}{d t}=\\frac{\\sin t\\cos t\\left(3 \\sin t-4\\sin^{3} t\\right)}{\\cos 2+\\sqrt{\\cos 2t}}$<\/p>\n\n\n\n

$\\frac{d x}{d t}=\\frac{\\sin t \\cos t\\sin 3t}{\\cos 2 t \\sqrt{\\cos 2 t}}$..(i)<\/p>\n\n\n\n

\u0905\u092c,<\/p>\n\n\n\n

$y=\\frac{\\cos 3 t}{\\sqrt{\\cos 2 t}}$<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{d y}{d t}=\\frac{3 \\cos ^{2} t \\cdot(-\\sin t) \\cdot \\sqrt{\\cos 2t}-\\cos ^{3} t \\cdot \\frac{1}{2 \\sqrt{\\cos 2 t}} \\times\\left(-\\sin 2 t\\right).2}{(\\sqrt{\\cos 2t})^{2}}$<\/p>\n\n\n\n

$=\\frac{\\frac{-3 \\cos ^{2} t \\sin t \\cdot \\cos 2 t+\\cos ^{3} t \\sin 2 t}{\\sqrt{\\cos 2 t}}}{\\cos 2 t}$<\/p>\n\n\n\n

$\\frac{d y}{d t}=\\frac{-3 \\cos ^{2} t \\sin t-\\left(2 \\cos ^{2} t-1\\right)+\\cos ^{3} t \\cdot 2 \\sin t \\cos t}{\\cos 2 t \\sqrt{\\cos 2 t}}$<\/p>\n\n\n\n

$\\frac{d y}{d t}=\\frac{-6 \\cos ^{4} t \\sin t+3 \\cos ^{2} t \\sin t+2 \\cos ^{4} t \\operatorname{sin} t}{\\cos 2t \\sqrt{\\cos 2t}}$<\/p>\n\n\n\n

$\\frac{d{y}}{d t}=\\frac{-4 \\cos ^{4} t {\\sin t}+3 \\cos^{2} t{\\sin } t}{\\cos 2 t \\sqrt{\\cos 2t}}$<\/p>\n\n\n\n

$\\frac{d y}{d t}=\\frac{-\\sin t \\cos \\left(4 \\cos ^{3} t-3 \\cos t\\right)}{\\cos 2 t \\sqrt{\\cos 2t}}$<\/p>\n\n\n\n

$\\frac{d y}{d t}=-\\frac{\\sin t \\cos t \\cdot \\cos 3 t}{\\cos 2 t \\sqrt{\\cos 2t}}$..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 (i) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

$\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}=\\frac{\\frac{-\\sin t \\cos t \\cdot \\cos 3 t}{\\cos 2t \\sqrt{\\cos 2t}}}{\\frac{\\sin t+\\cos t \\cdot \\sin 3 t}{\\cos {2} t \\sqrt{\\cos 2 t}}}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=-\\cot 3 t$<\/p>\n\n\n\n

$t=\\frac{\\pi}{6}$ \u092a\u0930 ,<\/p>\n\n\n\n

$\\left(\\frac{d y}{d x}\\right)_{t=\\frac{\\pi}{6}}=-\\cot 3 \\left( \\frac{\\pi}{6}\\right)$<\/p>\n\n\n\n

$=-\\cot \\frac{\\pi}{2}$<\/p>\n\n\n\n

=0<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 12 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question 1 $\\frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c[Find $\\frac{d y}{d x}$ when] (i) x=acos\u03b8 , y=asin\u03b8Sol :Differentiate w.r.t \u03b8 $\\frac{d x}{d \\theta}=-a \\sin \\theta$..(i) $\\frac{d y}{d \\theta}=a \\cos \\theta$..(ii) \u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 (ii) \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 , $\\frac{\\frac{d y}{d \\theta}}{\\frac{d x}{d \\theta}}=\\frac{a \\cos \\theta}{- a \\sin \\theta}$ =-cot\u03b8 (ii) x=acos\u03b8 , y=bcos\u03b8Sol :Differentiate w.r.t \u03b8 $\\frac{d […]<\/p>\n","protected":false},"author":302,"featured_media":626429,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[919],"tags":[],"boards":[],"class_list":["post-626434","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-11","entry"],"yoast_head":"\nKC Sinha: Exercise 11.5- Mathematics Solution Class 12 Chapter 11 \u0905\u0935\u0915\u0932\u0928 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1 $frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c (i) x=acos\u03b8 , y=asin\u03b8Sol :Differentiate w.r.t \u03b8 $frac{d x}{d theta}=-a sin theta$..(i) $frac{d y}{d\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 11.5- Mathematics Solution Class 12 Chapter 11 \u0905\u0935\u0915\u0932\u0928\" \/>\n<meta property=\"og:description\" content=\"Question 1 $frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c (i) x=acos\u03b8 , y=asin\u03b8Sol :Differentiate w.r.t \u03b8 $frac{d x}{d theta}=-a sin theta$..(i) $frac{d y}{d\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2023-09-11T12:38:05+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-09-11T12:39:28+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-38-scaled.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1600\" \/>\n\t<meta property=\"og:image:height\" content=\"901\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"10 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"KC Sinha: Exercise 11.5- Mathematics Solution Class 12 Chapter 11 \u0905\u0935\u0915\u0932\u0928\",\"datePublished\":\"2023-09-11T12:38:05+00:00\",\"dateModified\":\"2023-09-11T12:39:28+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/\"},\"wordCount\":3328,\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-38-scaled.jpg\",\"articleSection\":[\"class 11\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/\",\"name\":\"KC Sinha: Exercise 11.5- Mathematics Solution Class 12 Chapter 11 \u0905\u0935\u0915\u0932\u0928 - IndCareer Schools\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/#primaryimage\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-38-scaled.jpg\",\"datePublished\":\"2023-09-11T12:38:05+00:00\",\"dateModified\":\"2023-09-11T12:39:28+00:00\",\"description\":\"Question 1 $\\\\frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c (i) x=acos\u03b8 , y=asin\u03b8Sol :Differentiate w.r.t \u03b8 $\\\\frac{d x}{d \\\\theta}=-a \\\\sin \\\\theta$..(i) $\\\\frac{d y}{d\",\"breadcrumb\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/#primaryimage\",\"url\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-38-scaled.jpg\",\"contentUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-38-scaled.jpg\",\"width\":1600,\"height\":901,\"caption\":\"KC Sinha: Exercise 11.5- Mathematics Solution Class 12 Chapter 11 \u0905\u0935\u0915\u0932\u0928\"},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/www.indcareer.com\/schools\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"class 11\",\"item\":\"https:\/\/www.indcareer.com\/schools\/class-11\/\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"KC Sinha: Exercise 11.5- Mathematics Solution Class 12 Chapter 11 \u0905\u0935\u0915\u0932\u0928\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#website\",\"url\":\"https:\/\/www.indcareer.com\/schools\/\",\"name\":\"IndCareer Schools\",\"description\":\"School Admissions & Notices\",\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/www.indcareer.com\/schools\/?s={search_term_string}\"},\"query-input\":{\"@type\":\"PropertyValueSpecification\",\"valueRequired\":true,\"valueName\":\"search_term_string\"}}],\"inLanguage\":\"en-US\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\",\"name\":\"IndCareer\",\"url\":\"https:\/\/www.indcareer.com\/schools\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/logo\/image\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/06\/indcareer-logo2.png\",\"contentUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/06\/indcareer-logo2.png\",\"width\":512,\"height\":250,\"caption\":\"IndCareer\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/logo\/image\/\"},\"sameAs\":[\"https:\/\/www.facebook.com\/indcareer\",\"https:\/\/x.com\/indcareer\",\"https:\/\/www.youtube.com\/channel\/UC1liU3RZoBRuu8YcAuZMsOQ\"],\"email\":\"info@ebharat.in\",\"legalName\":\"IndCareer\",\"numberOfEmployees\":{\"@type\":\"QuantitativeValue\",\"minValue\":\"1\",\"maxValue\":\"10\"}},{\"@type\":\"Person\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\",\"name\":\"Pooja\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/350f7cfdfb6a23bcab67b56b5e77549db2a13b5d23e63175ac5bd07b5d44b720?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/350f7cfdfb6a23bcab67b56b5e77549db2a13b5d23e63175ac5bd07b5d44b720?s=96&d=mm&r=g\",\"caption\":\"Pooja\"}}]}<\/script>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"KC Sinha: Exercise 11.5- Mathematics Solution Class 12 Chapter 11 \u0905\u0935\u0915\u0932\u0928 - IndCareer Schools","description":"Question 1 $frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c (i) x=acos\u03b8 , y=asin\u03b8Sol :Differentiate w.r.t \u03b8 $frac{d x}{d theta}=-a sin theta$..(i) $frac{d y}{d","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/","og_locale":"en_US","og_type":"article","og_title":"KC Sinha: Exercise 11.5- Mathematics Solution Class 12 Chapter 11 \u0905\u0935\u0915\u0932\u0928","og_description":"Question 1 $frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c (i) x=acos\u03b8 , y=asin\u03b8Sol :Differentiate w.r.t \u03b8 $frac{d x}{d theta}=-a sin theta$..(i) $frac{d y}{d","og_url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/","og_site_name":"IndCareer Schools","article_publisher":"https:\/\/www.facebook.com\/indcareer","article_published_time":"2023-09-11T12:38:05+00:00","article_modified_time":"2023-09-11T12:39:28+00:00","og_image":[{"width":1600,"height":901,"url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-38-scaled.jpg","type":"image\/jpeg"}],"author":"Pooja","twitter_card":"summary_large_image","twitter_creator":"@indcareer","twitter_site":"@indcareer","twitter_misc":{"Written by":"Pooja","Est. reading time":"10 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/#article","isPartOf":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/"},"author":{"name":"Pooja","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e"},"headline":"KC Sinha: Exercise 11.5- Mathematics Solution Class 12 Chapter 11 \u0905\u0935\u0915\u0932\u0928","datePublished":"2023-09-11T12:38:05+00:00","dateModified":"2023-09-11T12:39:28+00:00","mainEntityOfPage":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/"},"wordCount":3328,"publisher":{"@id":"https:\/\/www.indcareer.com\/schools\/#organization"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/#primaryimage"},"thumbnailUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-38-scaled.jpg","articleSection":["class 11"],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/","url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/","name":"KC Sinha: Exercise 11.5- Mathematics Solution Class 12 Chapter 11 \u0905\u0935\u0915\u0932\u0928 - IndCareer Schools","isPartOf":{"@id":"https:\/\/www.indcareer.com\/schools\/#website"},"primaryImageOfPage":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/#primaryimage"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/#primaryimage"},"thumbnailUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-38-scaled.jpg","datePublished":"2023-09-11T12:38:05+00:00","dateModified":"2023-09-11T12:39:28+00:00","description":"Question 1 $\\frac{d y}{d x}$ \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u092c (i) x=acos\u03b8 , y=asin\u03b8Sol :Differentiate w.r.t \u03b8 $\\frac{d x}{d \\theta}=-a \\sin \\theta$..(i) $\\frac{d y}{d","breadcrumb":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/#primaryimage","url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-38-scaled.jpg","contentUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-38-scaled.jpg","width":1600,"height":901,"caption":"KC Sinha: Exercise 11.5- Mathematics Solution Class 12 Chapter 11 \u0905\u0935\u0915\u0932\u0928"},{"@type":"BreadcrumbList","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-11-5-mathematics-solution-class-12-chapter-11-avakalan\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/www.indcareer.com\/schools\/"},{"@type":"ListItem","position":2,"name":"class 11","item":"https:\/\/www.indcareer.com\/schools\/class-11\/"},{"@type":"ListItem","position":3,"name":"KC Sinha: Exercise 11.5- Mathematics Solution Class 12 Chapter 11 \u0905\u0935\u0915\u0932\u0928"}]},{"@type":"WebSite","@id":"https:\/\/www.indcareer.com\/schools\/#website","url":"https:\/\/www.indcareer.com\/schools\/","name":"IndCareer Schools","description":"School Admissions & Notices","publisher":{"@id":"https:\/\/www.indcareer.com\/schools\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/www.indcareer.com\/schools\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/www.indcareer.com\/schools\/#organization","name":"IndCareer","url":"https:\/\/www.indcareer.com\/schools\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/logo\/image\/","url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/06\/indcareer-logo2.png","contentUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/06\/indcareer-logo2.png","width":512,"height":250,"caption":"IndCareer"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/logo\/image\/"},"sameAs":["https:\/\/www.facebook.com\/indcareer","https:\/\/x.com\/indcareer","https:\/\/www.youtube.com\/channel\/UC1liU3RZoBRuu8YcAuZMsOQ"],"email":"info@ebharat.in","legalName":"IndCareer","numberOfEmployees":{"@type":"QuantitativeValue","minValue":"1","maxValue":"10"}},{"@type":"Person","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e","name":"Pooja","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/350f7cfdfb6a23bcab67b56b5e77549db2a13b5d23e63175ac5bd07b5d44b720?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/350f7cfdfb6a23bcab67b56b5e77549db2a13b5d23e63175ac5bd07b5d44b720?s=96&d=mm&r=g","caption":"Pooja"}}]}},"_links":{"self":[{"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/posts\/626434","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/users\/302"}],"replies":[{"embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/comments?post=626434"}],"version-history":[{"count":0,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/posts\/626434\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/media\/626429"}],"wp:attachment":[{"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/media?parent=626434"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/categories?post=626434"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/tags?post=626434"},{"taxonomy":"boards","embeddable":true,"href":"https:\/\/www.indcareer.com\/schools\/wp-json\/wp\/v2\/boards?post=626434"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}