RD Sharma Class 12 Ex 11.7 Solutions Chapter 11 Differentiation

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter11
Exercise11.7
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 11.7 Solutions Chapter 11 Differentiation

Question 1. Find \frac{dy}{dx} , when: x = at2 and y = 2at

Solution:

Given that x = at2, y = 2at

So,

\frac{dx}{dt}=\frac{d}{dt}(at^2)=2at\\ \frac{dy}{dt}=\frac{d}{dt}(2at)=2a

Therefore,

\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2a}{2at}=\frac{1}{t}

Question 2. Find \frac{dy}{dx} , when: x = a(θ + sinθ) and y = a(1 – cosθ)

Solution:

Here,

x = a(θ + sinθ)

Differentiating it with respect to θ,

\frac{dx}{dθ}=a(1+cosθ)\ \ \ ..........(1)

and,

y = a(1 – cosθ)

Differentiate it with respect to θ,

\frac{dy}{dθ}=a(θ+sinθ)\\ \frac{dy}{dθ}=asinθ\ \ \ \ .....(2)

Using equation (1) and (2),

\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}\\ =\frac{asinθ}{a(1-cosθ)}\\ =\frac{\frac{2sinθ}{2}\frac{cosθ}{2}}{\frac{2sin^2θ}{2}},\ \ \ \ \ \ \ \left\{Since,\ 1-cosθ=\frac{2sin2θ}{2},\frac{2sinθ}{2}\frac{cosθ}{2}=sinθ\right\}\\ =\frac{dy}{dx}=\frac{tanθ}{2}

Question 3. Find \frac{dy}{dx} , when: x = acosθ and y = bsinθ

Solution:

Then x = acosθ and y = bsinθ

Then,

\frac{dx}{dθ}=\frac{d}{dθ}(acosθ)=-asinθ\\ \frac{dy}{dθ}=\frac{d}{dθ}(bsinθ)=bcosθ

Therefore,

\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{bcosθ}{-asinθ}=-\frac{b}{a}cotθ

Question 4. Find \frac{dy}{dx} , when: x = aeΘ (sinθ -cosθ), y = aeΘ (sinθ +cosθ)

Solution:

Here,

x = aeΘ (sinθ – cosθ)

Differentiating it with respect to θ,

\frac{dx}{dθ}=a[e^θ\frac{d}{dθ}(sinθ-cosθ)+(sinθ-cosθ)\frac{d}{dθ}(e^θ)]\\ =a[e^θ(cosθ+sinθ)+(sinθ-cosθ)e^θ]\\ \frac{dx}{dθ}=a[2e^θsinθ]\ \ \ \ \ \ .......(1)

And,

y = aeΘ(sinθ+cosθ)

Differentiating it with respect to θ

\frac{dy}{dθ}=a[e^θ\frac{d}{dθ}(sinθ+cosθ)+(sinθ+cosθ)\frac{d}{dθ}(e^θ)]\\ =a[e^θ(cosθ-sinθ)+(sinθ+cosθ)e^θ]\\ \frac{dx}{dθ}=a[2e^θcosθ]\ \ \ \ \ \ .......(2)

Dividing equation (2) by equation (1)

\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{a(2e^θ cosθ)}{a(2e^θ sinθ)}\\ \frac{dy}{dx}=cotθ

Question 5. Find \frac{dy}{dx} , when: x = bsin2θ and y = acos2θ

Solution:

Here,

x = bsin2θ and y = acos2θ

Then,

\frac{d}{dθ}=\frac{d}{dθ}(bsin^2θ)=2bsinθ cosθ\\ \frac{dy}{dθ}=\frac{d}{dθ}(acos^2θ)=-2acosθ sinθ \\ \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{-2acosθ sinθ}{2bsinθ cosθ}=-\frac{a}{b}\\

Question 6. Find \frac{dy}{dx} , when: x = a(1 – cosθ) and y = a(θ +sinθ) at θ =\frac{\pi}{2}

Solution:

Here,

x = a(1 – cosθ) and y = a(θ + sinθ)

Then,

\frac{dx}{dθ}=\frac{d}{dθ}[a(1-cosθ)]=asinθ\\ \frac{dy}{dθ}=\frac{d}{dθ}[a(θ +sinθ)]=a(1+cosθ)

Therefore,

\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{a(1+cosθ)}{a(sinθ)}|_{θ =\frac{x}{2}}\\ =\frac{a(1+0)}{a}=1

Question 7. Find \frac{dy}{dx} , when: x=\frac{e^t+e^{-t}}{2} andy=\frac{e^t-e^{-t}}{2}

Solution:

Here,

x=\frac{e^t+e^{-t}}{2}

Differentiate it with respect to t,

\frac{dx}{dt}=\frac{1}{2}\left[\frac{d}{dt}(e^t)+\frac{d}{dt}(e^{-t})\right]\\ =\frac{1}{2}\left[e^t+e^{-t}\frac{d}{dt}(e^{-t})\right]\\ \frac{dx}{dt}=\frac{1}{2}(e^t-e^{-t})=y\ \ \ ......(1)

and,

y=\frac{e^t-e^{-t}}{2}

Differentiating it with respect to t,

\frac{dy}{dt}=\frac{1}{2}\left[\frac{d}{dt}(e^t)-\frac{d}{dt}(e^{-t})\right]\\ =\frac{1}{2}\left[e^t-e^{-t}\frac{d}{dt}(e^{-t})\right]\\ \frac{dx}{dt}=\frac{1}{2}(e^t+e^{-t})=x
\frac{dy}{dt}=\frac{1}{2}\left[\frac{d}{dt}(e^t)-\frac{d}{dt}(e^{-t})\right]\\ =\frac{1}{2}\left[e^t-e^{-t}\frac{d}{dt}(e^{-t})\right]\\ \frac{dx}{dt}=\frac{1}{2}(e^t+e^{-t})=x\ \ \ \ .....(2)

Dividing equation (2) and (1)

\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{x}{y}\\ \frac{dy}{dt}=\frac{x}{y}

Question 8. Find \frac{dy}{dx} , when: x=\frac{3at}{1+t^2} andy=\frac{3at^2}{1+t^2}

Solution:

Here,

x=\frac{3at}{1+t^2}

Differentiating it with respect to t using quotient rule,

\frac{dx}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(3at)-3at\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(3a)-3at(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{3a+3at^2-6at^2}{(1+t^2)^2}\right]\\ =\left[\frac{3a-3at^2}{(1-t^2)^2}\right]\\ \frac{dx}{dt}=\frac{3a(1-t^2)}{(1+t^2)^2}\ \ \ \ ....(1)

and,

y=\frac{3at^2}{1+t^2}

Differentiating it with respect to t using quotient rule,

\frac{dy}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(3at^2)-3at^2\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(6at)-(3at^2)(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{6at+6at^3-6at^3}{(1+t^2)^2}\right]\\ \frac{dy}{dt}=\frac{6at}{(1+t^2)^2}\ \ \ \ ....(2)

Dividing equation (2) by (1)

\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{6at}{(1+t^2)^2}\times\frac{(1+t^2)^2}{3a(1-t^2)}\\ \frac{dy}{dt}=\frac{2t}{1-t^2}

Question 9. If x and y are connected parametrically by the equation, without eliminating the parameter, find\frac{dy}{dx} when: x = a(cosθ +θsinθ), y = a(sinθ -θcosθ)

Solution:

The given equations are

x = a(cosθ +θ sinθ) and y = a(sinθ -θcosθ)

Then,

\frac{dx}{dθ}=a\left[\frac{d}{dθ}cosθ +\frac{d}{dθ}(θ sinθ)\right]\\ =a\left[-sinθ +θ \frac{d}{dθ}(sinθ)+sinθ \frac{d}{dθ}(θ)\right]

= a[-sinθ + θcosθ + sinθ] = aθcosθ

\frac{dy}{dθ}=a\left[\frac{d}{dθ}sinθ +\frac{d}{dθ}(θ cosθ)\right]\\ =a\left[cosθ -\{θ \frac{d}{dθ}(cosθ)+cosθ \frac{d}{dθ}(θ)\}\right]

= a[cosθ +θsinθ -cosθ]

= aθsinθ

Therefore,

\frac{dy}{dx}=\frac{\left(\frac{dy}{dθ}\right)}{\left(\frac{dx}{dθ}\right)}=\frac{aθsinθ}{aθ cosθ}=tanθ

Question 10. Find \frac{dy}{dx} , when: x=e^θ \left(θ +\frac{1}{θ}\right) andy=e^{-θ} \left(θ -\frac{1}{θ}\right)

Solution:

Here,

x=e^θ \left(θ +\frac{1}{θ}\right)

Differentiating it with respect to θ using product rule,

\frac{dx}{dθ}=e^θ \frac{d}{dθ}\left(θ +\frac{1}{θ}\right)+\left(θ +\frac{1}{θ}\right)\frac{d}{dθ}(e^θ)\\ =e^θ \left(1-\frac{1}{θ^2}\right)+\left(\frac{θ ^2+1}{θ}\right)e^θ \\ =e^θ \left(\frac{θ ^2-1+θ ^3+θ}{θ ^2}\right)\\ \frac{dx}{dθ}=\frac{e^θ (θ ^3+θ ^2+θ -1)}{θ ^2}\ \ \ .....(1)

and,

y=e^θ \left(θ -\frac{1}{θ}\right)

Differentiating it with respect to θ using product rule and chain rule

\frac{dy}{dθ}=e^{-θ} \frac{d}{dθ}\left(θ -\frac{1}{θ}\right)+\left(θ -\frac{1}{θ}\right)\frac{d}{dθ}(e^{-θ})\\ =e^{-θ} \left(1+\frac{1}{θ^2}\right)+\left(θ -\frac{1}{θ}\right)e^{-θ} \\ =e^{-θ} \left(1+\frac{1}{θ ^2}-θ +\frac{1}{θ}\right)\\ \frac{dy}{dθ}=e^{-θ}\left(\frac{θ ^2+1 -θ^3 +θ}{θ ^2}\right)\\ \frac{dy}{dθ}=e^{-θ}\left(\frac{-θ ^3+θ ^2+θ +1}{θ ^2}\right)\ \ \ \ ......(2)
Question 11. Find\frac{dy}{dx} , when x=\frac{2t}{1+t^2} andy=\frac{1-t^2}{1+t^2}
Solution:

Here,
x=\frac{2t}{1+t^2}
Differentiating it with respect to t using quotient rule,
\frac{dx}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(2t)-2t\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(2)-2t(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{2+2t^2-4t^2}{(1+t^2)^2}\right]\\ =\left[\frac{2+2t^2-4t^2}{(1+t^2)^2}\right]\\ =\left[\frac{2-2t^2}{(1+t^2)^2}\right]\\ \frac{dx}{dt}=\frac{2(1-t^2)}{(1+t^2)^2}\ \ \ \ \ ....(1)
and,
y=\frac{1-t^2}{1+t^2}
Differentiating it with respect to t using quotient rule,
\frac{dy}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(1-t^2)-(1-t^2)\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(-2t)-(1-t^2)(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{-2t-2t^3-2t+2t^3}{(1+t^2)^2}\right]\\ \frac{dy}{dt}=\frac{-4t}{(1+t^2)^2}\ \ \ \ \ ....(2)
Dividing equation (2) by (1)
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-4t}{(1+t^2)^2}\times\frac{(1+t^2)^2}{2(1-t^2)}\\ =\frac{-2t}{1-t^2}\\ \frac{dy}{dx}=-\frac{x}{y}\ \ \ \ \left[Since,\ \frac{x}{y}=\frac{2t}{1+t^2}\times\frac{1+t^2}{1-t^2}=\frac{2t}{1-t^2}\right]
Question 12. Find\frac{dy}{dx} , when x=cos^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right) andy=sin^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)
Solution:
Here,
x=cos^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)
Differentiating it with respect to t using chain rule,
\frac{dx}{dt}=\frac{-1}{\sqrt{1-\left(\frac{1}{1+t^2}\right)}^2}\frac{d}{dt}\left(\frac{1}{\sqrt{1+t^2}}\right)\\ =\frac{-1}{\sqrt{1-\frac{1}{(1+t^2)}}}\left[\frac{-1}{2(1+t^2)^{\frac{3}{2}}}\right]\frac{d}{dt}(1+t^2)\\ =\frac{(1+t^2)^{\frac{1}{2}}}{\sqrt{1+t^2-1}}\times\frac{-1}{2(1+t^2)^{\frac{3}{2}}}(2t)\\ =\frac{-t}{\sqrt{t^2}\times(1+t^2)}\\ \frac{dx}{dt}=\frac{-1}{1+t^2}\ \ \ \ \ ....(1)
Now,
y=sin^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)
Differentiating it with respect to t using chain rule,
\frac{dy}{dt}=\frac{1}{\sqrt{1-\frac{1}{(\sqrt{1+t^2})^2}}}\times\frac{d}{dt}\left(\frac{1}{\sqrt{1+t^2}}\right)\ \ \ \ .....(2)
Dividing equation (2) by (1)
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{1}{(1+t^2)}\times\frac{(1+t^2)}{-1}\\ \frac{dy}{dx}=1
Question 13. Find\frac{dy}{dx} , when x=\frac{1-t^2}{1+t^2} andy=\frac{2t}{1+t^2}
Solution:
Here,
x=\frac{1-t^2}{1+t^2}
Differentiating it with respect to t using quotient rule,
\frac{dx}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(1-t^2)-(1-t^2)\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(-2t)-(1-t^2)(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{-2t-2t^3-2t+2t^3}{(1+t^2)^2}\right]\\ \frac{dx}{dt}=\frac{-4t}{(1+t^2)^2}\ \ \ \ \ ....(1)
and,
y=\frac{2t}{1+t^2}
Differentiating it with respect to t using quotient rule,
\frac{dy}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(2t)-2t\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(2)-2t(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{2+2t^2-4t^2}{(1+t^2)^2}\right]\\ \frac{dx}{dt}=\frac{2(1-t^2)}{(1+t^2)^2}\ \ \ \ \ ....(2)
Question 14. If x = 2cosθ – cos2θ and y = 2sinθ – sin2θ, prove that\frac{dy}{dx}=tan\left(\frac{3\theta}{2}\right)
Solution:
Here,
x = 2cosθ – cos2θ
Differentiating it with respect to θ using chain rule,
\frac{dx}{dθ}=2(-sinθ )-(-sin2θ)\frac{d}{dθ} (2θ )\\ =-2sinθ +2sin2θ \\ \frac{dx}{dθ}=2(sin2θ -sinθ )\ \ \ \ \ ....(1)
and,
y = 2sinθ – sin2θ
Differentiating it with respect to θ using chain rule,
\frac{dy}{dθ}=2cosθ -cos2θ \frac{d}{dθ}(2θ )\\ =2cosθ -cos2θ (2)\\ =2cosθ -2cos2θ \\ \frac{dy}{dθ}=2(cosθ -cos2θ )\ \ \ \ \ ....(2)
Dividing equation (2) by equation (1),
\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{2(cosθ -cos2θ )}{2(sin2θ - sinθ )}\\ =\frac{cosθ -cos2θ}{sin2θ -sinθ}\\ \frac{dy}{dx}=\frac{-2sin\left(\frac{θ +2θ}{2}\right)sin\left(\frac{θ -2θ}{2}\right)}{2cos\left(\frac{2θ +θ}{2}\right)sin\left(\frac{2θ -θ}{2}\right)}\ \ \ \ \left[Since,\ sinA-sinB=2cos\left(\frac{A+B}{2}\right)sin\left(\frac{A-B}{2}\right),\ \ cosA-cosB=-2sin\left(\frac{A+B}{2}\right)sin\left(\frac{A-B}{2}\right)\right]\\ =\frac{-sin\left(\frac{3θ )}{2}\right)\left(sin\left(\frac{-θ}{2}\right)\right)}{cos\left(\frac{3θ}{2}\right)sin\left(\frac{θ}{2}\right)}\\ =\frac{sin\left(\frac{3θ}{2}\right)}{cos\left(\frac{3θ}{2}\right)}\\ \frac{dy}{dx}=tan\left(\frac{3θ}{2}\right)
Question 15. If x = ecos2t and y = esin2t prove that,\frac{dy}{dx}=\frac{y\ log\ x}{x\ log\ y}
Solution:
Here,
x = ecos2t
Differentiating it with respect to t using chain rule,
\frac{dx}{dt}=\frac{d}{dt}(e^{cos2t})\\ =e^{cos2t}\frac{d}{dt}(cos2t)\\ =e^{cos2t}(-sin2t)\frac{d}{dt}(2t)\\ \frac{dx}{dt}=-sin2te^{cos2t}\ \ \ \ \ .....(1)
and,
y = esin2t
Differentiating it with respect to t using chain rule,
\frac{dy}{dt}=\frac{d}{dt}(e^{sin2t})\\ =e^{sin2t}\frac{d}{dt}(sin2t)\\ =e^{sin2t}(cos2t)\frac{d}{dt}(2t)\\ =e^{sin2t}(cos2t)(2)\\ \frac{dy}{dt}=2cos2te^{sin2t}\ \ \ \ \ .....(2)
Dividing equation (2) by (1)
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2cos2te^{sin2t}}{-2sin2te^{cos2t}}\\ \frac{dy}{dx}=-\frac{ylogx}{xlogy}
\left[Since,\ x=e^{cos2t}\Rightarrow logx=cos2t\ \ y=e^{sin2t}\Rightarrow logy=sin2t\right]
Question 16. If x = cos t and y = sin t, prove that\frac{dy}{dx}=\frac{1}{\sqrt3}\ at\ t=\frac{2x}{3}
Solution:
Here,
x = cos t
Differentiating it with respect to t,
\frac{dx}{dt}=\frac{d}{dt}(cos\ t)\\ \frac{dx}{dt}=-sin\ t\ \ \ \ \ ....(1)
and,
y = sin t
Differentiating it with respect to t,
\frac{dx}{dt}=\frac{d}{dt}(sin\ t)\\ \frac{dx}{dt}=cos\ t\ \ \ \ \ ....(2)
Dividing equation (2) by (1),
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{cos\ t}{-sin\ t}\\ \frac{dy}{dx}=-cot\ t\\ \left(\frac{dy}{dx}\right)=-cot\left(\frac{2\pi}{3}\right)\\ =-cot\left(\pi-\frac{\pi}{3}\right)\\ =-\left[-cot\left(\frac{\pi}{3}\right)\right]\\ =cot\left(\frac{\pi}{3}\right)\\ \frac{dy}{dx}=\frac{1}{\sqrt3}
Question 17. Ifx=a\left(t+\frac{1}{t}\right) and y=a\left(t-\frac{1}{t}\right) , Prove that\frac{dy}{dx}=\frac{x}{y}
Solution:
Here,
x=a\left(t+\frac{1}{t}\right)
Differentiating it with respect to t,
\frac{dx}{dt}=a\frac{d}{dt}\left(t+\frac{1}{t}\right)\\ =a\left(1-\frac{1}{t^2}\right)\\ \frac{dx}{dt}=a\left(\frac{t^2-1}{t^2}\right)\ \ \ \ \ .....(1)
and,
y=a\left(t-\frac{1}{t}\right)
Differentiating it with respect to t,
\frac{dy}{dt}=a\frac{d}{dt}\left(t-\frac{1}{t}\right)\\ =a\left(1+\frac{1}{t^2}\right)\\ \frac{dy}{dt}=a\left(\frac{t^2+1}{t^2}\right)\ \ \ \ \ .....(2)
Dividing equation (2) by (1)
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=a\frac{(t^2+1)}{t^2}\times\frac{t^2}{a(t^2-1)}\\ \frac{dy}{dx}=\frac{t^2+1}{t^2-1}\\ \frac{dy}{dx}=\frac{x}{y}\ \ \ \ \ \ \left[Since,\ \frac{x}{y}=\frac{a(t^2+1)}{t}\times\frac{t}{a(t^2-1)}=\left(\frac{t^2+1}{t^2-1}\right)\right]
Question 18. Ifx=sin^{-1}\left(\frac{2t}{1+t^2}\right) andy =tan^{-1}\left(\frac{2t}{1-t^2}\right) , -1 < 1 < 1, prove that\frac{dy}{dx}=1
Solution:
Here,
x=sin^{-1}\left(\frac{2t}{1+t^2}\right)
Put t = tan θ
x=sin^{-1}\left(\frac{2tanθ}{1+tan^2θ}\right)\\ =sin^{-1}(sin2θ )\\ =2θ \ \ \ \ \ \left[Since,\ sin\ 2x=\frac{2tan\ x}{1+tan^2x}\right]\\ x=2(tan^{-1}t)\ \ \ \ \ [Since,\ t=sin\ θ ]
Differentiating it with respect to t,
\frac{dx}{dt}=\frac{2}{1+t^2}\ \ \ \ \ .....(1)
Further,
y =tan^{-1}\left(\frac{2t}{1-t^2}\right)
Put t = tan θ
y=tan^{-1}\left(\frac{2tanθ}{1+tan^2θ}\right)\\ =tan^{-1}(tan2θ )\\ =2θ \ \ \ \ \ \left[Since,\ tan\ 2x=\frac{2tan\ x}{1-tan^2x}\right]\\ y=2tan^{-1}t\ \ \ \ \ [Since,\ t=tan\ θ ]
Differentiating it with respect to t,
\frac{dy}{dt}=\frac{2}{1+t^2}\ \ \ \ \ \ ......(2)
Dividing equation (2) by (1),
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2}{1+t^2}\times\frac{1+t^2}{2}\\ \frac{dy}{dx}=1
Question 19. If x and y are connected parametrically by the equation, without eliminating the parameter, find\frac{dy}{dx}, when: x=\frac{sin^3t}{\sqrt{cos2t}} ,y=\frac{cos^3t}{\sqrt{cos2t}}
Solution:
Here, the given equations arex=\frac{sin^3t}{\sqrt{cos2t}} andy=\frac{cos^3t}{\sqrt{cos2t}}
Thus,
\frac{dx}{dt}\frac{d}{dt}\left[\frac{sin^3t}{\sqrt{cos2t}}\right]\\ =\frac{\sqrt{cos2t}.\frac{d}{dt}(sin^3t)-sin^3t.\frac{d}{dt}\sqrt{cos2t}}{cos2t}\\ =\frac{\sqrt{cos2t}.3sin^2t.\frac{d}{dt}(sin\ t)-sin^3t\times\frac{1}{2\sqrt{cos\ 2t}}.\frac{d}{dt}(cos2t)}{cos2t}\\ =\frac{3\sqrt{cos\ 2t}.sin^2t\ cos\ t-\frac{sin^3t}{2\sqrt{cos\ 2t}}.(-2sin\ 2t)}{cos\ 2t}\\ =\frac{3cos\ 2t\ sin^2tcos\ t+sin^3tsin\ 2t}{cos\ 2t\sqrt{cos\ 2t}}\\ \frac{dy}{dt}=\frac{d}{dt}\left[\frac{cos^3t}{\sqrt{cos\ 2t}}\right]\\ =\frac{\sqrt{cos\ 2t}.\frac{d}{dt}(cos^3t)-cos^3t.\frac{d}{dt}(\sqrt{cos\ 2t})}{cos\ 2t}\\ =\frac{-3cos\ 2t.cos^2t.sin\ t+cos^3t\ sin\ 2t}{cos\ 2t.\sqrt{cos\ 2t}}
Therefore,
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-3cos\ 2t.cos^2t. sin\ t+cos^3t\ sin\ 2t}{3cos\ 2t\ sin^2t\ cos\ t+sin^3t\ sin\ 2t}\\ =\frac{-3cos\ 2t\ cos^2t.sin\ t+cos^3t(2sin\ t\ cos\ t)}{3cos\ 2t\ sin^2t\ cos\ t+sin^3t(2sin\ t+2sin^3t)}\\ =\frac{[-3(2cos^2t-1)cos\ t+2cos^3t]}{[3(1-2sin^2t)sin\ t+2sin^3t]}\ \ \ \ \ \ \ [cos\ 2t=(2cos^2t-1),\ cos\ 2t=(1-2sin^2t)]\\ =\frac{-4cos^3t+3cos\ t}{3sin\ t-4sin^3\ t}\\ =\frac{-cos\ 3t}{sin\ 3t}\ \ \ \ \ \ [cos\ 3t=4cos^3t3cos\ t,\ sin\ 3t=3sin\ t-4sin^3t]\\ =-cot3t
Question 20. Ifx=\left(t+\frac{1}{t}\right)^a andy=a^{\left(t+\frac{1}{t}\right)} , find\frac{dy}{dx}
Solution:
Here,
x=\left(t+\frac{1}{t}\right)^a
Differentiating it with respect to t using chain rule,
\frac{dx}{dt}=\frac{d}{dt}\left(t+\frac{1}{t}\right)^a\\ =a\left(t+\frac{1}{t}\right)^{a-1}\frac{d}{dt}\left(t+\frac{1}{t}\right)\\ \frac{dx}{dt}=a\left(t+\frac{1}{t}\right)^{1-1}\left(1-\frac{1}{t^2}\right)\ \ \ \ \ .....(1)
And,
y=a^{\left(t+\frac{1}{t}\right)}
Differentiating it with respect to t using chain rule,
\frac{dy}{dt}=\frac{d}{dt}a^{\left(t+\frac{1}{t}\right)}\\ =a^{\left(t+\frac{1}{t}\right)}\times log\ a\ \frac{d}{dt}\left(t+\frac{1}{t}\right)\\ \frac{dy}{dt}=a^{\left(t+\frac{1}{t}\right)}\times log\ a\left(1-\frac{1}{t^2}\right)\ \ \ \ \ ......(2)
Dividing equation (2) by (1)
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{a^{\left(t+\frac{1}{t}\right)}\times log\ a\left(1-\frac{1}{t^2}\right)}{a\left(t+\frac{1}{t}\right)^{a-1}\left(1-\frac{1}{t^2}\right)}\\ \frac{dy}{dx}=\frac{a^{\left(t+\frac{1}{t}\right)}\times log\ a}{a\left(t+\frac{1}{t}\right)^{a-1}}

Question 21. If x=a\left(\frac{1+t^2}{1-t^2}\right)  and y=\frac{2t}{1-t^2}  , find \frac{dy}{dx}

Solution:

Here,

x=a\left(\frac{1+t^2}{1-t^2}\right)

Differentiate it with respect to t using chain rule,

\frac{dx}{dt}=a\left[\frac{(1+t^2)\frac{d}{dt}(1+t^2)-(1+t^2)\frac{d}{dt}(1-t^2)}{(1-t^2)^2}\right]\\ =a\left[\frac{(1-t^2)(2t)-(1+t^2)(-2t)}{(1-t^2)^2}\right]\\ =a\left[\frac{2t-2t^2+2t+2t^3}{(1-t^2)^2}\right]\\ \frac{dy}{dt}=\frac{4at}{(1-t^2)^2}\ \ \ \ \ ......(1)

And,

y=\frac{2t}{1-t^2}

Differentiate it with respect to t using quotient rule,

\frac{dy}{dt}=a\left[\frac{(1-t^2)\frac{d}{dt}(t)-(t)\frac{d}{dt}(1-t^2)}{(1-t^2)^2}\right]\\ =a\left[\frac{(1-t^2)(1)-(t)(-2t)}{(1-t^2)^2}\right]\\ =a\left[\frac{1-t^2+2t}{(1-t^2)^2}\right]\\ \frac{dy}{dt}=\frac{2(1+t^2)}{(1-t^2)}\ \ \ \ \ ......(2)

Question 22. Find \frac{dy}{dx} , if y = 12(1 – cos t), x = 10(t – sin t), -\frac{\pi}{2}<t<\frac{\pi}{2}

Solution:

It is given that, 

y = 12(1 – cos t),

x = 10(t – sin t)

Therefore,

\frac{dx}{dt}=\frac{d}{dt}[10(t-sin\ t)]\\ =10.\frac{d}{dt}(t-sin\ t)\\ =10(1- cos\ t)
\frac{dy}{dt}=\frac{d}{dt}[12(t-cos\ t)]\\ =12.\frac{d}{dt}(1-cos\ t)\\ =12(0- (-sin\ t)\\ =12sin\ t

Therefore,

\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{12sin\ t}{10(1-cos\ t)}\\ =\frac{12\times2sin\frac{t}{2}\times cos\frac{t}{2}}{10\times2\ sin^2\frac{t}{2}}\\ =\frac{6}{5}\ cot\ \frac{t}{2}

Question 23. If x = a(θ – sin θ) and y = a(1 – cos θ), find \frac{dy}{dx} , at θ = \frac{\pi}{3}

Solution:

Here,

x = a(θ – sin θ)

and

y = a(1 – cos θ)

Then,

\frac{dx}{dθ}=\frac{d}{dθ}[a(θ-sin\ θ]\\ =a(1-cos\ θ)
\frac{dy}{dθ}=\frac{d}{dθ}[a(1+cos\ θ]\\ =a(-sin\ θ)

Therefore,

\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{-asin\ θ}{a(1-cos\ θ)}|_{θ=\frac{\pi}{3}}\\ =-\frac{sin\frac{\pi}{3}}{1-cos\frac{\pi}{3}}\\ =\frac{\frac{\sqrt3}{2}}{1-\frac{1}{2}}=-\sqrt3

Question 24. If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), show that at t = \frac{\pi}{4},\ \frac{dy}{dx}=\frac{b}{a}

Solution:

Consider the given functions,

x = a sin 2t (1 + cos 2t)

and 

y = b cos 2t (1 – cos 2t)

Write again the functions,

x = a sin 2t + \frac{a}{2}  sin 4t

Differentiate the above function with respect to t,

\frac{dx}{dt}=2a\ cos\ 2t+2a\ cos\ 4t\ \ \ \ \ ....(1)\\

y = b cos 2t (1 – cos 2t)

y = b cos 2t – b cos2 2t

\frac{dy}{dx}=-2b\ sin\ 2t+2b\ cos\ 2t\ sin\ 2t\\=-2b\ sin\ 2t+b\ sin\ 4t\ \ \ \ ....(2)

From equation (1) and (2)

\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-2b\ sin\ 2t+b\ sin\ 4t}{2a\ cos\ 2t+2a\ cos\ 4t}\\ \therefore\frac{dy}{dx}|_{\frac{\pi}{4}}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}|_{t=\frac{\pi}{4}}=\frac{-2b}{-2a}=\frac{b}{a}

Question 25. If x = cos t (3 – 2cos2t) and y = sin t (3 – 2 sin2t), find the value of \frac{dy}{dx}  at t = \frac{\pi}{4}

Solution:

Here, the given function:

x = cos t (3 – 2cos2t)

x = cos t – 2cos3t

\frac{dx}{dt}=-3sin\ t+6cos^2t\ \ \ \ ......(1)

y = sin t (3 – 2 sin2t)

y = 3cos t – 2sin3t

\frac{dy}{dt}=3cos\ t-6sin^2tcos\ t\ \ \ \ .....(2)\\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{3cos\ t-6sin^tcos\ t}{-3sin\ t+6cos^2t\ sin\ t}\\ =\frac{3cos\ t(1-2sin^2t)}{3sin\ t(2cos^2t-1)}\\ =cot\ t\frac{(1-2(1-cos^2t))}{(2cos^2t-1)}\\ =cot\ t\\ \frac{dy}{dx}|_{\frac{\pi}{4}}=cot\frac{\pi}{4}=1

Question 26. If x=\frac{1+log\ t}{t^2} y=\frac{3+2log\ t}{t}  find \frac{dy}{dx}

Solution:

Here,

x=\frac{1+log\ t}{t^2}

 and

y=\frac{3+2log\ t}{t}
\frac{dx}{dt}=\frac{t^2\left(\frac{1}{t}\right)-(1+log\ t)(2t)}{t^4}\\ =\frac{t-2t-2tlog\ t}{t^4}\\ =\frac{-2log\ t-1}{t^3}
\frac{dy}{dt}=\frac{t\left(\frac{2}{t}\right)-(-3+2log\ t)(1)}{t^2}\\ =\frac{2-3-2tlog\ t}{t^2}\\ =\frac{-2log\ t-1}{t^2}
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{-2log\ t-1}{t^2}}{\frac{-2log\ t-1}{t^3}}=t

Question 27. If x = 3sin t – sin3t, y = 3cos t – cos3t, find \frac{dy}{dx}\ at\ t=\frac{\pi}{3}

Solution:

x = 3sin t – sin3t

and,

y = 3cos t – cos3t

\frac{dx}{dt}=3cos\ t-3cos3t\\ \frac{dy}{dt}=-3sin\ t+3sin3t\\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-3sin\ t+3sin3t}{3cos\ t-3cos3t}

When, t=\frac{\pi}{3}

\frac{dy}{dx}=\frac{-3sin(\frac{\pi}{3})+3sin(\pi)}{3cos(\frac{\pi}{3})-3cos(\pi)}=\frac{-3\times\frac{\sqrt3}{2}+0}{3\times\frac{1}{2}-3(-1)}=-\frac{1}{\sqrt3}

Question 28. If sin\ x=\frac{2t}{1+t^2} tan\ y=\frac{2t}{1-t^2}  find \frac{dy}{dx}

Solution:

sin\ x=\frac{2t}{1+t^2}

and,

tan\ y=\frac{2t}{1-t^2}
\Rightarrow x=sin^{-1}\left(\frac{2t}{1+t^2}\right)

and 

\Rightarrow y=tan^{-1}\left(\frac{2t}{1-t^2}\right)
\frac{dx}{dt}=\frac{1}{\sqrt{1-\left(\frac{2t}{1+t^2}\right)^2}}\times\frac{2(1+t^2)-(2t)(2t)}{(1+t^2)^2}\\ \frac{dx}{dt}=\frac{2}{1+t^2}\\ \frac{dy}{dt}=\frac{1}{\left(\frac{2t}{1-t^2}\right)2+1}\times\frac{2(1-t^2)-(2t)(-2t)}{(1-t^2)^2}\\ \frac{dy}{dt}=\frac{2}{1+t^2}
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{2}{(1+t^2)}}{\frac{2}{(1+t^2)}}=1

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