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  3. NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.6

NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.6

September 06, 2022

NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.6

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-5 (Continuity And Differentiability)Exercise 5.6 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.6

Exercise 5.6

If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find \frac{dy}{dx}

Question 1. x = 2at2, y = at4 

Solution:

Here, x = 2at2, y = at4 

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

= 2a \frac{d(t^2)}{dt}

= 2a (2t)

= 4at

And, now

\frac{dy}{dt} = \frac{d(at^4)}{dt}

= a \frac{d(t^4)}{dt}

= a (4t3)

= 4at3

Now, as

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

= \frac{4at^3}{4at}

\frac{dy}{dx} = t2

Question 2. x = a cos(θ), y = b cos(θ)

Solution:

Here, x = a cos(θ), y = b cos(θ)

\frac{dx}{dθ} = \frac{d(a cos(θ))}{dθ}

= a \frac{d(cos(θ))}{dθ}

= a (-sin(θ))

= – a sin(θ)

And, now

\frac{dy}{dθ} = \frac{d(b cos(θ))}{dθ}

= b \frac{d(cos(θ))}{dθ}

= b (-sin(θ))

= – b sin(θ)

Now, as

\frac{dy}{dx} = \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}

= \frac{- b sin(θ)}{- a sin(θ)}

\frac{dy}{dx} = \mathbf{\frac{b}{a}}

Question 3. x = sin(t), y = cos(2t) 

Solution:

Here, x = sin(t), y = cos(2t) 

\frac{dx}{dt} = \frac{d(sin(t))}{dt}

= cos(t) 

And, now

\frac{dy}{dt} = \frac{d(cos(2t) )}{dt}

= -sin(2t) \frac{d(2t)}{dt}

= – 2sin(2t)

Now, as

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

= \frac{- 2sin(2t)}{cos(t)}

= \frac{- 2(2 sin(t)cos(t))}{cos(t)}           (Using the identity: sin(2θ) = 2 sinθ cosθ)

\frac{dy}{dx} = – 4 sin(t)

Question 4. x = 4t, y = \frac{4}{t}

Solution:

Here, x = 4t, y = 4/t

\frac{dx}{dt} = \frac{d(4t)}{dt}

= 4 \frac{d(t)}{dt}

= 4

And, now

\frac{dy}{dt} = \frac{d(\frac{4}{t})}{dt}

= 4 \frac{d(\frac{1}{t})}{dt}

= 4 \frac{t\frac{d(1)}{dt} - 1\frac{d(t)}{dt}}{t^2}

= 4 \frac{t(0) - 1}{t^2}

= 4\frac{- 1}{t^2}

= \frac{- 4}{t^2}

Now, as

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

= \frac{\frac{- 4}{t^2}}{4}

\frac{dy}{dx} = \mathbf{\frac{- 1}{t^2}}

Question 5. x = cos(θ) – cos(2θ), y = sin(θ) – sin(2θ)

Solution:

Here, x = cos(θ) – cos(2θ), y = sin(θ) – sin(2θ)

\frac{dx}{dθ} = \frac{d(cos(θ) - cos(2θ))}{dθ}

= \frac{d(cos(θ))}{dθ} - \frac{d(cos(2θ))}{dθ}

= – sin(θ) – (-sin(2θ)) \frac{d(2θ)}{dθ}

= – sin(θ) + 2sin(2θ)

And, now

\frac{dy}{dθ} = \frac{d(sin(θ) - sin(2θ))}{dθ}

= \frac{d(sin(θ))}{dθ} - \frac{d(sin(2θ))}{dθ}

= cos(θ) – (cos(2θ)) \frac{d(2θ)}{dθ}

= cos(θ) – (2 cos(2θ)

Now, as

\frac{dy}{dx} = \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}

\frac{dy}{dx} = \mathbf{\frac{cos(θ) - (2 cos(2θ)}{2sin(2θ) - sin(θ) }}

Question 6. x = a (θ – sin(θ)), y = a (1 + cos(θ)) 

Solution:

Here, x = a (θ – sin(θ)), y = a (1 + cos(θ)) 

\frac{dx}{dθ} = \frac{d(a (θ - sin(θ)))}{dθ}

= a (\frac{d(θ)}{dθ} - \frac{d(sin(θ))}{dθ})

= a (1 – cos(θ))

And, now

\frac{dy}{dθ} = \frac{d(a (1 + cos(θ)))}{dθ}

= a (\frac{d(1)}{dθ} + \frac{d(cos(θ))}{dθ})

= a (0 + (- sin (θ)))

= – a sin (θ)

Now, as

\frac{dy}{dx} = \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}

=\frac{- a sin (θ)}{a (1 - cos(θ))}

= \frac{- 2 sin(θ/2) cos(θ/2)}{2sin^2(θ/2)}                  (Using identity: sin(2θ) = 2 sinθ cosθ and 1- cos(2θ) = 2 sin2θ)

\frac{dy}{dx} = – cot(θ/2)

Question 7. x = , y= \frac{sin^3(t)}{\sqrt{cos(2t)}}, y = \frac{cos^3(t)}{\sqrt{cos(2t)}}

Solution:

Here, x = \frac{sin^3(t)}{\sqrt{cos(2t)}}, y = \frac{cos^3(t)}{\sqrt{cos(2t)}} 

\frac{dx}{dt} = \frac{d(\frac{sin^3(t)}{\sqrt{cos(2t)}})}{dt}

= \frac{\sqrt{cos(2t)}(3sin^2t)\frac{d(sin(t))}{dt} - sin^3(t) \frac{1}{2}(cos(2t))^{-\frac{1}{2}} \frac{d(cos(2t))}{dt}}{cos(2t)}

= \frac{\sqrt{cos(2t)}(3sin^2t)(cos (t)) - sin^3(t) \frac{1}{2}(cos(2t))^{-\frac{1}{2}} (-2 sin(2t))} {cos(2t)}

=\frac{\sqrt{cos(2t)}(3sin^2t)(cos (t)) + \frac{sin^3(t)}{\sqrt{cos(2t)}} (sin(2t))} {cos(2t)}

= \frac{(cos(2t))(3sin^2t)(cos (t)) + sin^3(t) (sin(2t))} {(cos(2t))^{\frac{3}{2}}}

= \frac{(cos(2t))(3sin^2t)(cos (t)) + sin^3(t) (2 sin(t) cos(t))} {(cos(2t))^{\frac{3}{2}}}

= \frac{sin^2(t) cos (t)(3cos(2t) + 2sin^2(t))}{(cos(2t))^{\frac{3}{2}}}

And, now

\frac{dy}{dt} = \frac{d(\frac{cos^3(t)}{\sqrt{cos(2t)}})}{dt}

=\frac{\sqrt{cos(2t)}(3cos^2t)\frac{d(cos(t))}{dt} - cos^3(t) \frac{1}{2}(cos(2t))^{-\frac{1}{2}} \frac{d(cos(2t))}{dt}}{cos(2t)}

= \frac{\sqrt{cos(2t)}(3cos^2t)(- sin(t)) - sin^3(t) \frac{1}{2}(cos(2t))^{-\frac{1}{2}} (-2 sin(2t))} {cos(2t)}

= \frac{\sqrt{cos(2t)}(3cos^2t)(- sin (t)) + \frac{cos^3(t)}{\sqrt{cos(2t)}} (sin(2t))} {cos(2t)}

= \frac{- (cos(2t))(3cos^2t)(sin (t)) + cos^3(t) (sin(2t))} {(cos(2t))^{\frac{3}{2}}}

= \frac{- (cos(2t))(3cos^2t)(sin (t)) + cos^3(t) (2 sin(t) cos(t))} {(cos(2t))^{\frac{3}{2}}}

= \frac{cos^2(t) sin (t) (2cos^2(t) - 3cos(2t))}{(cos(2t))^{\frac{3}{2}}}

Now, as

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

= \frac{\frac{cos^2(t) sin (t) (2cos^2(t) - 3cos(2t))}{(cos(2t))^{\frac{3}{2}}}}{\frac{sin^2(t) cos (t)(3cos(2t) + 2sin^2(t))}{(cos(2t))^{\frac{3}{2}}}}

= \frac{cos(t) [2cos^2(t) - 3(2 cos^2(t) - 1)]}{sin(t) [3(1 - sin^2(t)) + 2 sin^2(t)]}

= \frac{cos(t) [3 - 4 cos^2(t)]}{sin(t) [3 - 4 sin^2(t)]}

= \frac{- (4 cos^3(t) - 3 cos(t))}{3 sin(t) - 4 sin^3(t)}

= \frac{- cos 3(t)}{sin 3(t)}

\frac{dy}{dx} = – cot 3(t)

Question 8. x = a (cos(t) + log tan\frac{t}{2}), y = a sin(t)

Solution:

Here, x = a (cos(t) + log tan \frac{t}{2}), y = a sin(t)

\frac{dx}{dt} = \frac{d(a (cos(t) + log tan \frac{t}{2})}{dt}

= a (\frac{d(cos(t))}{dt} + \frac{d(log tan \frac{t}{2})}{dt})

= a (-sin(t) + \frac{1}{tan \frac{t}{2}} . \frac{d(tan \frac{t}{2})}{dt})

= a (-sin(t) + \frac{1}{tan \frac{t}{2}} . sec^2\frac{t}{2}.\frac{1}{2})

= a (-sin(t) + \frac{cos \frac{t}{2}}{sin \frac{t}{2}} . \frac{1}{2 cos^2\frac{t}{2}})

= a (-sin(t) + \frac{1}{2 sin\frac{t}{2} cos\frac{t}{2}})

= a (-sin(t) + \frac{1}{sin(t)})              (Using identity: 2 sinθ cosθ = sin(2θ))

= a (\frac{1}{sin(t)} – sin(t))

= a (\frac{1 - sin^2(t)}{sin(t)})

= a (\frac{cos^2(t)}{sin(t)})

=\frac{a cos^2(t)}{sin(t)}

And, now

\frac{dy}{dt} = \frac{d(a sin(t))}{dt}

= a \frac{d(sin(t)}{dt}

= a cos(t)

Now, as

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

= \frac{a cos(t)}{(\frac{a cos^2(t)}{sin(t)})}

\frac{dy}{dx} = tan(t)

Question 9. x = a sec(θ), y = b tan(θ)

Solution:

Here, x = a sec(θ), y = b tan(θ)

\frac{dx}{dθ} = \frac{d(a \hspace{0.1cm}sec(θ))}{dθ}

= a (\frac{d(sec(θ))}{dθ})

= a (sec(θ) tan(θ))

= a sec(θ) tan(θ)

And, now

\frac{dy}{dθ} = \frac{d(b \hspace{0.1cm}tan(θ))}{dθ}

= b (\frac{d(tan(θ))}{dθ})

= b (sec2(θ))

Now, as

\frac{dy}{dx} = \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}

= \frac{b \hspace{0.1cm}sec^2(θ)}{a \hspace{0.1cm}sec(θ) \hspace{0.1cm}tan(θ)}

= \frac{b \hspace{0.1cm}sec(θ)}{a \hspace{0.1cm}tan(θ)}

= \frac{b \hspace{0.1cm}cos(θ)}{a \hspace{0.1cm}sin(θ) \hspace{0.1cm}cos(θ)}

\frac{dy}{dx} = \mathbf{\frac{b \hspace{0.1 cm}cosec(θ)}{a}}

Question 10. x = a (cos(θ) + θ sin(θ)), y = a (sin(θ) – θ cos(θ))

Solution:

Here, x = a (cos(θ) + θ sin(θ)), y = a (sin(θ) – θ cos(θ))

\frac{dx}{dθ} = \frac{d(a \hspace{0.1cm}(cos(θ) + θ sin(θ)))}{dθ}

= a (\frac{d(cos(θ))}{dθ} + \frac{d(θ sin(θ))}{dθ})

= a (- sin(θ) + (θ.\frac{d(sin(θ))}{dθ}) + sin(θ).\frac{d(θ)}{dθ})

= a (- sin(θ) + (θ.(cos(θ) + sin(θ).1))

= a (- sin(θ) + θ cos(θ) + sin(θ))

= aθ cos(θ)

And, now

\frac{dy}{dθ} = \frac{d(a \hspace{0.1cm}(sin(θ) - θ cos(θ)))}{dθ}

= a (\frac{d(sin(θ))}{dθ} - \frac{d(θ cos(θ))}{dθ})

= a (cos (θ) – (θ.\frac{d(cos(θ))}{dθ}) + cos(θ).\frac{d(θ)}{dθ})

= a (cos(θ) – (θ.(-sin (θ) + cos(θ).1))

= a (cos(θ) + θ sin(θ) – cos(θ))

= aθ sin(θ)

Now, as

\frac{dy}{dx} = \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}

= \frac{aθ \hspace{0.1cm}sin(θ)}{aθ \hspace{0.1cm}cos(θ)}

\frac{dy}{dx} = tan(θ)

Question 11. If x = \sqrt{a^{sin^{-1}t}}, y = \sqrt{a^{cos^{-1}t}}, show that \frac{dy}{dx} = -\frac{y}{x}

Solution:

Here, Let multiply x and y.

xy = ((\sqrt{a^{sin^{-1}t}})(\sqrt{a^{cos^{-1}t}})) 

= (\sqrt{a^{sin^{-1}t + cos^{-1}t}})

= (\sqrt{a^{\frac{π}{2}}})                           (Using identity: sin-1θ + cos-1θ = \mathbf{\frac{π}{2}})

Let’s differentiate w.r.t x,

\frac{d(xy)}{dx} = \frac{d(\sqrt{a^{\frac{π}{2}}})}{dx}

x.\frac{d(y)}{dx}+ y.\frac{d(x)}{dx} = 0

x.\frac{dy}{dx} + y = 0

\mathbf{\frac{dy}{dx} = -\frac{y}{x}}

Hence, Proved !!!

Chapter-5 (Continuity And Differentiability)

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