Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.6

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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 = $\frac{sin^3(t)}{\sqrt{cos(2t)}}$ , 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 !!!

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Last Updated : 18 Mar, 2021

Class 12 NCERT Solutions - Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.5 | Set 2

Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.7

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Applications of Derivatives

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Applications of Derivatives

Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.6

If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find

Question 1. x = 2at2, y = at4

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

Question 3. x = sin(t), y = cos(2t)

Question 4. x = 4t, y =

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

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

Question 7. x = , y =

Question 8. x = a (cos(t) + log tan), y = a sin(t)

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

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

Question 11. If x = , y = , show that

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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

Question 4. x = 4t, y = $\frac{4}{t}$

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

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

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