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

Solution:

Here, x = 2at², y = at⁴ 

= 2a 

= 2a (2t)

= 4at

And, now

= a 

= a (4t³)

= 4at³

Now, as

= 

 = t²

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

Solution:

Here, x = a cos(θ), y = b cos(θ)

= a 

= a (-sin(θ))

= – a sin(θ)

And, now

= b 

= b (-sin(θ))

= – b sin(θ)

Now, as

= 

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

Solution:

Here, x = sin(t), y = cos(2t) 

= cos(t) 

And, now

= -sin(2t) 

= – 2sin(2t)

Now, as

= 

=            (Using the identity: sin(2θ) = 2 sinθ cosθ)

 = – 4 sin(t)

Question 4. x = 4t, y = 

Solution:

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

= 4 

= 4

And, now

= 4 

= 4 

= 4 

= 4

= 

Now, as

= 

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

Solution:

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

= 

= – sin(θ) – (-sin(2θ)) 

= – sin(θ) + 2sin(2θ)

And, now

= 

= cos(θ) – (cos(2θ)) 

= cos(θ) – (2 cos(2θ)

Now, as

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

Solution:

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

= a ()

= a (1 – cos(θ))

And, now

= a ()

= a (0 + (- sin (θ)))

= – a sin (θ)

Now, as

=

=                   (Using identity: sin(2θ) = 2 sinθ cosθ and 1- cos(2θ) = 2 sin²θ)

 = – cot(θ/2)

Question 7. x = , y = 

Solution:

Here, x =

= 

= 

=

= 

= 

= 

And, now

=

= 

= 

= 

= 

= 

Now, as

= 

= 

= 

= 

 = – cot 3(t)

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

Solution:

Here, x = a (cos(t) + log tan ), y = a sin(t)

= a ()

= a (-sin(t) + )

= a (-sin(t) + )

= a (-sin(t) + )

= a (-sin(t) + )

= a (-sin(t) + )              (Using identity: 2 sinθ cosθ = sin(2θ))

= a ( – sin(t))

= a ()

= a ()

=

And, now

= a 

= a cos(t)

Now, as

= 

 = tan(t)

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

Solution:

Here, x = a sec(θ), y = b tan(θ)

= a ()

= a (sec(θ) tan(θ))

= a sec(θ) tan(θ)

And, now

= b ()

= b (sec²(θ))

Now, as

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))

= a ()

= a (- sin(θ) + (θ.) + sin(θ).)

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

= a (- sin(θ) + θ cos(θ) + sin(θ))

= aθ cos(θ)

And, now

= a ()

= a (cos (θ) – (θ.) + cos(θ).)

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

= a (cos(θ) + θ sin(θ) – cos(θ))

= aθ sin(θ)

Now, as

= 

 = tan(θ)

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

Solution:

Here, Let multiply x and y.

xy = () 

= ()

= ()                           (Using identity: sin^-1θ + cos^-1θ = )

Let’s differentiate w.r.t x,

x.+ y. = 0

x. + y = 0

Hence, Proved !!!