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 !!!