Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6

Question 1.
x = 2at², y = at⁴
Solution:
x = 2at², y = at⁴
So, (frac { dx }{ dt }) = 4at, (frac { dy }{ dt }) = 4at³.
∴ (frac { dy }{ dx }) = (frac{frac{d y}{d t}}{frac{d x}{d t}}) = (frac{4 a t^{3}}{4 a t}) = t².

Question 2.
x = a cos θ, y = b cos θ .
Solution:
x = x = a cos θ, y = b cos θ .
So, (frac { dx }{ dθ }) = – a sin θ, (frac { dy }{ dθ }) = – b sin θ.
∴ (frac { dy }{ dx }) = (frac{frac{d y}{d θ}}{frac{d x}{d θ}}) = (frac{-b sin theta}{-a sin theta}) = (frac { b }{ a }).

Question 3.
x = sin t, y = cos 2t
Solution:
x = sin t, y = cos 2t
So, (frac { dx }{ dt }) = cos t, (frac { dy }{ dt }) = – 2 sin t = – 4 sin tcos t.
∴ (frac { dy }{ dx }) = (frac{frac{d y}{d t}}{frac{d x}{d t}}) = (frac{-4 sin t cos t}{cos t}) = – 4 sin t.

Question 4.
x = 4t, y = (frac { 4 }{ t })
Solution:
x = 4t, y = (frac { 4 }{ t })
∴ (frac { dx }{ dt }) = 4, (frac { dy }{ dt }) = (frac{-4}{t^{2}})
∴ (frac { dy }{ dx }) = (frac{frac{d y}{d t}}{frac{d x}{d t}}) = (frac{-4}{frac{t^{2}}{4}}) = – (frac{1}{t^{2}})

Question 5.
x = cos θ – cos 2θ, y = sin θ – sin 2θ.
Solution:

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

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

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

Question 9.
x = a sec θ, y = b tan θ
Solution:
x = a sec θ, y = b tan θ
So, (frac { dx }{ dθ }) = a sec θ tan θ, (frac { dy }{ dθ }) = b sin² θ.
∴ (frac { dy }{ dx }) = (frac{frac{d y}{d θ}}{frac{d x}{d θ}}) = (frac{b sec ^{2} theta}{a sec theta tan theta}) = (frac { b }{ a }) . (frac{1}{cos theta frac{sin theta}{cos theta}})
= (frac { b }{ a })cosec θ.

Question 10.
x = a(cos θ + θ sin θ), y = a (sin θ – θ cos θ)
Solution:
So, (frac { dx }{ dθ }) = a(- sin θ + 1 . sin θ + θ cos ) = a θ cos θ.
(frac { dy }{ dθ }) = a(cos θ – 1. cos θ – θ . ( – sin θ)] = a θ sin θ
∴ (frac { dy }{ dθ }) = (frac{frac{d y}{d θ}}{frac{d x}{d θ}}) = (frac{a theta sin theta}{a theta cos theta}) = tan θ.

Question 11.
If x = (sqrt{a^{sin ^{-1} cdot t}}) and y = (sqrt{a^{cos ^{-1} cdot t}}), show that (frac { dy }{ dx }) =(frac { – y }{ x })
Solution: