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:
