# GSEB Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6

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: