Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.7 | Set 1

Question 1. Find , when: x = at² and y = 2at

Solution:

Given that x = at², y = 2at

So,

Therefore,

Question 2. Find , when: x = a(θ + sinθ) and y = a(1 – cosθ)

Solution:

Here,

x = a(θ + sinθ)

Differentiating it with respect to θ,

and,

y = a(1 – cosθ)

Differentiate it with respect to θ,

Using equation (1) and (2),

Question 3. Find , when: x = acosθ and y = bsinθ

Solution:

Then x = acosθ and y = bsinθ

Then,

Therefore,

Question 4. Find , when: x = ae^Θ (sinθ -cosθ), y = ae^Θ (sinθ +cosθ)

Solution:

Here,

x = ae^Θ (sinθ – cosθ)

Differentiating it with respect to θ,

And,

y = ae^Θ(sinθ+cosθ)

Differentiating it with respect to θ

Dividing equation (2) by equation (1)

Question 5. Find , when: x = bsin²θ and y = acos²θ

Solution:

Here,

x = bsin²θ and y = acos²θ

Then,

Question 6. Find , when: x = a(1 – cosθ) and y = a(θ +sinθ) at θ =

Solution:

Here,

x = a(1 – cosθ) and y = a(θ + sinθ)

Then,

Therefore,

Question 7. Find , when: and

Solution:

Here,

Differentiate it with respect to t,

and,

Differentiating it with respect to t,

Dividing equation (2) and (1)

Question 8. Find , when: and

Solution:

Here,

Differentiating it with respect to t using quotient rule,

and,

Differentiating it with respect to t using quotient rule,

Dividing equation (2) by (1)

Question 9. If x and y are connected parametrically by the equation, without eliminating the parameter, find when: x = a(cosθ +θsinθ), y = a(sinθ -θcosθ)

Solution:

The given equations are

x = a(cosθ +θ sinθ) and y = a(sinθ -θcosθ)

Then,

= a[-sinθ + θcosθ + sinθ] = aθcosθ

= a[cosθ +θsinθ -cosθ]

= aθsinθ

Therefore,

Question 10. Find , when: and

Solution:

Here,

Differentiating it with respect to θ using product rule,

and,

Differentiating it with respect to θ using product rule and chain rule