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

• Last Updated : 20 May, 2021

### 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 = aeΘ (sinθ -cosθ), y = aeΘ (sinθ +cosθ)

Solution:

Here,

x = aeΘ (sinθ – cosθ)

Differentiating it with respect to θ, And,

y = aeΘ(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 My Personal Notes arrow_drop_up