# RD Sharma Class 12 Ex 11.7 Solutions Chapter 11 Differentiation

Here we provide RD Sharma Class 12 Ex 11.7 Solutions Chapter 11 Differentiation for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 11.7 Solutions Chapter 11 Differentiation book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 11.7 Solutions Chapter 11 Differentiation

### 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

### Question 21. If  and  , find

Solution:

Here,

Differentiate it with respect to t using chain rule,

And,

Differentiate it with respect to t using quotient rule,

### Question 22. Find , if y = 12(1 – cos t), x = 10(t – sin t),

Solution:

It is given that,

y = 12(1 – cos t),

x = 10(t – sin t)

Therefore,

Therefore,

Solution:

Here,

x = a(θ – sin θ)

and

y = a(1 – cos θ)

Then,

Therefore,

### Question 24. If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), show that at t =

Solution:

Consider the given functions,

x = a sin 2t (1 + cos 2t)

and

y = b cos 2t (1 – cos 2t)

Write again the functions,

x = a sin 2t + sin 4t

Differentiate the above function with respect to t,

y = b cos 2t (1 – cos 2t)

y = b cos 2t – b cos2 2t

From equation (1) and (2)

### Question 25. If x = cos t (3 – 2cos2t) and y = sin t (3 – 2 sin2t), find the value of  at t =

Solution:

Here, the given function:

x = cos t (3 – 2cos2t)

x = cos t – 2cos3t

y = sin t (3 – 2 sin2t)

y = 3cos t – 2sin3t

Solution:

Here,

and

### Question 27. If x = 3sin t – sin3t, y = 3cos t – cos3t, find

Solution:

x = 3sin t – sin3t

and,

y = 3cos t – cos3t

When,

### Question 28. If ,  find

Solution:

and,

and

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