Differentiation of Parametric Functions MCQ Quiz in मल्याळम - Objective Question with Answer for Differentiation of Parametric Functions - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

CONCEPT:

If x = f(t), y = g(t), where t is a parameter, then 

CALCULATION:

Given: x = a (θ + sin θ) and y = a (1 - cos θ)

Here, we have to find 

So, first we have to find dx/dθ and dy/dθ

⇒ 

⇒ 

As we know that, if x = f(t), y = g(t), where t is a parameter, then 

⇒ 

Hence, correct option is 1.

Concept:

​Parametric Form:

If f(x) and g(x) are the functions in x, then 

 =  

Calculation:

Let f(x) = and g(x) = ln x

 = (2x)

 = 2x 

Also

 =

Now Differentiation of f(x) with respect to g(x) is 

 =  

 = 

 = 2x²

Concept:

​Parametric Form:

If f(x) and g(x) are the functions in x, then 

 =  

Calculation:

Given f(x) = sin² x - cos² x and g(x) = 2 tan x

 = 2 sin x cos x - 2 cos x (-sin x)

 = 4 sin x cos x = 2 sin 2x

Also

 = 2 sec² x

Now Differentiation of f(x) with respect to g(x) is 

 =  

 = 

 = sin 2x cos² x 

Calculation:

We have, dy/dθ = a(θ sin θ ) and dx/dθ = a θ cos θ

Hence, option (3) is correct.

Concept:

Chain Rule of Derivatives: For two functions u and v of x, we have: .

Calculation:

Using the chain rule of derivatives, we have:

= 

= 

= 

Concept:

Using the Chain Rule:

Calculation:

Using equation of circle: x² + y² = a²

Concept:

Differentiation of parametric functions:

If x = f(t), y = g(t), where t is a parameter, then 

Calculation:

Given function is y = 3sin2 θ , x = 2 cos θ ?

We differentiate the function with respect to θ first

⇒  

As we know that, if x = f(t), y = g(t), where t is a parameter, then 

⇒ 

⇒ 

∵ x = 2cos θ ⇒ cosθ = x/2

⇒ 

Hence, option 4 is correct.

Explanation:

x = cos t 

Differentiating w.r.t. t, we get,

y = sin t

Differentiating w.r.t. t, we get,

Thus, 

Concept:

If x = f(t), y = f(t) where t be the parameter then to find 

Use chain rule 

Calculations:

Given, 

Here x = f(t), y = f(t).

To find , use chain rule ....(1)

Now,  and 

⇒  and 

Put these values in equation (1) 

 = 

Multiply and divide the above equation by t.

⇒

⇒

⇒

Concept:

Derivatives of Trigonometric Functions:

Chain Rule of Derivatives:

• .
• .

Calculation:

Let v = f(tan x) and u = g(sec x).

Now,  = f'(tan x).sec² x and  = g'(sec x).tan x.sec x.

And 

⇒ 

= 

Substituting the given values f'(1) = 2 and g'(√2) = 4, we get:

 = 1/√2.