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

CONCEPT:

• $\frac{d\left(\sin x\right)}{dx}=\cos x$
• $\frac{d\left(\cos x\right)}{dx}=-\sin x$

If x = f(t), y = g(t), where t is a parameter, then $\frac{dy}{dx}=\frac{g^{\prime }\left(t\right)}{f^{\prime }\left(t\right)}$

CALCULATION:

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

Here, we have to find $\frac{dy}{dx}$

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

⇒ $\frac{dx}{d\theta }=a+acos\theta$

⇒ $\frac{dy}{d\theta }=asin\theta$

As we know that, if x = f(t), y = g(t), where t is a parameter, then $\frac{dy}{dx}=\frac{g^{\prime }\left(t\right)}{f^{\prime }\left(t\right)}$

⇒ $\frac{dy}{dx}=\frac{asin\theta }{a(1+cos\theta )}=tan\frac{\theta }{2}$

Hence, correct option is 1.

Concept:

​Parametric Form:

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

$\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{g}(\mathrm{x})}$ = $\frac{\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{x}}}{\frac{\mathrm{d}\mathrm{g}(\mathrm{x})}{\mathrm{d}\mathrm{x}}}$ 

Calculation:

Let f(x) = ${\mathrm{e}}^{1+{\mathrm{x}}^2}$ and g(x) = ln x

$\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{x}}$ = ${\mathrm{e}}^{1+{\mathrm{x}}^2}$(2x)

$\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{x}}$ = 2x ${\mathrm{e}}^{1+{\mathrm{x}}^2}$

Also

$\frac{\mathrm{d}\mathrm{g}(\mathrm{x})}{\mathrm{d}\mathrm{x}}$ = $\frac{1}{\mathrm{x}}$

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

$\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{g}(\mathrm{x})}$ = $\frac{\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{x}}}{\frac{\mathrm{d}\mathrm{g}(\mathrm{x})}{\mathrm{d}\mathrm{x}}}$ 

$\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{g}(\mathrm{x})}$ = $\frac{2\mathrm{x}{\mathrm{e}}^{1+{\mathrm{x}}^2}}{\frac{1}{\mathrm{x}}}$

Concept:

​Parametric Form:

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

$\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{g}(\mathrm{x})}$ = $\frac{\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{x}}}{\frac{\mathrm{d}\mathrm{g}(\mathrm{x})}{\mathrm{d}\mathrm{x}}}$ 

Calculation:

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

$\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{x}}$ = 2 sin x cos x - 2 cos x (-sin x)

$\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{x}}$ = 4 sin x cos x = 2 sin 2x

Also

$\frac{\mathrm{d}\mathrm{g}(\mathrm{x})}{\mathrm{d}\mathrm{x}}$ = 2 sec² x

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

$\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{g}(\mathrm{x})}$ = $\frac{\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{x}}}{\frac{\mathrm{d}\mathrm{g}(\mathrm{x})}{\mathrm{d}\mathrm{x}}}$ 

$\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{g}(\mathrm{x})}$ = $\frac{2\sin 2\mathrm{x}}{2{\sec }^2\mathrm{x}}$

Calculation:

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

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\tan \theta$

$\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right)$

$=\frac{\mathrm{d}}{\mathrm{d}\theta }\left(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right)\times \frac{\mathrm{d}\theta }{\mathrm{d}\mathrm{x}}$

$=\frac{\mathrm{d}}{\mathrm{d}\theta }\left(\tan \theta \right)\times \frac{1}{\mathrm{a}\theta \cos \theta }$

$={\sec }^2\theta \times \frac{\sec \theta }{\mathrm{a}\theta }$

$=\frac{{\sec }^3\theta }{\mathrm{a}\theta }$

Explanation:

x = cos t 

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

$\frac{dx}{dt}=-sint$

y = sin t

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

$\frac{dy}{dt}=cost$

Thus, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$=\frac{cost}{-sint}=-cott$

Concept:

Chain Rule of Derivatives: For two functions u and v of x, we have: $\frac{\mathrm{d}\mathrm{u}}{\mathrm{d}\mathrm{v}}=\frac{\mathrm{d}\mathrm{u}}{\mathrm{d}\mathrm{x}}\times \frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{v}}$.

Calculation:

Using the chain rule of derivatives, we have:

$\frac{\mathrm{d}{\mathrm{x}}^2}{\mathrm{d}{\mathrm{x}}^3}=\frac{\mathrm{d}{\mathrm{x}}^2}{\mathrm{d}\mathrm{x}}\times \frac{\mathrm{d}\mathrm{x}}{\mathrm{d}{\mathrm{x}}^3}$

= $\frac{\mathrm{d}{\mathrm{x}}^2}{\mathrm{d}\mathrm{x}}\times \frac{1}{\frac{\mathrm{d}{\mathrm{x}}^3}{\mathrm{d}\mathrm{x}}}$

= $\frac{2\mathrm{x}}{3{\mathrm{x}}^2}$

= $\frac{2}{3\mathrm{x}}$

Concept:

Using the Chain Rule: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{g}\left(\mathrm{x}\right)\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$

$\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right)$

Calculation:

$\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right)=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(-\frac{\mathrm{x}}{\mathrm{y}}\right)=-\mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\frac{1}{\mathrm{y}}\right)+\frac{1}{\mathrm{y}}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(-\mathrm{x}\right)=-\mathrm{x}\left(-\frac{1}{{\mathrm{y}}^2}\right)\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\left(-\frac{1}{\mathrm{y}}\right)$

$\Rightarrow \frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}=\frac{\mathrm{x}}{{\mathrm{y}}^2}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}-\frac{1}{\mathrm{y}}$

$\mathrm{R}\mathrm{e}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{c}\mathrm{i}\mathrm{n}\mathrm{g}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=-\frac{\mathrm{x}}{\mathrm{y}}\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{v}\mathrm{i}\mathrm{o}\mathrm{u}\mathrm{s}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}$

$\Rightarrow \frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}=\frac{\mathrm{x}}{{\mathrm{y}}^2}\left(-\frac{\mathrm{x}}{\mathrm{y}}\right)-\frac{1}{\mathrm{y}}=-\frac{{\mathrm{x}}^2}{{\mathrm{y}}^3}-\frac{1}{\mathrm{y}}=-\frac{{\mathrm{x}}^2+{\mathrm{y}}^2}{{\mathrm{y}}^3}$

Using equation of circle: x² + y² = a²

$\Rightarrow \frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}=\frac{-{\mathrm{a}}^2}{{\mathrm{y}}^3}$

Concept:

Differentiation of parametric functions:

If x = f(t), y = g(t), where t is a parameter, then $\frac{dy}{dx}=\frac{g^{\prime }\left(t\right)}{f^{\prime }\left(t\right)}and\frac{dx}{dy}=\frac{f^{\prime }\left(t\right)}{g^{\prime }\left(t\right)}$

Calculation:

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

We differentiate the function with respect to θ first

⇒  $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\theta }=6\mathrm{s}\mathrm{i}\mathrm{n}\theta \cdot \mathrm{c}\mathrm{o}\mathrm{s}\theta ,\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\theta }=-2\mathrm{s}\mathrm{i}\mathrm{n}\theta$

As we know that, if x = f(t), y = g(t), where t is a parameter, then $\frac{dy}{dx}=\frac{g^{\prime }\left(t\right)}{f^{\prime }\left(t\right)}and\frac{dx}{dy}=\frac{f^{\prime }\left(t\right)}{g^{\prime }\left(t\right)}$

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{6\mathrm{s}\mathrm{i}\mathrm{n}\theta \mathrm{c}\mathrm{o}\mathrm{s}\theta }{-2\mathrm{s}\mathrm{i}\mathrm{n}\theta }$

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=-3\mathrm{c}\mathrm{o}\mathrm{s}\theta$

∵ x = 2cos θ ⇒ cosθ = x/2

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{-3\mathrm{x}}{2}$

Hence, option 4 is correct.

Concept:

If x = f(t), y = f(t) where t be the parameter then to find ${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}$

Use chain rule ${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}={\displaystyle \frac{{\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{t}}}}{{\displaystyle \frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}}}}$

Calculations:

Given, $\mathrm{x}=\mathrm{a}(\mathrm{t}-{}^1╱{}_{\mathrm{t}}),\mathrm{y}=\mathrm{a}(\mathrm{t}+{}^1╱{}_{\mathrm{t}})$

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

To find ${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}$, use chain rule ${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}={\displaystyle \frac{{\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{t}}}}{{\displaystyle \frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}}}}$....(1)

Now, $\mathrm{x}=\mathrm{a}(\mathrm{t}-{\displaystyle \frac{1}{\mathrm{t}}})$ and $\mathrm{y}=\mathrm{a}(\mathrm{t}+{\displaystyle \frac{1}{\mathrm{t}}})$

⇒ ${\displaystyle \frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}}=\mathrm{a}(1+{\displaystyle \frac{1}{{\mathrm{t}}^2}})$ and ${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{t}}}=\mathrm{a}(1-{\displaystyle \frac{1}{{\mathrm{t}}^2}})$

Put these values in equation (1) 

${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}={\displaystyle \frac{{\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{t}}}}{{\displaystyle \frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}}}}$ = ${\displaystyle \frac{\mathrm{a}(1-{\displaystyle \frac{1}{{\mathrm{t}}^2}})}{\mathrm{a}(1+{\displaystyle \frac{1}{{\mathrm{t}}^2}})}}$

Multiply and divide the above equation by t.

⇒${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}={\displaystyle \frac{\mathrm{a}\mathrm{t}(1-{\displaystyle \frac{1}{{\mathrm{t}}^2}})}{\mathrm{a}\mathrm{t}(1+{\displaystyle \frac{1}{{\mathrm{t}}^2}})}}$

⇒${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}={\displaystyle \frac{\mathrm{a}(\mathrm{t}-{\displaystyle \frac{1}{\mathrm{t}}})}{\mathrm{a}(\mathrm{t}+{\displaystyle \frac{1}{\mathrm{t}}})}}$

⇒${\displaystyle \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}}={\displaystyle \frac{\mathrm{x}}{\mathrm{y}}}$

Concept:

Derivatives of Trigonometric Functions:

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\sin \mathrm{x}=\cos \mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\cos \mathrm{x}=-\sin \mathrm{x} \\ \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\tan \mathrm{x}={\sec }^2\mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\cot \mathrm{x}=-{\csc }^2\mathrm{x} \\ \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\sec \mathrm{x}=\tan \mathrm{x}\sec \mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\csc \mathrm{x}=-\cot \mathrm{x}\csc \mathrm{x} \end{gathered}$$

Chain Rule of Derivatives:

• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{f}(\mathrm{g}(\mathrm{x}))=\frac{\mathrm{d}}{\mathrm{d}\mathrm{g}(\mathrm{x})}\mathrm{f}(\mathrm{g}(\mathrm{x}))\times \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{g}(\mathrm{x})$.
• $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{u}}\times \frac{\mathrm{d}\mathrm{u}}{\mathrm{d}\mathrm{x}}$.

Calculation:

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

Now, $\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{x}}$ = f'(tan x).sec² x and $\frac{\mathrm{d}\mathrm{u}}{\mathrm{d}\mathrm{x}}$ = g'(sec x).tan x.sec x.

And $\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{u}}=\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{x}}\times \frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{u}}=\frac{{\mathrm{f}}^{\prime }(\tan \mathrm{x}).{\sec }^2\mathrm{x}}{{\mathrm{g}}^{\prime }(\sec \mathrm{x}).\tan \mathrm{x}.\sec \mathrm{x}}=\frac{{\mathrm{f}}^{\prime }(\tan \mathrm{x}).\sec \mathrm{x}}{{\mathrm{g}}^{\prime }(\sec \mathrm{x}).\tan \mathrm{x}}=\frac{{\mathrm{f}}^{\prime }(\tan \mathrm{x})}{{\mathrm{g}}^{\prime }(\sec \mathrm{x}).\sin \mathrm{x}}$

⇒ ${\left(\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{u}}\right)}_{\mathrm{x}=\frac{\pi }{4}}=\frac{{\mathrm{f}}^{\prime }(\tan \frac{\pi }{4})}{{\mathrm{g}}^{\prime }(\sec \frac{\pi }{4}).\sin \frac{\pi }{4}}$

= $\frac{{\mathrm{f}}^{\prime }(1)}{{\mathrm{g}}^{\prime }(\sqrt{2}).\left(\frac{1}{\sqrt{2}}\right)}$

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

${\left(\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{u}}\right)}_{\mathrm{x}=\frac{\pi }{4}}=\frac{2}{4\left(\frac{1}{\sqrt{2}}\right)}=\frac{\sqrt{2}}{2}$ = 1/√2.