Download App

CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If $\text{x}=\text{a}\cos\theta,\text{y}=\text{b}\sin\theta$ Show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$ 
✓

Answer

Given,
$\text{x}=\text{a}\cos\theta\dots\text{ eq. }1$ 
$\text{y}=\text{b}=\sin\theta\dots\text{ eq. }2$
To prove: $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
Let's find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, let's first find $\frac{\text{dy}}{\text{dx}}$ using parametric form and defferentiate it again.
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\text{a}\cos\theta=-\text{a}\sin\theta\dots\text{ eq. 3}$
Similarly, $\frac{\text{dy}}{\text{d}\theta}=\text{b}\cos\theta\dots\text{ eq. 4}$
$\Big[\because\frac{\text{d}}{\text{dx}}\cos\text{x}=-\sin\text{x}\tan\text{x},\frac{\text{d}}{\text{dx}}\sin\text{x}=\cos\text{x}\Big]$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=-\frac{\text{b}\cos\theta}{\text{a}\sin\theta}=-\frac{\text{b}}{\text{a}}\cot\theta$
Differentiating again w.r.t. x:
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\text{d}}{\text{dx}}\Big(-\frac{\text{b}}{\text{a}}\cot\theta\Big)$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{b}}{\text{a}}\text{cosec}^2\theta\frac{\text{d}\theta}{\text{dx}}\dots\text{ eq. 5}$
$\Big[$Using chain rule and $\frac{\text{d}}{\text{dx}}\cot\text{x}=-\text{cosec}^2\text{x}\Big]$
From equation 3:
$\frac{\text{dx}}{\text{d}\theta}=-\text{a}\sin\theta$
$\therefore\frac{\text{d}\theta}{\text{dx}}=\frac{-1}{\text{a}\sin\theta}$
Putting the value in equation 5:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}}{a}\text{cosec}^2\theta\frac{1}{\text{a}\sin\theta}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{b}}{\text{a}^2\sin^3\theta}$
From equation 1:
$\text{y}=\text{b}\sin\theta$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{b}}{\frac{\text{a}^2\text{y}^3}{\text{b}^3}}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}\dots\text{proved.}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "4 Marks" from CONTINUITY AND DIFFERENTIABILITY→CONTINUITY AND DIFFERENTIABILITY — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Evaluate the following integrals as limit of sum:
$\int\limits^2_{0}\big(\text{x}^2+4\big)\text{dx}$
→
A population grows at the rate of 5% per year. How long does it take for the population to double?
→
If $\text{y}=(\sin^{-1}\text{x})^2,$ prove that $(1-\text{x}^2)\text{y}_2-\text{xy}_1-2=0$
→
Let $\vec{\text{a}}=\text{x}^2\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and  $\vec{\text{c}}=\text{x}^2\hat{\text{i}}+5\hat{\text{j}}-4\hat{\text{k}}$  be three vectors. Find the valuse of x for which the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is acute and the angle between $\vec{\text{b}}$ and $​​\vec{\text{c}}$ is obtuse.
→
Evaluate the following intregals:
$\int\frac{\sin2\text{x}}{(1+\sin\text{x})(2+\sin\text{x})}\text{ dx}$
→
If P is a point and ABCD is a quadrilateral and $\overrightarrow{\text{AP}}+\overrightarrow{\text{PB}}+\overrightarrow{\text{PD}}=\overrightarrow{\text{PC}}$, show that ABCD is a parallelogram.
→
Differentiate $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$ with respect to $\sec^{-1}\Big(\frac{1}{\sqrt{1+\text{x}^2}}\Big),$ if:
$\text{x}\in\Big(\frac{1}{\sqrt{2}},1\Big)$
→
If A = diag(a, b, c), show that An = diag(an, bn, cn) for all positive integer n.
→
Represent the following families of curves by forming the corresponding differential equation:
$\text{x}^2+\text{y}^2=\text{ax}^3$
→
By using the properties of definite integrals, evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}$
→