Download App

CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\text{x}=\text{a}\sec\theta,\text{y}=b\tan\theta$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
✓

Answer

Here,
$\text{x}=\text{a}\sec\theta,\text{y}=b\tan\theta$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\sec\theta\tan\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=b\sec^2\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{d}\theta}\times\frac{\text{d}\theta}{\text{dx}}=\frac{\text{b}\sec^2\theta}{\text{a}\sec\theta\tan\theta}=\frac{\text{b}\ \text{cosec}\theta}{\text{a}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{b}}{\text{a}}\times-\text{cosec}\theta\cot\theta\times\frac{\text{d}\theta}{\text{dx}} $
$=-\frac{\text{b}}{\text{a}}\times\text{cosec}\theta\cot\theta\times\frac{1}{\text{a}\sec\theta\tan\theta}$
$=\frac{-\text{b}}{\text{a}^2}\times\frac{1}{\tan^3\theta}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{b}}{\frac{\text{a}^2\text{y}^3}{\text{b}^3}}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
Hence 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 "5 Marks Questions" 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

A man 160cm tall, walks away from a source of light situated at the top of a pole 6m high, at the rate of 1.1m/ sec. How fast is the length of his shadow increasing when he is 1m away from the pole?
→
Find the equation of the plane passing through the point (2, 1, -1) and (-1, 3, 4) and perpendicular to the plane x - 2y + 4z = 10.
→
For each of the differential equations given in find a particular solution satisfying the given condition:
$\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}=\sin 2\text{x};\ \text{y}=2\ \text{when x}=\frac{\pi}{2}$
→
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.
→
Find the angle between line $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-1}=\frac{\text{z}+1}{1}$ and the plane $2x + y - z = 4$.
→
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}0&\text{x}&\text{y}\\-\text{x}&0&\text{z}\\-\text{y}&-\text{z}&0\end{vmatrix}$
→
Using properties of determinants, prove that:
$\begin{vmatrix}3\text{a}&-\text{a+b}&-\text{a+c}\\-\text{b+a}&3\text{b}&-\text{b+c}\\-\text{c+a}&\text{c+b}&3\text{c}\end{vmatrix}=3 (\text{a + b + c}) (\text{ab + bc + ca})$
→
Evaluate the following integrals:
$\int\frac{1}{\cos\text{x}(5-4\sin\text{x})}\ \text{dx}$
→
Find the area of the region bounded by the parabola $y^2 = 2x$ and the straight line $x - y = 4$
→
Find the equations of the tangent and normal to the curve $x = a \sin^3 \theta$ and $y = a \cos^3  \theta$ at $\theta = \frac{\pi}{4}.$
→