Download App

Higher Order Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives5 Marks
Question
If $\text{x}=\text{a}(\theta+\sin\theta)\ \text{and}\ \text{y}=\text{a}(1+\cos\theta)$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{a}}{\text{y}^2}.$
✓

Answer

Here
$\text{x}=\text{a}(\theta+\sin\theta)\ \text{and}\ \text{y}=\text{a}(1+\cos\theta)$
Differentiating w.r.t.$\theta$, we get
$\frac{\text{dx}}{\text{d}\theta}=\text{a}+\text{a}\cos\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=-\text{a}\sin\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{-\text{a}\sin\theta}{\text{a}+\text{a}\cos\theta}=\frac{-\sin\theta}{1+\cos\theta}$
Differentiating w.r.t.$\theta$, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\Big\{\frac{(1+\cos\theta)\cos\theta+\sin^2\theta}{(1+\cos\theta)^2}\Big\}\frac{\text{d}\theta}{\text{dx}}$
$=\frac{-\cos\theta-\cos^2\theta-\sin^2\theta}{(1+\cos\theta)^2}\times\frac{1}{\text{a}+\text{a}\cos\theta}$
$=\frac{-(1+\cos\theta)}{\text{a}(1+\cos\theta)^3}$
$=\frac{-1}{\text{a}(1+\cos\theta)^2}$
$=\frac{-\text{a}}{\text{y}^2}\ [\because\text{y}=\text{a}(1+\cos\theta)]$
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 Higher Order Derivatives→Higher Order Derivatives — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Find the points of local maxima or local minima and corresponding local maximum and local minimum values of the following functions. Also, find the points of inflection,
$\text{f}'(\text{x})=\text{x}^{4}-62\text{x}^{2}+120\text{x}+9$
→
Integrate the rational function $\frac{3 x+5}{x^{3}-x^{2}-x+1}$
→
Find the maximum and the minimum values, if any, without using derivaives of the following functions:f(x) = |sin4x + 3| on R.
→
Prove that:
$\begin{vmatrix}\text{a}^2&\text{a}^2-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&\text{b}^2-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&\text{c}^2-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
→
Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$
→
For the matrix $\text{A}=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}.$ Show that $A^{-3} - 6A^2 + 5A + 11I_3 = 0$ Hence, find $A^{-1}$.
→
Write the values of 'a' for which the following distribution of probabilities becomes a probability distributioin:
$\text{X}=\text{x}_\text{i}:$ $-2$ $-1$ $0$ $1$
$\text{P}(\text{X}=\text{x}_\text{i}):$ $\frac{1-\text{a}}{4}$ $\frac{1+2\text{a}}{4}$ $\frac{1-2\text{a}}{4}$ $\frac{1+\text{a}}{4}$
→
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}})+9=0.$
→
By using properties of determinants, show that:
$\begin{vmatrix}1+a^2-b^2&2ab&-2b\\2ab&1-a^2+b^2&2a\\2b&-2a&1-a^2-b^2\end{vmatrix}=(1+a^2+b^2)^3$
→
The two adjecent sides of a parallelogram are $2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$ and $\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}.$ Find the unit vector parallel to one of its diagonals. Also, find its area.
→