Advertisements
Advertisements
Question
If \[x = 2 \cos \theta - \cos 2 \theta \text{ and y} = 2 \sin \theta - \sin 2 \theta\], prove that \[\frac{dy}{dx} = \tan \left( \frac{3 \theta}{2} \right)\] ?
Advertisements
Solution
\[\Rightarrow \frac{dx}{d\theta} = 2\left( - \sin\theta \right) - \left( - \sin2\theta \right)\frac{d}{d\theta}\left( 2\theta \right)\]
\[ \Rightarrow \frac{dx}{d\theta} = - 2\sin\theta + 2 \sin2\theta\]
\[ \Rightarrow \frac{dx}{d\theta} = 2\left( \sin2\theta - \sin\theta \right) . . . \left( i \right)\]
\[\text{ and }, \]
\[y = 2 \sin\theta - \sin2\theta\]
\[\Rightarrow \frac{dy}{d\theta} = 2 \cos\theta - \cos2\theta\frac{d}{d\theta}\left( 2\theta \right)\]
\[ \Rightarrow \frac{dy}{d\theta} = 2 \cos\theta - \cos2\theta\left( 2 \right)\]
\[ \Rightarrow \frac{dy}{d\theta} = 2 \cos\theta - 2 \cos2\theta\]
\[ \Rightarrow \frac{dy}{d\theta} = 2\left( \cos\theta - \cos2\theta \right) . . . \left( ii \right)\]
\[\text{ Dividing equation } \left( ii \right) \text{ by equation } \left( i \right), \]
\[\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2\left( \cos\theta - \cos2\theta \right)}{2\left( \sin2\theta - \sin\theta \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\cos\theta - \cos2\theta}{\sin2\theta - \sin\theta}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 2\sin\left( \frac{\theta + 2\theta}{2} \right)\sin\left( \frac{\theta - 2\theta}{2} \right)}{2\cos\left( \frac{2\theta + \theta}{2} \right)\sin\left( \frac{2\theta - \theta}{2} \right)} ............[\because \sin A - \sin B = 2 \cos\left( \frac{A + B}{2} \right)\sin\left( \frac{A - B}{2} \right)\text{ and } \cos A - \cos B = - 2\sin\left( \frac{A + B}{2} \right)\sin\left( \frac{A - B}{2} \right)]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- \sin\left( \frac{3\theta}{2} \right)\sin\left( \frac{- \theta}{2} \right)}{\cos\left( \frac{3\theta}{2} \right)\sin\left( \frac{\theta}{2} \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- \sin\left( \frac{3\theta}{2} \right)\left( - \sin\frac{\theta}{2} \right)}{\cos\left( \frac{3\theta}{2} \right)\sin\left( \frac{\theta}{2} \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sin\left( \frac{3\theta}{2} \right)}{\cos\left( \frac{3\theta}{2} \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \tan\left( \frac{3\theta}{2} \right)\]
APPEARS IN
RELATED QUESTIONS
Differentiate the following functions from first principles e−x.
Differentiate tan (x° + 45°) ?
Differentiate \[e^{\sin^{- 1} 2x}\] ?
Differentiate \[x \sin 2x + 5^x + k^k + \left( \tan^2 x \right)^3\] ?
Differentiate \[\frac{x^2 \left( 1 - x^2 \right)}{\cos 2x}\] ?
Differentiate \[\log \sqrt{\frac{x - 1}{x + 1}}\] ?
If \[y = \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}\] , prove that \[\left( 1 - x^2 \right) \frac{dy}{dx} = x + \frac{y}{x}\] ?
If \[y = \left( x - 1 \right) \log \left( x - 1 \right) - \left( x + 1 \right) \log \left( x + 1 \right)\] , prove that \[\frac{dy}{dc} = \log \left( \frac{x - 1}{1 + x} \right)\] ?
If \[x \sqrt{1 + y} + y \sqrt{1 + x} = 0\] , prove that \[\left( 1 + x \right)^2 \frac{dy}{dx} + 1 = 0\] ?
If \[xy \log \left( x + y \right) = 1\] ,Prove that \[\frac{dy}{dx} = - \frac{y \left( x^2 y + x + y \right)}{x \left( x y^2 + x + y \right)}\] ?
Differentiate \[{10}^{ \log \sin x }\] ?
Differentiate \[\sin \left( x^x \right)\] ?
find \[\frac{dy}{dx}\] \[y = \frac{\left( x^2 - 1 \right)^3 \left( 2x - 1 \right)}{\sqrt{\left( x - 3 \right) \left( 4x - 1 \right)}}\] ?
Find \[\frac{dy}{dx}\] \[y = \left( \tan x \right)^{\cot x} + \left( \cot x \right)^{\tan x}\] ?
If \[y = \sin \left( x^x \right)\] prove that \[\frac{dy}{dx} = \cos \left( x^x \right) \cdot x^x \left( 1 + \log x \right)\] ?
If \[x^x + y^x = 1\], prove that \[\frac{dy}{dx} = - \left\{ \frac{x^x \left( 1 + \log x \right) + y^x \cdot \log y}{x \cdot y^\left( x - 1 \right)} \right\}\] ?
If \[x^m y^n = 1\] , prove that \[\frac{dy}{dx} = - \frac{my}{nx}\] ?
If \[\left( \sin x \right)^y = \left( \cos y \right)^x ,\], prove that \[\frac{dy}{dx} = \frac{\log \cos y - y cot x}{\log \sin x + x \tan y}\] ?
If \[y = \left( \sin x - \cos x \right)^{\sin x - \cos x} , \frac{\pi}{4} < x < \frac{3\pi}{4}, \text{ find} \frac{dy}{dx}\] ?
If \[y = e^{x^{e^x}} + x^{e^{e^x}} + e^{x^{x^e}}\], prove that \[\frac{dy}{dx} = e^{x^{e^x}} \cdot x^{e^x} \left\{ \frac{e^x}{x} + e^x \cdot \log x \right\}+ x^{e^{e^x}} \cdot e^{e^x} \left\{ \frac{1}{x} + e^x \cdot \log x \right\} + e^{x^{x^e}} x^{x^e} \cdot x^{e - 1} \left\{ x + e \log x \right\}\]
If \[x = \cos t \text{ and y } = \sin t,\] prove that \[\frac{dy}{dx} = \frac{1}{\sqrt{3}} \text { at } t = \frac{2 \pi}{3}\] ?
If \[x = 10 \left( t - \sin t \right), y = 12 \left( 1 - \cos t \right), \text { find } \frac{dy}{dx} .\] ?
\[\text { If }x = \cos t\left( 3 - 2 \cos^2 t \right), y = \sin t\left( 3 - 2 \sin^2 t \right) \text { find the value of } \frac{dy}{dx}\text{ at }t = \frac{\pi}{4}\] ?
If \[y = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] write the value of \[\frac{dy}{dx}\text { for } x > 1\] ?
If \[f\left( x \right) = \log \left\{ \frac{u \left( x \right)}{v \left( x \right)} \right\}, u \left( 1 \right) = v \left( 1 \right) \text{ and }u' \left( 1 \right) = v' \left( 1 \right) = 2\] , then find the value of `f' (1)` ?
Given \[f\left( x \right) = 4 x^8 , \text { then }\] _________________ .
If \[3 \sin \left( xy \right) + 4 \cos \left( xy \right) = 5, \text { then } \frac{dy}{dx} =\] _____________ .
If \[f\left( x \right) = \left| x^2 - 9x + 20 \right|\] then `f' (x)` is equal to ____________ .
Find the second order derivatives of the following function x3 log x ?
If y = 2 sin x + 3 cos x, show that \[\frac{d^2 y}{d x^2} + y = 0\] ?
If y = (sin−1 x)2, prove that (1 − x2)
\[\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] ?
If log y = tan−1 x, show that (1 + x2)y2 + (2x − 1) y1 = 0 ?
If y = cot x show that \[\frac{d^2 y}{d x^2} + 2y\frac{dy}{dx} = 0\] ?
If x = 4z2 + 5, y = 6z2 + 7z + 3, find \[\frac{d^2 y}{d x^2}\] ?
If y = sin (log x), prove that \[x^2 \frac{d^2 y}{d x^2} + x\frac{dy}{dx} + y = 0\] ?
If y = x + ex, find \[\frac{d^2 x}{d y^2}\] ?
If x = f(t) cos t − f' (t) sin t and y = f(t) sin t + f'(t) cos t, then\[\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 =\]
If x = sin t and y = sin pt, prove that \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] .
The number of road accidents in the city due to rash driving, over a period of 3 years, is given in the following table:
| Year | Jan-March | April-June | July-Sept. | Oct.-Dec. |
| 2010 | 70 | 60 | 45 | 72 |
| 2011 | 79 | 56 | 46 | 84 |
| 2012 | 90 | 64 | 45 | 82 |
Calculate four quarterly moving averages and illustrate them and original figures on one graph using the same axes for both.
