Let \(\theta\) denote the angle at instant t
\(\frac{d\theta}{dt}=2\frac{d}{dt}(\sin \theta)\)
\(\frac{d\theta}{dt}=2\cos\theta.(\frac{d\theta}{dt})\)
\(1=2\cos\ \theta\)
\(2\cos\theta=1 cos\theta=\frac{1}{2}\)
\(\Rightarrow \theta=\cos^{-1}(\frac{1}{2})\)
Hence required angles is \(\frac{\pi}{3}\)

Let x=0.04
\(x+\Delta x=0.037\)
Then   \(\Delta x=0.037-0.040\)
\(\Rightarrow \Delta x=-0.003 \)
\(y=x^{1/2} \)
\(\Rightarrow (0.04)^{1/2}=0.2 \)
\(y=x^{1/2}\)
\(\Rightarrow \frac{dy}{dx} =\frac{1}{2\sqrt x}\)
\(\Rightarrow \Delta y=\frac{\Delta x}{2\sqrt x}\)
\(\Rightarrow \Delta y=\frac{-0.003}{2\sqrt {0.04}}=\frac{-0.003}{2\times 0.2}\)
\(=\frac{-3}{4\times 100}=\frac{-0.75}{100}\)
\(\Rightarrow \Delta y=-0.0075 \)
\(y+\Delta y=0.2-0.0075\)
\(=0.1925\)
\(\Rightarrow 0.037=0.1925\)

Let \(y=x^{1/3}\)
Let  \(x=125 \ and \ x+\Delta x=127\)
\(\Rightarrow \Delta x=2\)
\(y=(125)^{1/3}\Rightarrow5\)
\(y=x^{1/3}\)
\(\frac{dy}{dx}=\frac{1}{3}(x)^{-2/3}\)
\(\Rightarrow \Delta y=\frac{x^{-2/3}}{3}\times \Delta x \)
\(=\frac{2}{3(5)^2}=\frac{2}{75}=0.0266 \)
\(\therefore\ y+\Delta y=5+0.0266\Rightarrow5.0266 \)
\( \Rightarrow 3\sqrt{127}=5.0266\)

Minimum value=-2

Maximum value = 10.
Minimum value = nil.

30.255 Concern for environment;Responsibility for pollution free envirnment     

We have : \(f(x)\)  = tan x - 4x.
\(f'(x)=sec^{ 2 }x-4\)
When  \((\frac { -\pi }{ 3 } )\)
Thus for \( (\frac { -\pi }{ 3 } )\)  
Hence 'f' is strictly decreasing on \(\left( -\frac { \pi }{ 3 } ,\frac { \pi }{ 3 } \right) \)

The given curve is \(ay^{ 2 }=x^{ 3 }\)
\(\Rightarrow y=a^{ -\frac { 1 }{ 2 } }x^{ \frac { 3 }{ 2 } }\)
\(\therefore \frac { dy }{ dx } =a^{ -\frac { 1 }{ 2 } }\left( \frac { 3 }{ 2 } \right) x^{ \frac { 1 }{ 2 } }\)
\(At(am^{ 2 },am^{ 3 }),\frac { dy }{ dx } =\frac { 3 }{ 2a^{ \frac { 1 }{ 2 } } } (am^{ 2 })^{ \frac { 1 }{ 2 } }=\frac { 3m }{ 2 } \)
\(\therefore \) The equation of the normal is:
\((y-y^{ ' })\frac { dy }{ dx } +(x-x^{ ' })=0\)
\(\Rightarrow (y-am^{ 3 })\frac { 3m }{ 2 } +(x-am^{ 2 })=0\)
\(\Rightarrow 2x+3my=3am^{ 4 }+2am^{ 2 }.\) 

Radius (say r) of sphere 
\(=\frac { 1 }{ 2 } (diameter)\)
\(=\frac { 1 }{ 2 } ,\frac { 3 }{ 2 } (2x+3)\)
\(=\frac { 3 }{ 4 } (2x+3)\)
Let V be the volume of the sphere 
Then\(V=\frac { 4 }{ 3 } \pi r^{ 3 }=\frac { 4 }{ 3 } \pi \left( \frac { 3 }{ 4 } (2x+3 \right) ^{ 3 }\)
\(=\frac { 9 }{ 16 } \pi (2x+3)^{ 3 }\)
\(\therefore \) Rate of change of volume w,r.t.x
\(=\frac { dv }{ dx } =\frac { 9 }{ 16 } \pi .3(2x+3)^{ 2 }.2\)
\(=\frac { 27 }{ 8 } \pi (2x+3)^{ 2 }\)

\(V=\frac{4}{3} 4 \pi r^{3} \Rightarrow \frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t} \)
\(\left.\Rightarrow \frac{d r}{d t}=\frac{900}{4 \pi r^{2}} \Rightarrow \frac{d r}{d t}\right]_{r=30}=\frac{1}{4 \pi} \mathrm{cm} / \mathrm{s} ; \)
\(S=4 \pi r^{2} \Rightarrow \frac{d S}{d t}=8 \pi r \frac{d r}{d t}=\frac{8 \pi r \times 900}{4 \pi r^{2}} \)
\(\left.\Rightarrow \frac{d S}{d t}\right]_{r=30}=\frac{1800}{30}=60 \mathrm{~cm}^{2} / \mathrm{s} \)

Let length be x m and breadth be y m.
༜ Length of fence L = x + 2y
Let given area \(=a\Rightarrow xy=a\ or\ y=\frac{a}{x}\)
\(\Rightarrow L=x+\frac{2a}{a}\)
\(\Rightarrow \frac{dL}{dx}=1-\frac{2a}{x^2}\)
For maxima or minima,
\(\frac{dL}{dx}=0\Rightarrow x^2=2a, \ \therefore x=\sqrt{2a}\)
\(\frac{d^2L}{dx^2}=\frac{2a}{x^3}>2\)
For minimum length,
\(L=\sqrt{2a}+\frac{2a}{\sqrt{2a}}\)
\(=2\sqrt{2a}\)
\(x=\sqrt{2a},y=\frac{a}{\sqrt{2a}}=\frac{\sqrt{2a}}{2}=\frac{1}{2}x\)
\(\therefore x=2y\)