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

Given, \(y=x^2-ax+7\)
\(\Rightarrow \frac{dy}{dx}=2x-a\)
\(m_1=2x-a\)
Line  \(2x-y+9=0\)
\(\Rightarrow 2-\frac{dy}{dx}=0\)
\(\Rightarrow \frac{dy}{dx}=2=m_2\)
for parallel, \(m_1=m_2\)
\(\therefore 2x-a=2\)
\(\Rightarrow 2(-1)-a=2\)
\(\Rightarrow -2-2=a\)
\(\Rightarrow a=-4\)

Given, \(y=x^2-7x+18\)
\(\frac{dy}{dx}=2x-7\)
\(\frac{dy}{dx} \ at\ (3,0)=2(3)-7 \)
\(=6-7=-1\)
\(\frac{dy}{dx}\ at\ (4,0)=2(4)-7\)
\(=8-7=1\)
\(m_1=-1, m_2=1\)
\(m_1\times m_2=-1\times 1=-1\)
Hence tangent are at right angles to each other,

Given, \(x^2+2y+y^2=0\)
\(2x+2\frac{dy}{dx}+2y\frac{dy}{dx}=0\)
\(\frac{dy}{dx}(2+2y)=-2x\)
\(\frac{dy}{dx}=\frac{-2x}{2(1+y)}=-\frac{x}{1+y}\)
Slope of tangent at (-1,2)
\(\frac{-1(-1)}{1+2}=\frac{1}{3}\)
Slope of normal at (-1,2)
\(=-\frac{3}{1}=-3\)

\(y^2 =8x + 3\)
\(2y\frac{dy}{dt}=8\frac{dx}{dt}\)
\(\frac{dy}{dt}=8\frac{dx}{dt}\)
\(2y\cdot8\frac{dx}{dt}=8\frac{dx}{dt}\)
\(\Rightarrow y=\frac{8}{16}=\frac{1}{2}\)
\(y=\frac{1}{2},y^2=8x+3\)
\((\frac{1}{2})^2=8x+3\)
\(\frac{1}{4}-3=8x\)
\(\Rightarrow x=\frac{1-12}{4}\times \frac{1}{8}=-\frac{11}{32}\)
Points  \((-\frac{11}{32},\frac{1}{2})\)

Let x denote the edge of cube, v denote the volume and s denotes the surface area of cube at instant t.
\(\frac{dv}{dt}=9\ cm^2/sec.\ \ \ x=10cm\)
\(v=x^3\)
\(​​\frac{dv}{dt}=3x^2\frac{dx}{dt}\)
\(\frac{9}{3x^2}=\frac{dx}{dt}\)
Now, s=6x² 
\(\frac{ds}{dt}=12x\times\frac{dx}{dt}\)
\(\frac{ds}{dt}=12\times10 \times \frac{9}{3\times 10\times 10}\)
\(\frac{12\times 9}{3\times 10}=\frac{36}{10}\)
\(=3.6\ cm^2/sec\)

Let A denote the area of rectangle at instant t
∴  A = xy  (area of rectangle),
\(\frac { dx }{ dt } =-5cm/min\)
\(\frac { dx }{ dt } =4cm/min\)
\(\frac { dx }{ dt } =x\frac{dy}{dt}+y\frac{dx}{dt}\)
\(\Rightarrow \frac{dA}{dt}=5 \times 4+8\times -5\)
\(\Rightarrow \frac{dA}{dt}=20-40\)
\(\Rightarrow \frac{dA}{dt}=-20\ cm^2/min\)
(-) ve sign shows that area is decreasing at the rate of 20cm²/min.

Let x denote the side and A denote the area of the equilateral triangle at instant t.
\(\frac{dx}{dt}=5 \ and\ x=10\ cm\)
\(A=\frac{\sqrt3}{4}x^2\)
\(\frac{dA}{dt}=\frac{\sqrt3}{4}2x\frac{dx}{dt}\)
\(\frac{dA}{dt}=\frac{\sqrt3}{4}\times 2\times5\times 10\)
\(\frac{dA}{dt}=\sqrt3\times 5\times 5\)
\(=25\sqrt3\)
Hence required rate of change
\(=25\sqrt3 \ cm^2/sec\)

\(y-\log\ x,\ x=4,\ \delta x=\cdot01\)
\(\frac{\delta y}{\delta x}=(\frac{dy}{dx})_{x=4}\)
\(\delta y=(\frac{dy}{dx})_{x=4} \times \delta x\)
\(=\frac{1}{400}=\cdot0025\)