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

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

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 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
\(=25\sqrt3 \ cm^2/sec\)

Minimum value=-2

0.75

f′(x) = (x−1) ex + ex = x⋅e^x, \(e^x>0 \forall x \text { and } x>0\)
∴ f′(x) > 0, for x > 0
Hence, function is increasing.

m < 0

30.255 Concern for environment;Responsibility for pollution free envirnment     

Let 'r' be the radius and 'h' the length of the cylinder
By the question \(V=\pi r^{ 2 }h\) ..(1) [Given]
Now \(S=2\pi rh+\pi r^{ 2 }\)
\(=2\pi r\frac { V }{ \pi r^{ 2 } } +\pi r^{ 2 }\)  [Using (1)]
=\(2\frac { V }{ r } +\pi r^{ 2 }\)
\(\therefore \)  \(\frac { ds }{ dr } =-\frac { 2V }{ r^{ 2 } } =2\pi r\)
and   \(\frac { d^{ 2 }S }{ dr^{ 2 } } =\frac { 4V }{ r^{ 3 } } +2r>0\)
S is minimum when  \(\frac { dS }{ dr } =0and\frac { d^{ 2 }S }{ dr^{ 2 } } >0\)
Now \(\frac { dS }{ dr } =0\Rightarrow -\frac { 2V }{ r^{ 2 } } +2\pi r=0\)
\(\Rightarrow \)  \(V=\pi r^{ 3 }\)
Putting (1), \(\pi r^{ 3 }=\pi r^{ 2 }h\Rightarrow h=r\)
Hence S in maximum when the height is equal to the radius of the base.   

Let 'r' and 'h' be the radius and height respectively of the cylinder.
\(\therefore \) S, the surface area = \(\pi r^{ 2 }+2\pi rh\)
\(\Rightarrow \)  \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)....(1)
And V,the volume = \(\pi r^{ 2 }h\)
i.e. \(V=\pi r^{ 2 }\left( \frac { S-\pi r^{ 2 } }{ 2\pi r } \right) \)  [Using (1)]
\(\Rightarrow \) \(V=\frac { r }{ 2 } (S-\pi r^{ 2 })\)
\(\Rightarrow \) \(V=\frac { r }{ 2 } (Sr-\pi r^{ 2 })\)
\(\therefore \)  \(\frac { dV }{ dr } =\frac { 1 }{ 2 } (S-3\pi r^{ 2 })\)...(2)
and   \(\frac { d^{ 2 }V }{ dr } =-\frac { 3\pi }{ 2 } (2r)=-3\pi r\)   (3)
For greatest volume \(\frac { dV }{ dr } =0\Rightarrow -\frac { 1 }{ 2 } (S-3\pi r^{ 2 })=0\)
\(\Rightarrow \) \(S=3\pi r^{ 2 }\Rightarrow r=\sqrt { \frac { S }{ 3\pi } } \)
Putting in (3),
\(\frac { d^{ 2 }V }{ dr^{ 2 } } =-3\pi \sqrt { \frac { S }{ 3\pi } } =-\sqrt { 3\pi S } \) which is -ve.
Thus V is the greatest when \(r=\sqrt { \frac { s }{ 3\pi } } \)
And from (1),
\(h=\frac { S-\pi \left( \frac { S }{ 3\pi } \right) }{ 2\pi \sqrt { \frac { S }{ 3\pi } } } =\frac { \frac { 2S }{ 3 } }{ \frac { 2 }{ \sqrt { 3 } } \sqrt { \pi S } } =\sqrt { \frac { s }{ 3\pi } } \)
Hence, height = Radius of the base.                   

The given curve is : x = 1 \(cos\theta ,y-=\theta -sin\theta \)
\(\therefore \frac { dx }{ d\theta } =sin\theta ,\frac { dy }{ d\theta } =1-cos\theta \)
\(\therefore \frac { dy }{ dx } =\frac { dy/d\theta }{ dx/d\theta } =\frac { 1-cos\theta }{ sin\theta } =\frac { 2sin^{ 2 }\frac { \theta }{ 2 } }{ 2sin\frac { \theta }{ 2 } cos\frac { \theta }{ 2 } } \)
\(=\frac { sin\frac { \theta }{ 2 } }{ cos\frac { \theta }{ 2 } } =tan\frac { \theta }{ 2 } \)
At \(\theta \) \(=\frac { \pi }{ 4 } \)    \(x=1-cos\frac { \pi }{ 4 } =1-\frac { 1 }{ \sqrt { 2 } } \)
\(y=\frac { \pi }{ 4 } -sin\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)
and  \(\frac { dy }{ dx } =tan\frac { \pi }{ 8 } \)

\(\text { The given function is } f(x)=x^{2}-x+1 \text { . }\)
\(\therefore f^{\prime}(x)=2 x-1 \)
\(\text {Now, } f^{\prime}(x)=0 \Rightarrow x=\frac{1}{2} . \)
\(\text {The point } \frac{1}{2} \text { divides the interval ( }-1,1 \text { ) into two disjoint intervals i.e., }\left(-1, \frac{1}{2}\right) \text { and }\left(\frac{1}{2}, 1\right) \text { . }\)
\(\text {Now, in interval }\left(-1, \frac{1}{2}\right), f^{\prime}(x)=2 x-1<0 \text { . }\)
 Therefore, f is strictly decreasing in interval (−1,12) . " id="MathJax-Element-1011-Frame" role="presentation" style="position: relative;" tabindex="0">Therefore, f is strictly decreasing in interval \( \left(-1, \frac{1}{2}\right) \text { . }\)
\(\text {However, in interval }\left(\frac{1}{2}, 1\right), f^{\prime}(x)=2 x-1>0 \text { . }\)
 Therefore, f is strictly increasing in interval (12,1) . " id="MathJax-Element-1013-Frame" role="presentation" style="position: relative;" tabindex="0"> Therefore, f is strictly increasing in interval \( \left(\frac{1}{2}, 1\right) \text { . }\)
Hence, f is neither strictly increasing nor decreasing in interval  (-1,1) 

r=2x \(\Rightarrow \) radius of cone =  diameter of base of cylinder.

8.95

Let point be (x, y), then \(\frac{d y}{d t}=-\frac{d x}{d t}\)
Differentiating both sides of the ellipse
16x² + 9y² = 400 w.r.t. t, we get
\(16 \cdot 2 x \cdot \frac{d x}{d t}+18 y \frac{d y}{d t}=0 \)
\(\Rightarrow \frac{d y}{d t}=-\frac{16 x}{9 y} \cdot \frac{d x}{d t} \Rightarrow 16 x=9 y \quad \text { \{using (i) } \)
Substituting in curve, we get
\(16 x^{2}+\frac{256}{9} x^{2}=400 \)
\(\Rightarrow x^{2}=9 \rightarrow x=\pm 3, \text { then } y=\pm \frac{16}{3}\)
Hence, point are \((3,{16\over 3}),(-3,{-16\over 3})\)

Let r be the radius, h be the height and V be the volume of the sand cone
Also given that, \(\frac{d V}{d t}=12 \mathrm{~cm}^{3} / \mathrm{s}, h=\frac{1}{6} r\) 
\(\Rightarrow r=6 h \text { and } h=4 \mathrm{~cm}\) 
Volume of sand cone,
\(V=\frac{1}{3} \pi r^{2} h\) 
\( \Rightarrow V=\frac{1}{3} \pi(6 h)^{2} h \)
\(\Rightarrow V=\frac{1}{3} \pi \times 36 h^{2} \times h=12 \pi h^{3} \) 
On differentiating both sides w.r.t. t, we get
\( \frac{d V}{d t}=12 \pi \times 3 h^{2} \frac{d h}{d t}=36 \pi h^{2} \frac{d h}{d t} \)
\(\Rightarrow 12=36 \pi(4)^{2} \frac{d h}{d t} \)
\(\Rightarrow \frac{d h}{d t}=\frac{12}{36 \pi \times 16}=\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s} \) 
Hence, the height of the sand cone is increasing at the rate of \(\frac{1}{48 \pi} \mathrm{cm} / \mathrm{s}\). when the height is 4 cm,

Given equations are
\(x=\frac{\sin ^3 t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^3 t}{\sqrt{\cos 2 t}}\)
Now, differentiate both w.r.t t
We get,
\(\frac{d x}{d t}=\frac{d\left(\frac{\sin ^3 t}{\sqrt{\cos 2 t}}\right)}{d t}=\frac{\sqrt{\cos 2 t} \cdot \frac{d\left(\sin ^3 t\right)}{d t}-\sin ^3 t \cdot \frac{d(\sqrt{\cos 2 t})}{d t}}{(\sqrt{\cos 2 t})^2}=\frac{3 \sin ^2 t \cos t \cdot \sqrt{\cos 2 t}-\sin ^3 t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{\cos 2 t}\)
\(=\frac{3 \sin ^2 t \cos t \cdot \cos 2 t+\sin ^3 t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}} \)
\(=\frac{\sin ^3 t \sin 2 t(3 \cot t \cot 2 t+1)}{\cos 2 t \sqrt{\cos 2 t}} \quad\left(\because \frac{\cos x}{\sin x}=\cot x\right)\)
Similarly,
\(\frac{d y}{d t}=\frac{d\left(\frac{\cos ^3 t}{\sqrt{\cos 2 t})}\right.}{d t}=\frac{\sqrt{\cos 2 t} \cdot \frac{d\left(\cos ^3 t\right)}{d t}-\cos ^3 t \cdot \frac{d(\sqrt{\cos 2 t})}{d t}}{(\sqrt{\cos 2 t})^2}=\frac{3 \cos ^2 t(-\sin t) \cdot \sqrt{\cos 2 t}-\cos ^3 t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{(\sqrt{\cos 2 t})^2}\)
\(=\frac{-3 \cos ^2 t \sin t \cos 2 t+\cos ^3 t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}} \)
\( =\frac{\sin 2 t \cos ^3 t(1-3 \tan t \cot 2 t)}{\cos 2 t \sqrt{\cos 2 t}} \)
\( \text { Now, } \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{\sin 2 t \cos ^3 t(1-3 \tan t \cot 2 t)}{\cos 2 t \sqrt{\cos 2 t}}}{\frac{\sin ^3 t \sin 2 t(3 \cot t \cot 2 t+1)}{\cos 2 t \sqrt{\cos 2 t}}}=\frac{\cot ^3 t(1-3 \tan t \cot 2 t)}{(3 \cot t \cot 2 t+1)} \)
\(=\frac{\cos ^3 t\left(1-3 \cdot \frac{\sin t}{\cos t} \cdot \frac{\cos 2 t}{\sin 2 t}\right)}{\sin ^3 t\left(3 \cdot \frac{\cos t}{\sin t} \cdot \frac{\cos 2 t}{\sin 2 t}+1\right)}=\frac{\cos ^2 t(\cos t \sin 2 t-3 \sin t \cos 2 t)}{\sin ^2 t(3 \cos t \cos 2 t+\sin t \sin 2 t)} \)
\( =\frac{\cos ^2 t\left(\cos t \cdot 2 \sin t \cos t-3 \sin t\left(2 \cos ^2 t-1\right)\right)}{\sin ^2 t\left(3 \cos t\left(1-2 \sin ^2 2 t\right)+\sin t \cdot 2 \sin t \cos t\right)} \)
\( \left(\because \sin 2 x=2 \sin x \cos x \text { and } \cos 2 x=2 \cos ^2 x-1 \text { and } \cos 2 x=1-2 \sin ^2 x\right) \)
\( =\frac{\cos ^2 t\left(2 \sin t \cos ^2 t-6 \sin t \cos ^2 t+3 \sin t\right)}{\sin ^2 t\left(3 \cos t-6 \cos t \sin ^2 t+2 \sin ^2 \cos t\right)} \)
\( =\frac{\sin t \cos t\left(-4 \cos ^3 t+3 \cos t\right)}{\sin t \cos t\left(3 \sin t-4 \sin ^3 t\right)} \)
\( \frac{d y}{d x}=\frac{-4 \cos ^3 t+3 \cos t}{3 \sin t-4 \sin ^3 t}=\frac{-\cos 3 t}{\sin 3 t}=-\cot 3 t \)
\( \left(\because \sin 3 t=3 \sin t-4 \sin ^3 t \text { and } \cos 3 t=4 \cos ^3 t-3 \cos t\right) \)
Therefore, the answer is \(\frac{d y}{d x}=-\cot 3 t\)

f(x) = sin3x - cos3x, 0 < x < π 
\(\Rightarrow f'(x)=3\cos3x+3\sin3x\)
\(=3(\cos3x+\sin3x)\)
Put \(f'(x)=0\)
\(\Rightarrow \cos3x+\sin3x=0\)
\(\Rightarrow \sin3x=-\cos3x\)
\(\Rightarrow -\tan3x=1\)
\(\Rightarrow \tan3x=-1\)
As 0\(\pi\), 0 < 3x > 3\(\pi\)
ஃ tan 3x is negative for the following values:
\(3x=\frac{3\pi}{4}\)
\(\Rightarrow x=\frac{\pi}{4}\)
\(3x=\pi+\frac{3\pi}{4}=\frac{7\pi}{4}\)
\(\Rightarrow x=\frac{7\pi}{12}\)
\(3x=\frac{7\pi}{4}+\pi=\frac{11\pi}{4}\)
\(\Rightarrow x=\frac{11\pi}{12}\)
Hence we have intervals:
Hence, \(f(x)-\sin3x-\cos3x\) is strictly increasing in the intervals \((0,\frac{\pi}{4})\cup(\frac{7\pi}{12},\frac{11\pi}{12}) \) and strictly decreasing in intervals \((\frac{\pi}{4},\frac{7\pi}{12})\cup(\frac{11\pi}{12},\pi)\)