Ncert Solutions Maths class 12th

Let 'x' cm be the length of edge of the cube which is a fxⁿ of time t then,

$\frac{dx}{dt}$ = 3cm/s as it is increasing.

Now, volume v of the cube is v = x³

Ø Rate of change of volume of the cube $\frac{dv}{dt}=\frac{d{q}^3}{dt}.$

$=\frac{d{x}^3}{dx}\frac{dx}{dt}$

= 3x².3

= 9x² ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

When x = 10cm.

$\frac{dv}{dt}$ = 9 x (10)²= 900 ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

Let 'r' cm be the radius of the circle which is afxn of time.

Then,  $\frac{dy}{dt}$ = 8 3cm/s as it is increasing.

Now, the area A of the circle is A = πr2.

So, the rate at which the area of the circle change $\frac{d}{dt}=\frac{d}{dt}$ πr2.

$=\frac{d}{dr}\pi r^2·\frac{dr}{dt}$

= 2πr 3

= 6πr. ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

When r = 10cm,

$\frac{d}{dt}$ = 6.π * 10 = 60π ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

Let x be the length of edge,v be the value and s be the surface area of the cube then,

y = x3.

and S = 6x2, where x is a fxn of time.

Now, $\frac{dY}{dF}=8c{m}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow \frac{d}{dt}$ (x3) = 8

$\frac{d{x}^3}{dx}·\frac{dx}{dt}=8$ (by chain rule)

$\Rightarrow$ 3x2 $\frac{dx}{dx}=8.$

$\frac{dx}{dt}=\frac{8}{3x^2.}$

Now, $\frac{dS}{dt}=\frac{d}{dt}$ (bx2) = $\frac{d\left(6x^2\right)}{dx}-\frac{dx}{dx}$ = 12x $\frac{8}{3x^2}=\frac{32}{x}$ $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

When x = 12 cm,

$\frac{dS}{dt}=\frac{32}{12}=\frac{8}{3}$ $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

(a) r = 3cm

When r = 3 cm,

$\frac{d}{dr}$ = 2 * π * 3 cm = 6π.

Thus, the area of the circle is changing at the rate of 6π cm $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

(b) r = 4cm

whenr = 4cm,

$\frac{d}{dr}$ = 2 * π * 4cm = 8π.

Thus, the area of the circle is changing at the rate of 8 $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

Kindly go through the solution

Kindly go through the solution

A $\left|\begin{array}{ccc}\hfill x+2\hfill & \hfill x+3\hfill & \hfill x+2a\hfill \\ \hfill x+3\hfill & \hfill x+4\hfill & \hfill x+2b\hfill \\ \hfill x+4\hfill & \hfill x+5\hfill & \hfill x+2c\hfill \end{array}\right|=0$

Given equations are :-

$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$

$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$

$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

This system of equation can be written, in matrix form, as AX= B, Where

⇒ x = 2, y = 3, z = 5.

$\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill sin\alpha \hfill & \hfill cos\alpha \hfill & \hfill cos\left(\alpha +\delta \right)\hfill \\ \hfill sin\beta \hfill & \hfill cos\beta \hfill & \hfill cos\left(\beta +\delta \right)\hfill \\ \hfill sin\gamma \hfill & \hfill cos\gamma \hfill & \hfill cos\left(\gamma +\delta \right)\hfill \end{array}\right|\\ =\frac{1}{sin\delta cos\delta }\left|\begin{array}{ccc}\hfill sin\alpha sin\delta \hfill & \hfill cos\alpha cos\delta \hfill & \hfill cos\left(\alpha +\delta \right)\hfill \\ \hfill sin\beta sin\delta \hfill & \hfill cos\beta cos\delta \hfill & \hfill cos\left(\beta +\delta \right)\hfill \\ \hfill sin\gamma sin\delta \hfill & \hfill cos\gamma cos\delta \hfill & \hfill cos\left(\gamma +\delta \right)\hfill \end{array}\right|\end{array}$

Applying $cos\left(A+B\right)=cosAcosB-sinAsinB$ in $c_3$

$\begin{array}{l}{c}_1\to c_1+c_3\\ ?=\frac{1}{sin\delta cos\delta }\left|\begin{array}{ccc}\hfill cos\alpha cos\delta \hfill & \hfill cos\alpha cos\delta \hfill & \hfill cos\alpha cos\delta -sin\alpha sin\delta \hfill \\ \hfill cos\beta cos\delta \hfill & \hfill cos\beta cos\delta \hfill & \hfill cos\beta cos\delta -sin\beta sin\delta \hfill \\ \hfill cos\gamma cos\delta \hfill & \hfill cos\gamma cos\delta \hfill & \hfill cos\gamma cos\delta -sin\gamma sin\delta \hfill \end{array}\right|\\ c_1=c_2\\ \therefore ?=0=R.H.S.\end{array}$

$\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1+p\hfill & \hfill 1+p+q\hfill \\ \hfill 2\hfill & \hfill 3+2p\hfill & \hfill 4+3p+2q\hfill \\ \hfill 3\hfill & \hfill 6+3p\hfill & \hfill 10+6p+3q\hfill \end{array}\right|\\ R_2\to R_2-2R_1\\ R_3\to R_3-3R_1\\ =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1+p\hfill & \hfill 1+p+q\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2+p\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill 7+3p\hfill \end{array}\right|\\ R_3\to R_3-3R_2\\ =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1+p\hfill & \hfill 1+p+q\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2+p\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|\end{array}$

Expanding along $c_1$

$?=\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2+p\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=1=R.H.S.$