Ncert Solutions Maths class 12th

Let r cm and h cm be the radius and the height of the cone. Then,

h =$\frac{1}{6}$ r. H = 6h

So, volume, V of the cone =$\frac{1}{3}$ πr²h

$=\frac{1}{3}\pi {(6h)}^2*h.$ $\frac{4d}{dt}.$

= 12 * h³

Rate of change of volume of the cone wrt the height is

$\frac{\mathrm{dV}}{dh}=\frac{d}{dh}\left(12\pi h^3\right)$ = 12 * π * 3 * h².

As the sand is pouring from the pipe at rate of 12${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

we have

$\frac{d}{dt}=12$

$\Rightarrow \frac{dv}{dh}*\frac{dh}{dt}=12$

$\Rightarrow 36\pi h^2\frac{dh}{dt}=12.$

$\Rightarrow \frac{dh}{dt}=\frac{12}{36\pi h^2}=\frac{1}{3\pi h^2}.$

${\therefore \frac{dh}{dt}|}_{h=4 }=\frac{1}{3\pi *{(4)}^2}=\frac{1}{48\pi }.$

Hence, the height is increasing at the rate of$\frac{1}{48x}$ cm/s.

Given, diameter of the spherical balloon = $\frac{3}{2}$ (2x + 1)

So, radius of the spherical r = $\frac{1}{2}*\frac{3}{2}(2x+1)$

$=\frac{3}{4}(2x+1)$

Then, volume of the spherical V = $\frac{4}{9}\pi r^3$

$=\frac{4}{3}\pi *[\frac{3}{4}{\left(\frac{3}{(2x+1)}\right]}^3$

$=\frac{9\pi }{16}{(2x+1)}^3.$

Q Rate of change of volume wrt.tox, $\frac{dV}{dx}=\frac{d}{dx}\left[\frac{\pi }{16}{(2x+1)}^3\right]$

$=\frac{9\pi }{16}*3*{(2x+1)}^2·\frac{d}{dx}(2x+1)$

$=\frac{27\pi }{16}{(2x+1)}^2*2=\frac{27\pi }{8}{(2x+1)}^2.$

Let x be the radius of the bubble with volume .V. then,

$\frac{dr}{dt}=\frac{1}{2}$ cm/s

andV = $\frac{4}{3}\pi n^3$

Rate of change of volume $\frac{dV}{dt}=\frac{d}{dt}\left(\frac{4}{3}{\pi }^3\right)$ = $\frac{4}{3}\pi \frac{d}{d{t}^4}{r}^3$

$=\frac{4}{3}\pi *3r^2\frac{dr}{dt}.$

= 4πr² *$\frac{1}{2}$

= 2πr².

${\frac{dv}{dt}|}_{r=cm}$ = 2x (1)² 2π. ${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

Given eqⁿ of the curve is 6y = x³ + 2.______ (1)

Wheny coordinate change s 8 times as fast as x-coordinate

$\frac{dy}{dx}$ = 8 _____ (2)

Now, differentiating eqⁿ (1) wrt.x we get,

$\frac{d}{dx}(6y)=\frac{d}{dx}\left(x^3+2\right)$

$\Rightarrow 6\frac{dy}{dx}=3x^2+0.$

6 * 8 = 3x² (using eqⁿ (2)

$\Rightarrow x^2=\frac{48}{3}=16$

$\Rightarrow +\surd x=\pm \surd 16$

x = $\pm$4.

When x = 4, we have, 6y = 4³+ 2 = 64 + 2 + 66

$\Rightarrow y=\frac{66}{6}$ y =11.

And when x = -4, we have, 6y = ( -4)3 + 2 = -64 + 2 = -62

$\Rightarrow y=\frac{62}{6}=\frac{31}{3}$

The tequired point s are (4, 11) and $\left(-4,\frac{31}{3}\right)$

Since, the bottom of ground is increasing with time t,

$\frac{dx}{dt}$ = 2cm/s

From fig, Δ ABC, by Pythagorastheorem

AB² + BC² = AC²

$\Rightarrow$ x² + y² = 5²

x² + y² = 25 ____ (1)

Differentiating eqⁿ (1) w. r. t. time t we get,

$\frac{d}{dt}\left(x^2+y^2\right)=\frac{d}{dt}(25)$

$\Rightarrow 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0.$

$\Rightarrow 2x*2+2y\frac{2y}{dy}=0$

$\Rightarrow \frac{dy}{dt}=-\frac{2x}{y}$ m/s

When x = 4m, the rate at which its height on the wall decreases is

$\frac{dy}{dt}=-\frac{2*4}{3}$ $\{\begin{array}{l}\therefore 4^2+y^2=5^2\\ \Rightarrow y^2=25-16\\ \Rightarrow y\text{\hspace{0.17em}√9}\left(L\text{\hspace{0.17em}engthcan'tbenegetive}\right)\\ \Rightarrow y=3\end{array}$

$\Rightarrow \frac{dy}{dt}=-\frac{8}{3}$ room

The volume v of a spherical balloon with radius r is V.$\frac{4}{3}\pi r^3.$

with respect its radius.

Then, the rate of change of volume $|\frac{\mathrm{dV}}{dr}=\frac{d}{dr}\left(\frac{4}{3}·\pi r^3\right)$

$=\frac{4}{3}\pi \left(\frac{d}{dr}{r}^3\right)$

$=\frac{4}{3}\pi *3*{\pi }^2$

= 4$\pi$ r²

Whenx = 10 cm,

$\frac{\mathrm{dV}}{dr}$ = 4 $\pi$10)² = 400 $\pi$cm³/cm

Let 'r' cm be the radius of volume V. measured Then,

$V\; =\frac{4}{3}\pi r^3$

Now, rate at which balloon is being inflated = 900${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow \frac{dv}{dr}=900$ ${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$ = 900 ${\mathrm{cm}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow \frac{d}{dr}\left(\frac{4}{3}*r^3\right)\frac{dr}{dt}=900$

$\frac{4}{3}$$\pi$ * 3 * r² $\frac{dr}{dr}$ = 900.

$\Rightarrow \frac{dr}{dt}=\frac{900}{4\pi r^2}.$

When r = 15cm,

${\frac{ds}{dt}|}_{r=15}=\frac{900}{4\pi *{(15)}^2}$ = $\frac{900}{900\pi }=\frac{1}{\pi }$ cm/s.

Q Radius of balloon increases by $\frac{1}{\pi }$ per second.

Since the length x is decreasing and the widthy is increasing with respect to time we have

$\frac{dx}{dt}$ = 5cm/min and $\frac{dy}{dt}$ = A cm/min

(a) The perimeter P of a rectangle will be, P = 2 (x + y)

Q  Rate of change of perimeter, $\frac{dP}{dt}=\frac{d^2}{dt}(x+y)$

$=2\left(\frac{dx}{dt}+\frac{dy}{dt}\right)$

= 2 ( -5 + 4)

= -2 cm/min

(b) The area A of the rectangle is A = x. y.

Rate of change of area is $\frac{\mathrm{dA}}{dt}=\frac{d}{dt}(x·y)$

$=x\frac{dy}{dt}+\frac{dx}{dt}·y.$

= 4x - 5y

So,  $\frac{\mathrm{dA}}{dt}$ |x = 8ay = 6cm = 4 (8) - 5 (6) = 32 - 30 = 2 cm²/min spherical balloon

The circumference C of the circle with radius r is C = 2πr.

Then, rate of change in circumference is $\frac{\mathrm{dC}}{dt}=\frac{d}{dt}2\pi r$ = $2\pi \frac{dx}{df}.$

Q Radius of circle increases at rate 0.7 cm/s we get,

$\frac{dy}{dt}=0.7$ cm/s

So,  $\frac{\mathrm{dC}}{dt}$ = 2.* 0.7 cm/s = 1.4 * cm/s.

The area A of the circle with radius π is A = πr²

Then, rate of change in area $\frac{\mathrm{dA}}{dt}=\frac{d\stackrel{?}{x}{r}^2}{dt}$ = $\frac{dx{r}^2}{dr}·\frac{dr}{dt}$

= 2πr $\mathrm{dr}$
${\frac{}{dt}}_{.}$

Q The wave moves at a rate 5cm/s we have,

$\frac{dr}{\mathrm{dF}}$ = 5cm/s

So,  $\frac{\mathrm{dA}}{dt}$ =r. 5 = 10πr. ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

When r = 8 cm.

$\frac{d}{dt}$ = 10.π.8 cm ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$ = 80 * cm ${\mathrm{cm}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$