Class 12th

The area A of the isle with radius r is given by with respect to radius r A = πr².

Then, rate of change of area of the circle $\frac{d·A}{dx}=\frac{d\pi r^2}{dr}$

= 2πr.

When r = 6 cm

$\frac{\mathrm{dA}}{dt}=2\pi *6=12\pi .$

Q option (B) is correct.

Given, R (x) = 13x² + 26x + 15.

Marginal revenue is the rate of change of total revenue with respect to the number of units sold Marginal revenue (MR) = $\frac{dR(x)}{dx}$

$=\frac{d}{d}\left(13x^2+26x+15\right)$

= 13 * 2x + 26

= 26x + 26

When x = 7,

MR = 26 * 7 + 26 = 182 + 26 = 208.

Hence, the required marginal reverse = ' 208.

Choose the correct answer for questions 17 and 18.

Given, c (x) = 0.007 x³- 0.003x² + 15x + 400.

Since the marginal cost is the rate of change of total cost wrt the output we have,

Marginal cost, MC, =$\frac{\mathrm{dC}}{dx}(x)$

= 0.007 * 3x²- 0.003 * 2x + 15.

When x = 17,

Then, MC = 0.007 * 2. (17)² - 0.003 2 (17) + 15.

= 6.069 - 0.102 + 15.

= 20.967

Hence, the required marginal cost = ' 20, 97.

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.