Maths M.C.Q (1 Marks)

1. $8\sqrt{3}\text{cm}^{2}/\text{hr}$

Solution:

$\text{A}=\frac{\sqrt{3}}{4}\times2$

$\frac{\text{dA}}{\text{dt}}=\frac{\sqrt{3}}{2}\times\frac{\text{dx}}{\text{dt}}$

$\frac{\text{dA}}{\text{dt}}=\frac{\sqrt{3}}{2}\times2\times8$

$=8\sqrt{3}\text{cm}^{2}/\text{hr}$ 

4. $3, \frac{1}{3}$

Solution:

Let, $\text{y}=\text{x}^{3}-5\text{x}^{2}+5\text{x}+8$ 

Differentiating with respect yo t, 

$\frac{\text{dy}}{\text{dt}}=(3\text{x}^{2}-10\text{x}+5)\frac{\text{dx}}{\text{dt}}$ 

Given that twice of rate of increase in x euals rate of increase in x, 

$\Rightarrow\frac{\text{dx}}{\text{dt}}=2\frac{\text{dx}}{\text{dt}}$

$\Rightarrow(3\text{x}^{2}-10\text{x}+5)\frac{\text{dx}}{\text{dt}}=2\frac{\text{dx}}{\text{dt}}$

$\Rightarrow3\text{x}^{2}-10\text{x}+5=2$

$\Rightarrow\text{x}=3$ or $\text{x}=\frac{1}{3}$ 

3. Surface area times the rete of radius. 

Solution:

Let r be the redius V be the volume of sphere at any time t. Then, $\text{V}=\frac{4}{3}\pi\text{r}^{3}$

$\Rightarrow\frac{\text{dv}}{\text{dt}}=\frac{4}{3}(3\pi\text{r}^{2})\Big(\frac{\text{dr}}{\text{dt}}\Big)$

$\Rightarrow\frac{\text{dv}}{\text{dt}}=4\pi\text{r}^{2}\Big(\frac{\text{dr}}{\text{dt}}\Big)$

Thus, the rate of change of volume is surface area times the rate of change of the radius.

2. $\frac{1}{\pi}\ \text{unit}$

Solution:

$\text{A}=\pi\text{r}^{2}$

$\frac{\text{dA}}{\text{dt}}=2\pi\text{r}\frac{\text{dr}}{\text{dt}} ...(\text{i})$

$\text{D}=2\text{r}$ (D is diameter of the circle.)

$\frac{\text{dD}}{\text{dt}}=2\frac{\text{dr}}{\text{dt}} ...(\text{ii})$

Given that $\frac{\text{dA}}{\text{dt}}=\frac{\text{dD}}{\text{dt}}$

$2\pi\text{r}\frac{\text{dr}}{\text{dt}}=2\frac{\text{dD}}{\text{dt}}$

$2\pi\text{r}=2$

$\text{r}=\frac{1}{\pi}\ \text{unit}$ 

3. $6\text{ft}/\text{sec}.$

Solution:

$\frac{15}{6}=\frac{\text{u+v}}{\text{u}}$

$\Rightarrow\frac{15}{6}=\frac{\text{v}}{\text{u}}+1$

$\Rightarrow\frac{\text{v}}{\text{u}}=\frac{3}{2}$

$\Rightarrow\text{u}=\frac{2\text{v}}{3}$

$\Rightarrow\frac{\text{du}}{\text{du}}=\frac{2}{3}\frac{\text{dv}}{\text{dt}}$

$\Rightarrow\frac{\text{du}}{\text{dt}}=\frac{2}{3}\times9$

$=6\text{ft}/\text{sec}.$

1. $ 54\pi\text{cm}^{2}/\text{min}$

Solution:

Given that $\frac{\text{dr}}{\text{dt}}=3\text{cm}/\text{min}, \frac{\text{dh}}{\text{dt}}=-4\text{cm}/\text{min}$

$\text{h}=24\text{cm}, \text{r}=7\text{cm}$

$\text{l}^{2}=\text{h}^{2}+\text{r}^{2}$

$\Rightarrow\text{l}^{2}=24^{2}+7^{2}$

$\Rightarrow\text{l}=25$

$\text{s}=\pi\text{r}\text{l}$

$\Rightarrow\text{s}^{2}=\pi\text{r}^{2}\text{l}^{2}$

$\Rightarrow\text{s}^{2}=\pi\text{r}^{2}(\text{h}^{2}+\text{r}^{2})$

$\Rightarrow\text{s}=\pi\text{r}^{2}\text{h}^{2}+\pi\text{r}^{4}$

$\Rightarrow2\text{s}\frac{\text{ds}}{\text{dt}}=2\pi^{2}\text{r}^{2}\text{h}\frac{\text{dh}}{\text{dt}}+2\pi^{2}\text{h}^{2}\text{r}\frac{\text{dr}}{\text{dt}}+4\pi^{2}\text{r}^{3}\frac{\text{dr}}{\text{dt}}$

$\Rightarrow2\pi\text{r}\text{l}\frac{\text{ds}}{\text{dt}}=2\pi^{2}\text{r}\text{h}\Bigg(\text{r}+\frac{\text{dh}}{\text{dt}}+\text{h}\frac{\text{dr}}{\text{dt}}+\frac{2\text{r}^{2}}{\text{h}}\frac{\text{dr}}{\text{dt}}\bigg)$

$\Rightarrow\pi\text{r}\text{l}\frac{\text{ds}}{\text{dt}}=\pi^{2}\text{r}\text{h}\Bigg(\text{r}+\frac{\text{dh}}{\text{dt}}+\text{h}\frac{\text{dr}}{\text{dt}}+\frac{2\text{r}^{2}}{\text{h}}\frac{\text{dr}}{\text{dt}}\Bigg)$

$\Rightarrow25\frac{\text{ds}}{\text{dt}}=\pi\text{r}\text{h}\Bigg(\text{r}+\frac{\text{dh}}{\text{dt}}+\text{h}\frac{\text{dr}}{\text{dt}}+\frac{2\text{r}^{2}}{\text{h}}\frac{\text{dr}}{\text{dt}}\Bigg)$

$\Rightarrow25\frac{\text{ds}}{\text{dt}}=\pi\text{h}\Bigg(\text{r}+\frac{\text{dh}}{\text{dt}}+\text{h}\frac{\text{dr}}{\text{dt}}+\frac{2\text{r}^{2}}{\text{h}}\frac{\text{dr}}{\text{dt}}\Bigg)$

$\Rightarrow25\frac{\text{ds}}{\text{dt}}=24\pi\Bigg(7\times(-4)+24\times3+\frac{2\times(7)^{2}}{24}\times3\bigg)$

$\Rightarrow\frac{\text{ds}}{\text{dt}}=54\pi\text{cm}^{2}/\text{sec}$

2. $10\sqrt{3}\text{cm}^2/\sec.$

Solution:

$\text{A}=\frac{\sqrt{3}}{4}\text{x}^2$

$\frac{\text{dA}}{\text{dt}}=\frac{\sqrt{3}}{2}\text{x}\frac{\text{dx}}{\text{dt}}$

$\Rightarrow\frac{\text{dA}}{\text{dt}}=\frac{\sqrt{3}}{2}\times2\times10=10\sqrt{3}\text{cm}^2/\sec.$

2.  $\frac{\pi}{3}+{\sqrt{3}}\text{m}/\text{sec}.$

Solution:

$\text{s}=2\text{t}^{2}+\sin2\text{t}$

$\text{v}=\frac{\text{ds}}{\text{dt}}=4\text{t}+2\cos2\text{t}$

$\text{a}=\frac{\text{d}^{2}\text{s}}{\text{dt}^{2}}=4-4\sin2\text{t}$ 

Given that $\text{a}=2\text{m}/\text{sec}^{2}$

$\Rightarrow4-4\sin2\text{t}=2$

$\Rightarrow2-2\sin2\text{t}=1$

$\Rightarrow2\sin2\text{t}=1$

$\Rightarrow\sin2\text{t}=\frac{1}{2}$

$\Rightarrow2\text{t}=\frac{\pi}{6}$

$\Rightarrow\text{t}=\frac{\pi}{12}$

$\text{v}=4\text{t}+2\cos2\text{t}$ at $\text{t}=\frac{\pi}{12},$ 

$\text{v}=4\text{t}\times\frac{\pi}{12}+2\cos\frac{\pi}{6}=\frac{\pi}{3}+\sqrt{3}\text{m}/\text{sec}.$ 

4. $-\frac{-16}{3}\ \text{unit}/\text{sec}.$

Solution:

According to the question, 

$\text{s}=\text{t}^{3}-4\text{t}^{2}+5$

$\Rightarrow\frac{\text{ds}}{\text{dt}}=3\text{t}^{2}-8\text{t}$

$\Rightarrow\frac{\text{d}^{2}\text{s}}{\text{dt}^{2}}=6\text{t}-8$

$\Rightarrow6\text{t}-8=0$ $\Big[$As velocity diminish, then $\frac{\text{d}^{2}\text{s}}{\text{dt}^{2}}=0\Big]$

$\Rightarrow\text{t}=\frac{4}{3}$

Now, $\Big(\frac{\text{ds}}{\text{dt}}\Big)_{\text{t}=\frac{4}{3}}=3\Big(\frac{4}{3}\Big)^{2}-8\Big(\frac{4}{3}\Big)$

$\Rightarrow\frac{\text{ds}}{\text{dt}}=\frac{16}{3}-\frac{32}{3}$

$\Rightarrow\frac{\text{ds}}{\text{dt}}=-\frac{16}{3}\ \text{unit}/\text{sec}.$ 

3. $0.24\pi\text{cm}^{2}/\text{sec}.$

Solution:

$\text{A}=\pi\text{r}^{2}$

$\frac{\text{dA}}{\text{dt}}=2\pi\text{r}\frac{\text{dr}}{\text{dt}}$

$\Rightarrow\frac{\text{dA}}{\text{dt}}=2\pi\times12\times0.01$

$=0.24\pi\text{cm}^{2}/\text{sec}.$ 

4. $\frac{1}{2\sqrt{\pi}}\text{unit}$

Solution: 

Let r be the radius and V be the volume of the sphere at any time t. Then,$\text{v}=\frac{4}{3}\pi\text{r}^{3}$  

$\Rightarrow\frac{\text{dv}}{\text{dt}}=\frac{4}{3}(3\pi\text{r}^{2})\frac{\text{dr}}{\text{dt}}$

$\Rightarrow\frac{\text{dv}}{\text{dt}}=4\pi\text{r}^{2}\frac{\text{dr}}{\text{dt}}$

$\Rightarrow4\pi\text{r}^{2}=1\ [\therefore\frac{\text{dv}}{\text{dt}}=\frac{\text{dr}}{\text{dt}}]$

$\Rightarrow\text{r}^{2}=\frac{1}{4\pi}$

$\Rightarrow\text{r}=\sqrt{\frac{1}{4\pi}}$

$\Rightarrow\text{r}=\frac{1}{2\sqrt{\pi}}\text{unit}$

1. $9\ \text{sec}.$

Solution: 

$\text{s}=45\text{t}+11\text{t}^{2}-\text{t}^{3}$

$\frac{\text{ds}}{\text{dt}}=45+22\text{t}-3\text{t}^{2}$

Given that particle moves in a straight line.

$\Rightarrow\frac{\text{ds}}{\text{dt}}=0$

$\Rightarrow3\text{t}^{2}-22\text{t}-45=0$

$\Rightarrow\text{t}=9 \ \text{or}\ \text{t}\neq\frac{-5}{3}$

$\text{t}=9\ \text{sec}.$ 

2. $\frac{3}{16\pi}\text{cm}/\text{sec}.$ 

Solution:

Let r be the radius and V be the volume of the sphere at any time t. Then, $\text{v}=\frac{4}{3}\pi\text{r}^{3}$   

$\Rightarrow\frac{\text{dv}}{\text{dt}}=4\pi\text{r}^{2}\frac{\text{dr}}{\text{dt}}$

$\Rightarrow\frac{\text{dr}}{\text{dt}}=\frac{1}{4\pi\text{r}^{2}}\frac{\text{dv}}{\text{dt}}$ 

$\Rightarrow\frac{\text{dr}}{\text{dt}}=\frac{3}{4\pi(2)^{2}}$

$\Rightarrow\frac{\text{dr}}{\text{dt}}=\frac{3}{16\pi}\text{cm}/\text{sec}.$

3. $\pi^{2}\text{cm}^{2}/\text{sec}.$

Solution:

Let D be the diameter and A be the area of the cricle at any time t. Then, 

$\text{A}=\pi\text{r}^{2}$ (where r is the radius of the circle)

$\Rightarrow\text{A}=\pi\frac{\text{D}^{2}}{4} \ \Big[\therefore\text{r}=\frac{\text{D}}{2}\Big]$

$\Rightarrow\frac{\text{dA}}{\text{dt}}=2\pi\frac{\text{D}}{4}\frac{\text{dD}}{\text{dt}}$ $\Big[\therefore\frac{\text{dD}}{\text{dt}}=1\text{cm}/\text{sec}\Big]$ 

$\Rightarrow\frac{\text{dA}}{\text{dt}}=\pi^{2}\text{cm}^{2}/\text{sec}.$ 

3. $\frac{160}{3}\text{cm}/\text{sec}.$

Solution:

Let r be the radius, h be the height and $\alpha$ be the semi-vertical angle of the cone. 

Then, $\tan\alpha = \text{rh}$

$\Rightarrow\sec2\alpha\text{d}\alpha\text{dt}=\text{dr}\text{h }\text{dt}$

$\Rightarrow\text{dr }\text{dt}=\text{h}\times\sec2\alpha\text{d}\alpha\text{dt}$

$\Rightarrow\text{dr }\text{dt}=20\times\sec230\times2 $

$\therefore\text{h}=20, \alpha=30^{\circ}$ and per second

$\Rightarrow\text{dr }\text{dt}=40\times232$

$\Rightarrow\text{dr }\text{dt}=\frac{160}{3}\text{cm}/\text{sec}.$ 

3. $160\pi\text{ cm}^2/\sec.$

Solution:

Let r be the radius nad S be the surface area of the sphere at any time t. Then,

$\text{S}=4\pi\text{r}^2$

$\Rightarrow\frac{\text{dS}}{\text{dt}}=8\pi\text{r}\frac{\text{dr}}{\text{dt}}$

$\Rightarrow\frac{\text{dS}}{\text{dt}}=8\pi(200)(0.1)$

$\Rightarrow\frac{\text{dS}}{\text{dt}}=160\pi\text{ cm}^2/\sec.$

2. -42

Solution:

$\text{x}=\text{t}^{3}-12\text{t}^{2}+6\text{t}+8$

$\Rightarrow\frac{\text{dx}}{\text{dt}}=3\text{t}^{2}-24\text{t}+8$

$\Rightarrow\frac{\text{d}^{2}\text{x}}{\text{dt}^{2}}=6\text{t}-24$

$\Rightarrow6\text{t}-24=0 $ $[\therefore$ acceleration is zero$]$

$\Rightarrow\text{t}=4$

So, Velocity at $\text{t}=4$

$\Rightarrow\frac{\text{dx}}{\text{dt}}=3(4)^{2}-24\times4+6$

$\Rightarrow\frac{\text{dx}}{\text{dt}}=48-96+6$

$\Rightarrow\frac{\text{dx}}{\text{dt}}=-42$

3. $\frac{1}{36}$

Solution:

$\text{v}=\frac{4}{3}\pi\text{r}^{3}$

$\frac{4}{3}\pi\text{r}^{3}=288\pi$

$\text{r}^{3}=288\times\frac{3}{4}$

$\text{r}^{3}=216$

$\text{v}=\frac{4}{3}\pi\text{r}^{3}$

$\frac{\text{dv}}{\text{dt}}=4\pi \text{r}^{2}\frac{\text{dr}}{\text{dt}}$

$\frac{\text{dv}}{\text{dt}}=4\pi(6)^{2}\frac{\text{dr}}{\text{dt}}$

$4\pi=144\pi\frac{\text{dr}}{\text{dt}}$

$\frac{\text{dr}}{\text{dt}}=\frac{4}{144}$

$=\frac{1}{36}$

1. $(3, \frac{16}{3})$

Solution:

According to the question,

$\frac{\text{dy}}{\text{dt}}=\frac{-\text{dx}}{\text{dt}}$

$16\text{x}^{2}+9\text{y}^{2}=400$

$\Rightarrow32\text{x}\frac{\text{dx}}{\text{dt}}+18\text{y}\frac{\text{dy}}{\text{dt}}=0$

$\Rightarrow32\text{x}\frac{\text{dx}}{\text{dt}}=-18\text{y}\frac{\text{dy}}{dt}$

$\Rightarrow32\text{x}=18\text{y}$

$\Rightarrow\text{x}=\frac{9\text{y}}{16} ...(\text{i}) $

Now, 

$16\Big(\frac{9\text{y}}{16}\Big)^{2}+9\text{y}^{2}=400$ 

$\Rightarrow\frac{81\text{y}^{2}}{16}+9\text{y}^{2}=400$

$\Rightarrow81\text{y}^{2}+144\text{y}^{2}=6400$

$\Rightarrow225\text{y}^{2}=6400$

$\Rightarrow\text{y}^{2}=\frac{6400}{225}$

$\Rightarrow\text{y}=\sqrt{\frac{6400}{225}}$

$\Rightarrow\text{y}=\frac{16}{3}$ and $-\frac{16}{3}$ 

So, $\text{x}=\frac{9}{16}\times\frac{16}{3}$ [Using (1)]

Or $\text{x}=-\frac{9}{16}\times\frac{16}{3}$

$\Rightarrow\text{x}=3 \text{ or} -3$

So, the required point is $(3, \frac{16}{3}).$

4. $8\pi$ times the rate of change of radius.

Solution:

$\text{S}=4\pi\text{r}^{2}$

$\frac{\text{dS}}{\text{dt}}=8\pi\frac{\text{dr}}{\text{dt}}$

The rate of surface area is $8\pi$ times the rate of change of the radius.

4. 0.002cm/ sec.

Solution:

$\text{V} = \frac{1}{3} \pi \text{r}^{2}\text{h}$

Given that height is equals diamiter.

$\Rightarrow \text{h} = 2\text{r}$

$\text{V} = \frac{1}{3} \pi\text{r}^{2}\text{2r}$

$\text{V} = \frac{2}{3} \pi\text{r}^{3}$

$\Rightarrow \frac{\text{dV}}{\text{dt}} = 2\pi\text{r}^{2} \frac{\text{dr}}{\text{dt}}$

$\Rightarrow \frac{\text{dr}}{\text{dt}} = \frac{1}{2\times10^{4}} \times40 \ \ \begin{pmatrix}\because \pi\text{r}^{2} = 1\text{m}^{2}\\\Rightarrow1\text{m}^{2} = 10^{4} \text{cm}^{2} \end{pmatrix}$

$\Rightarrow \frac{\text{dr}}{\text{dt}} = \frac{1}{2\times10^{4}} \times40 = 0.002\text{cm}/\sec.$

1. $1\text{m}/\text{hr}$

Solution:

$\text{V}=\pi\text{r}^{2}\text{h}$

$\Rightarrow\frac{\text{dV}}{\text{dt}}=\pi\text{r}^{2}\frac{\text{dh}}{dt}$

$\Rightarrow314=3.14\times100\times\frac{\text{dh}}{\text{dt}}$

$\Rightarrow\frac{\text{dh}}{\text{dt}}=1\text{m}/\text{hr}$

1. 1m/ minute

Solution:

$\text{V}=\pi\text{r}^2\text{h}$

$\Rightarrow\frac{\text{dV}}{\text{dt}}=\pi\text{r}^2\frac{\text{dh}}{\text{dt}}$

$\Rightarrow\frac{\text{dh}}{\text{dt}}=\frac{1}{\pi\text{r}^2}\frac{\text{dV}}{\text{dt}}$

$\Rightarrow\frac{\text{dh}}{\text{dt}}=\frac{1}{\pi(0.5)^2}\times(0.5)^2\pi$

$\Rightarrow\frac{\text{dh}}{\text{dt}}=1\text{m}/\sec.$

2.  $180\pi\ \text{cm}^{3}/\text{sec} .$

Solution:

$\text{v}=\frac{4}{3}\pi\text{r}^{3}$

$\Rightarrow\frac{\text{dv}}{\text{dt}}=4\pi\text{r}^{2}\frac{\text{dr}}{\text{dt}}$

$\Rightarrow\frac{\text{dv}}{\text{dt}}=4\pi(15)^{2}\times0.2$

$=180\pi \ \text{cm}^{3}/\text{sec}.$

4. $3.2\text{km}/\text{hr}$

Solution:

Since, $\triangle\text{P}\text{Q}\text{R}$ and $\triangle\text{X}\text{Y}\text{R}$ are similar. 

$\Rightarrow\frac{\text{PQ}}{\text{XY}}=\frac{\text{QR}}{\text{YR}}$

$\Rightarrow\frac{5}{2}=\frac{\text{u+v}}{\text{u}}$

$\Rightarrow\frac{5}{2}=1-\frac{\text{v}}{\text{u}}$

$\Rightarrow\frac{\text{v}}{\text{u}}=\frac{3}{2}$

$\Rightarrow\text{u}=\frac{2}{3}\text{v}$

$\Rightarrow\frac{\text{du}}{\text{dt}}=\frac{2}{3}\frac{\text{dv}}{\text{dt}}$

$\Rightarrow\frac{\text{du}}{\text{dt}}=\frac{2}{3}\times4.8$

$=3.2\text{km}/\text{hr}$

2. $4\pi$

Solution:

Given: $\text{V}=\frac{4}{3}\pi\text{r}^3,\text{and } \frac{\text{dr}}{\text{dt}}=0.01$

$\frac{\text{dV}}{\text{dt}}=4\pi\text{r}^2\frac{\text{dr}}{\text{dt}}$

$\frac{\text{dV}}{\text{dt}}=4\pi(10)^2\times0.01$

$\frac{\text{dV}}{\text{dt}}=4\pi$