Topics

\(\theta\) is very small and it must be in radian when you are taking approximation.

\[\begin{matrix} & sin\theta \simeq \theta,tan\theta \simeq \theta \\ & sin\theta \simeq tan\theta. \\ & cos\theta \simeq 1 \end{matrix}\]

Practice Exercise

Q. 1 Find the value of
(i) \(cos\left( - 30^{\circ} \right)\)
(ii) \(sin120^{\circ}\)
(iii) \(sin135^{\circ}\)
(iv) \(cos120^{\circ}\)
(v) \(sin270^{\circ}\)
(vi) \(cos270^{\circ}\)

Ans.
(i) \(\frac{1}{\sqrt{3}}\)
(ii) \(\frac{\sqrt{3}}{2}\)
(iii) \(\frac{1}{\sqrt{2}}\)
(iv) \(- \frac{\sqrt{3}}{2}\)
(v) -1
(vi) 0

Introduction:

Suppose on your \(8^{\text{th}\text{~}}\) birthday at midnight you measured your height and found it to be 120 cm . One year later at midnight on your Eight birthday you again measure your height & find it to be 132 cm . Now if we say rate of increase of your height is \(12\text{ }cm/\) year it does not mean that at the exact 11:59:59 PM you were 120 cm and as soon as clock moved to 12:00 midnight you streched and became 132 cm .
Now if we express your rate of growth as \(1\text{ }cm/\) month we are not implying that every month at midnight of \(1^{\text{st}\text{~}}\) you suddenly strech by \(1\text{ }cm/\) month is just a unit and the growth is a continuous process. If we assume you grow uniformly at all times the rate simply implies that if you grow \(\Delta\) h in time \(\Delta\) t at some point of time, the ratio \(\left( \frac{\Delta h}{\Delta t} \right)\) will be \(12\text{ }cm/\) year or \(1\text{ }cm/month\) or whatever unit you may choose.
So we can calculate the rate of growth of height in a finite time internal. If we calculate rate of growth \(\left( \frac{\Delta h}{\Delta t} \right)\) in different time interval like
(i) Rate of growth per day \(\left( \frac{\Delta h}{\Delta t} \right) = \frac{1}{30}\text{ }cm/\) day
(ii) Rate of growth per hour \(\left( \frac{\Delta h}{\Delta t} \right) = \frac{1}{30 \times 24}\text{ }cm/\) hour
(iii) Rate of growth per min \(\left( \frac{\Delta h}{\Delta t} \right) = \frac{1}{30 \times 24 \times 60}\text{ }cm/min\)

Similarly if we want to calculate the rate of growth exactly at 12 O 'clock mid night then we take time interval \(\Delta t\) is zero. So in that time \(\Delta h\) is also zero becasue nothing can change in zero time. So, rate of growth becomes \(\frac{0}{0}\), which is meaningless.

Now the question is how two find the rate of growth exactly at 12 O 'clock or at any instant of time.

Differential calculus play an important role to find the rate of growth at any instant. The rate of growth at any instant is known as instantaneous rate. In many situations in physics, it is sometimes necessary to use the concept of instantaneous rate. The basic tool for this is calculus, invented by Newton and Liebnitz independently. The use of calculus is fundamental in the treatment of various problems in physics.

Numerical interpretation:

1. Power rule: \(\frac{d}{dt}\left( t^{n} \right) = nt^{n - 1}\)

2. \(\frac{d}{dt}(t) = 1\)

3. \(\frac{d}{dt}(sint) = cost\)

4. \(\frac{d}{dt}(cost) = - sint\)

5. \(\ \frac{d}{dt}\left( e^{t} \right) = e^{t}\)

6. \(\frac{d}{dt}\left( \log_{e}t \right) = \frac{1}{t};\ \left\lbrack \log_{e}t \equiv lnt \right\rbrack\)

7. \(\frac{d}{dt}(tant) = \sec^{2}t\)

8. \(\frac{d}{dt}(sect) = secttant\)

Differentiation Rules : \(u\) and \(v\) are functions of \(t\) i.e. \(u = f(t)\) and \(v = g(t)\)

1. \(\frac{d}{dt}(c) = 0\)

2. Constant multiple rule: \(\frac{d}{dt}(cu) = c\frac{du}{dt}\)
   where \(c\) : constant

3. Sum rule: \(\frac{d}{dt}(u + v) = \frac{du}{dt} + \frac{dv}{dt}\) 4. Difference rule: \(\frac{d}{dt}(u - v) = \frac{du}{dt} - \frac{dv}{dt}\)

4. Product rule \(:\frac{d}{dt}(uv) = u\frac{dv}{dt} + v\frac{du}{dt}\) 6. Quotient rule: \(\frac{d}{dt}\left( \frac{u}{v} \right) = \frac{v\frac{du}{dt} - u\frac{dv}{dt}}{v^{2}}\)

Example : Differentiate w.r.t. time.
(i) \(y = t^{2}\)
(ii) \(x = t^{3/2}\)
(iii) \(y = \frac{1}{\sqrt{t}}\)
(iv) \(x = 4t^{3}\)
(v) \(y = 2\sqrt{t}\)
(vi) \(y = 2t^{2} + t - 1\)
(vii) \(y = 3\sqrt{t} + \frac{2}{\sqrt{t}}\)
(viii) \(y = t^{3}sint\)
(ix) \(x = te^{t}\)
(x) \(x = \sqrt{t}(1 - t)\)

Sol. (i) \(\frac{dy}{dt} = 2t\)
(ii) \(\frac{dx}{dt} = \frac{3}{2}t^{1/2}\)
(iii) \(\frac{dy}{dt} = - \frac{1}{2}t^{- 3/2}\)
(iv) \(\frac{dx}{dt} = 12t^{2}\)
(v) \(\frac{dy}{dt} = 2\left( \frac{1}{2}t^{- 1/2} \right) = t^{- 1/2}\)
(vi) \(\frac{dy}{dt} = 4t + 1 + 0 = 4t + 1\)
(vii) \(\frac{dy}{dt} = 3t^{- 1/2} + 2\left( - \frac{1}{2}t^{- 3/2} \right) = \frac{3}{\sqrt{t}} - t^{- 3/2}\)
(viii) \(\frac{dy}{dt} = 3t^{2}sint + t^{3}cost\)
(ix) \(\frac{dx}{dt} = e^{t} + te^{t} = e^{t}(1 + t)\)
(x) \(\frac{dx}{dt} = \frac{1}{2}t^{- 1/2} - \frac{3}{2}t^{1/2} = \frac{1}{2\sqrt{t}} - \frac{3}{2}\sqrt{t}\)

Example : Differentiate with respect to x
(i) \(y = xlnx\)
(ii) \(y = x^{2}e^{x}\)
(iii) \(y = \frac{sinx}{x}\)
(iv) \(y = \frac{3x^{2} + 2\sqrt{x}}{x}\)

Sol.
(i) \(\frac{dy}{dx} = x\left( \frac{1}{x} \right) + \mathcal{l}nx = 1 + \mathcal{l}nx\)
(ii) \(\frac{dy}{dx} = 2xe^{x} + x^{2}e^{x} = xe^{x}(2 + x)\)
(iii) \(y = x^{- 1}sinx\)

\[\frac{dy}{dx} = \left( - x^{- 2} \right)sinx + x^{- 1}cosx\]

(iv) \(y = 3x + 2x^{- 1/2}\)

\[\frac{dy}{dx} = 3 + 2\left( - 1/2x^{- 3/2} \right) = 3 - x^{- 3/2}\]

Suppose you are asked to differentiate the function

\[y = \sqrt{x^{2} + 1}\]

The differentiation formula you learned in the previous section of this chapter do not enable you to calculate \(\frac{dy}{dx}\)

In fact, if we let \(y = \sqrt{u}\) and let \(u = x^{2} + 1\),
Then we can evaluate \(\frac{dy}{du},\frac{du}{dx}\) easily using the formula that we have learned in previous section.
\[\frac{dy}{du} = \frac{1}{2\sqrt{u}}\ \frac{du}{dx} = 2x \]but our aim is to calculate \(\frac{dy}{dx}\) so we can write.

\[\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}\]

and its value will become

\[\begin{matrix} \frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot 2x\ \text{~}\text{Substituting value of 'u' we will get}\text{~} \\ \frac{dy}{dx} = \frac{1}{2\sqrt{x^{2} + 1}} \cdot 2x = \frac{x}{\sqrt{x^{2} + 1}} \end{matrix}\]

Ex. \(1\ x = (at + b)^{n}\) where \(a,b\) and \(n\) are a real number then \(\frac{dx}{dt} = an(at + b)^{n - 1}\).
Sol. \(\ u = at + b\) and \(x = u^{n}\)

\[\begin{matrix} & \frac{dx}{du} = nu^{n - 1} \\ & \frac{du}{dt} = a \\ & \frac{dx}{dt} = \frac{dx}{du} \times \frac{du}{dt} = n(u)^{n - 1} \times a = an(at + b)^{n - 1} \end{matrix}\]

Ex. 3 It \(x = \sin^{2}\theta\), then find \(\frac{dx}{dt}\) where \(\frac{d\theta}{dt} = \omega\)
Sol. \(\frac{dx}{d\theta} = 2(sin\theta)(cos\theta) = sin2\theta\)

\[\begin{matrix} & \frac{dx}{dt} = \frac{dx}{\text{ }d\theta} \cdot \frac{\text{ }d\theta}{dt} \\ & \frac{dx}{dt} = sin2\theta \cdot \omega \end{matrix}\]

\[velocity(v) = \frac{dx}{dt},\ \text{~}\text{acceleration}\text{~}(a) = \frac{dv}{dt};\ \left( a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = \frac{dv}{dx}v = v\frac{dv}{dx} \right)\]

Force \((F) =\) rate of change of momentum \(= \frac{dp}{dt}\);

Current \((i) = \frac{dq}{dt};\ \) angular speed \((\omega) = \frac{d\theta}{dt};\)
angular acceleration \((\alpha) = \frac{d\omega}{dt}\ ;\left( \alpha = \frac{d\omega}{dt} = \frac{d\omega}{d\theta}\frac{d\theta}{dt} = \omega\frac{d\omega}{d\theta} \right)\)

Ex. The radius of a circle is increasing at a rate \(\frac{dr}{dt} = \alpha\). Find the rate at which its area is increasing when radius is equal to 3 m .
Sol area of cricle \((A) = \pi r^{2}\)

\[\begin{matrix} & \frac{dA}{dt} = \pi(2r)\frac{dr}{dt} = 2\pi r\left( \frac{dr}{dt} \right) = 2\pi r(\alpha) \\ & \left( \frac{dA}{dt} \right)_{atr = 3\text{ }m} = 6\pi\alpha{\text{ }m}^{2}/sec. \end{matrix}\]

\[\overrightarrow{v} = \frac{d\overrightarrow{r}}{dt},\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt};\overrightarrow{F} = \frac{d\overrightarrow{p}}{dt};\overrightarrow{\alpha} = \frac{d\overrightarrow{\omega}}{dt}\]

Ex. \(\overrightarrow{r} = 2t\widehat{i} + 3t^{2}\widehat{j}\). Find \(\overrightarrow{v}\) and \(\overrightarrow{a}\) where \(\overrightarrow{v} = \frac{d\overrightarrow{r}}{dt},\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt}\)
Sol. \(\overrightarrow{v} = \frac{d\overrightarrow{r}}{dt} = 2\widehat{i} + 6t\widehat{j}\text{ }m/s\)

\[\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt} = 0 + 6\widehat{j}\text{ }m/s^{2}\]

Physical meaning of \(\frac{\mathbf{dy}}{\mathbf{dx}}\)

1. The ratio of small change in the function \(y\) and the varibalbe \(x\) is called the average rate of change of \(y\) w.r.t.x. For example, if a body cover a small distance \(\Delta s\) in small time \(\Delta t\), then average velocity of the body, \(v_{av} = \frac{\Delta s}{\Delta t}\) Also, if the velocity of a body changes by a small amount \(\Delta v\) in small time \(\Delta t\), then average acceleration of the body, \(a_{av} = \frac{\Delta v}{\Delta t}\)

2. When \(\Delta x \rightarrow 0\) The limiting value of \(\frac{dy}{dx}\) is \({Lim}_{\Delta x \rightarrow 0}\frac{\Delta y}{\Delta x} = \frac{dy}{dx} = tan\theta\)

\[= \text{~}\text{slope of the tangent}\text{~}\]

It is called the instantaneous rate change of \(y\) w.r.t.x.
The differentiation of a function w.r.t. a veriable imples the instantaneous rate change of the funcaiton w.r.t. that variable.

Like wise, instantaneous velocity of the body, \((v) = {Lim}_{\Delta t \rightarrow 0}\frac{\Delta s}{\Delta t} = \frac{ds}{dt}\)
and instantaneous acceleration of the body \(\ \) (a) \(= {Lim}_{\Delta t \rightarrow 0}\frac{\Delta v}{\Delta t} = \frac{dv}{dt}\)

Maxima & minima of a function \(y = f(x)\)
for maximum value \(\frac{dy}{dx} = 0\ \) and \(\frac{d^{2}y}{{dx}^{2}} =\) negative
for minimum value \(\frac{dy}{dx} = 0\ \) and \(\frac{d^{2}y}{dx^{2}} =\) positive
Ex. Find minimum value of \(y = 25x^{2} - 10x + 5\).
Sol. For maximum/minimum value \(\frac{dy}{dx} = 0 \Rightarrow 50x - 10 \Rightarrow x = \frac{1}{5}\)
Now at \(x = \frac{1}{5},\frac{{\text{ }d}^{2}y}{{dx}^{2}} = 50\), which is positive
So \(y_{\text{min}\text{~}} = 25\left( \frac{1}{5} \right)^{2} - 10\left( \frac{1}{5} \right) + 5 = 1 - 2 + 5 = 4\)
Ex. A body is moving vertically upwards under gravity such that its position from ground is given as \(y = ut - \frac{1}{2}{gt}^{2}\). Find the max height reached by body.

Sol. \(\frac{dy}{dt} = u - gt = 0,t = \frac{u}{g}\)
\(\frac{d^{2}y}{{dx}^{2}} = - g < 0\) (which is negative). so we get maximum height \(y_{\max}\) at \(t = \frac{u}{g}\sec\).
\[{y_{\text{max}\text{~}} = u\left( \frac{u}{g} \right) - \frac{1}{2}\text{ }g\left( \frac{u}{g} \right)^{2} }{= \frac{u^{2}}{2\text{ }g}}\]

It is path traversed by a particle, independent of time parameter.
Ex. Find the equation of trajectory for the particle moving in circular path as shown.

\[x = rcos\theta\ y = rsin\theta\ x^{2} + y^{2} = r^{2}\ \Rightarrow \ x^{2} + y^{2} = 1\]

So far we have studied as how to find velocity of a particle when its position is given. Now if we know the velocity of a particle and we might wish to know its position at any given time or an engineer who can measure the variable rate at which water is leaking from a tank wants to know the amount leaked over a certain time period. This reverse process is called antiderivative of integration.

There are two types of integration :

(a) Indefinite integration (b) Definite integration.

(a) Indefinite integration

Since we know that, \(\Delta t = t_{2} - t_{1}\)
Now, when time \(t_{2}\) approaches \(t_{1};\Delta t \rightarrow dt\)
To remind you, \(v = \underset{\Delta t \rightarrow 0}{Lim}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}\)
So, If we integrate \(dt\), we should get \(t_{2} - t_{1}\); i.e. \(\int_{}^{}\ dt = t_{2} - t_{1}\)
Similarly, \(\int_{}^{}\ dx = x_{2} - x_{1}\)
but since the differentiation of a constant becomes zero, so i.e. why we can not retrive the value of that constant.
\(\therefore\) We should write \(\int_{}^{}\ dx = x + c\).
Where 'c' is integration constant.

Integration Formulae :

1. \(\int x^{n}dx = \frac{x^{n + 1}}{n + 1} + c\lbrack n \neq - 1\rbrack\)

2. \(\ \int(ax + b)^{n}dx = \frac{1}{a}\frac{(ax + b)^{n + 1}}{n + 1} + c\)

3. \(\int cosxdx = sinx + c\)

4. \(\int sinxdx = - cosx + c\)

5. \(\ \int e^{x}dx = e^{x} + c\)

6. \(\ \int\frac{1}{x}dx = ln|x| + c,\lbrack x \neq 0\rbrack\)

7. \(\ \int cos(ax + b)dx = \frac{1}{a}sin(ax + b) + c\)

8. \(\ \int sin(ax + b)dx = - \frac{1}{a}cos(ax + b) + c\)

9. \(\int e^{(ax + b)}dx = \frac{1}{a}e^{(ax + b)} + c\)

10. \(\ \int\frac{1}{(ax + b)}dx = \frac{1}{a}ln|(ax + b)| + c\)

Rules of Integration :

1. \(\ \int kudx = k\int udx\) where is k constant.

2. \(\int(u + v)dx = \int udx + \int vdx\)

Ex. 1 Evaluate indefinite integration.
(i) \(x = \int dt\)
(ii) \(x = \int tdt\)
(iii) \(x = \int(2t)dt\)
(iv) \(x = \int\left( t^{2} \right)dt\)
(v) \(x = \int\left( - \frac{2}{t^{3}} \right)dt\)

Sol.
(i) \(x = t + c\)
(ii) \(x = \frac{t^{2}}{2} + c\)
(iii) \(x = \frac{2t^{2}}{2} + c = t^{2} + c\)
(iv) \(x = \frac{t^{3}}{3} + c\)
(v) \(x = - 2\int t^{- 3}dt \Rightarrow x = - 2\left( \frac{t^{- 2}}{- 2} \right) + c = t^{- 2} + c\)

Where 'c' is integration constant.
(b) Definite integration.

When a function is integrated between a lower limit and an upper limit, it is called a definite integral.
If \(\frac{d}{dx}(f(x)) = g(x)\)
then \(\int g(x)dx\) is called indefinite integral and \(\int_{a}^{b}\mspace{2mu} g(x)dx = \lbrack f(b) - f(a)\rbrack\) is called definite integral.
Here, a and b are called lower and upper limits of the variable x .
After carrying out integration, the result is evaluated between upper and lower limits as explained below :
\(\int_{x_{1}}^{x_{2}}\mspace{2mu} g(x)dx = f\left( x_{2} \right) - f\left( x_{1} \right) =\) Area under the curve.

Ex. Evaluate the integral :
(i) \(\int_{1}^{5}\mspace{2mu} x^{2}dx\)
(ii) \(\int_{0}^{1}\mspace{2mu} t^{2}dt\)
(iii) \(\int_{3}^{5}\mspace{2mu} tdt\)
(iv) \(\int_{0}^{1}\mspace{2mu} t^{x_{1}}dx\)

Sol. (i) \(\int_{1}^{5}\mspace{2mu} x^{2}dx = \left\lbrack \frac{x^{3}}{3} \right\rbrack_{1}^{5} = \frac{1}{3}\left\lbrack x^{3} \right\rbrack_{1}^{5} = \frac{1}{3}\left( (5)^{3} - (1)^{3} \right) = \frac{1}{3}(125 - 1) = \frac{124}{3}\)
(ii) \(\int_{0}^{1}\mspace{2mu} t^{2}dt = \left\lbrack \frac{t^{3}}{3} \right\rbrack_{0}^{1} = \frac{1}{3}\)
(iii) \(\int_{3}^{5}\mspace{2mu} tdt = \left\lbrack \frac{t^{2}}{2} \right\rbrack_{3}^{5} = \frac{5^{2} - 3^{2}}{2} = 8\)
(iv) \(\int_{0}^{1}\mspace{2mu} t^{3/2}dx = \left\lbrack \frac{t^{5/2}}{5/2} \right\rbrack_{0}^{1} = \frac{2}{5}\)

Practise Exercise :
(i) \(\int_{R}^{\infty}\mspace{2mu}\frac{GMm}{x^{2}}dx\)
(ii) \(\int_{u}^{v}\mspace{2mu} Mvdv\)
(iii) \(\int_{0}^{\pi/2}\mspace{2mu} cosxdx\)

Ans.
(i) \(\frac{GMm}{R}\)
(ii) \(\frac{1}{2}M\left( v^{2} - u^{2} \right)\)
(iii) 1

Ex. 3 Find displacement of a particle in 1-D if its velocity is \(v = (2t - 5)m/s\), from \(t = 0\) to \(t = 4\) sec.
Sol. \(\frac{dx}{dt} = 2t - 5\)
\[{\int_{x_{1}}^{x_{2}}\mspace{2mu} dx = \int_{0}^{4}\mspace{2mu}(2t - 5)dt }{x_{2} - x_{1} = \left( t^{2} - 5t \right)_{0}^{4} }\]displacement \(= 16 - 20 = - 4\text{ }m\)

Since \(\frac{dx}{dt} = v \Rightarrow cdx = vdt\ \int dx = \int vdt\)

\[\begin{array}{r} \therefore\ x_{2} - x_{1} = \int_{}^{}\ vdt\#(1) \end{array}\]

Now, we have to understand the meaning of \(\int_{}^{}\ vdt\)
Graphically, it is equivalent to finding the area under a curve. Suppose the \(v - t\) graph for a particle is as shown in Fig. given below. We want to find the displacement for time interval \(t_{i} - t_{f}\). Let us divide the time interval \(t_{i} - t_{f}\) into many small intervals, each of duration \(\Delta t\) and if \(\Delta t \rightarrow dt\). We can find the displacement ' dx ' of the particle during any small interval, such as the one shaded in figure given below, by \(dx = vdt\), where ' v ' is the velocity in that interval. Therefore, the displacement during this small interval is simply the area of the shaded rectangle. The total displacement for the interval \(t_{i} - t_{f}\) is the sum of the areas of all the rectangles from \(t_{i}\) to \(t_{f}\). But by adding up it means integrating.
Thus, Area under curve \(= \int_{}^{}\ v\) dt ... (2)
Hence from (1) & (2), A rea under curve \(= x_{2} - x_{1} =\) displacement.

Ex. 1 From the v versus t graph of figure (a) the time(s) at which the particle is at rest (b) at what time, if any does the particle reverse the direction of its motion? (c) The distance and displacement of the particle from \(t = 0\text{ }s\) to \(t = 6\text{ }s\).

Sol. (a) \(t = 0,3sec\)
(b) \(t = 3sec\)
(c) displacement \(= - \frac{1}{2}(3 \times 4) + \frac{1}{2}(3 + 2)\)

\[\begin{array}{r} = - 6 + 10 = 4\text{ }m\#(4) \end{array}\]

distance \(= \left| - \frac{1}{2}(3 \times 4) \right| + \left| \frac{1}{2}(3 + 2)(4) \right|\)

\[= 6 + 10 = 16\text{ }m\]

Lets take velocity vector \(\overrightarrow{v} = v_{x}\widehat{i} + v_{y}\widehat{j} + v_{z}\widehat{k}\)
\[{\Rightarrow v_{x} = \frac{dx}{dt};v_{y} = \frac{dy}{dt};y_{z} = \frac{dz}{dt} }{\therefore\ x = \int v_{x}dt;y = \int v_{y}dt\ z = \int v_{z}dt}\]

Ex. 1 Find the equation of trajectory of a particle whose velocity components are \(v_{x} = 2x + 1,v_{y} = 2y + 3\)
Given that particle starts from rest from origin.

Sol. \(\ v_{x} = 2x + 1\),
\[{\frac{dx}{dt} = 2x + 1 }{\int_{0}^{x}\mspace{2mu}\frac{dx}{2x + 1} = \int_{0}^{t}\mspace{2mu} dt}\]

\[\begin{array}{r} \frac{1}{2}ln(2x + 1) = t\#(1) \end{array}\]

\[{v_{y} = 2y + 3 }{\int_{0}^{y}\mspace{2mu}\frac{dy}{2y + 3} = \int_{0}^{t}\mspace{2mu} dt}\]

\[\begin{array}{r} \frac{1}{2}ln\frac{(2y + 3)}{3} = t\#(2) \end{array}\]

from (1) and (2)
\[{\frac{1}{2}ln(2x + 1) = \frac{1}{2}ln\frac{(2y + 3)}{3} }{2x + 1 = \frac{(2y + 3)}{3} }{y = 3x}\]

For \(x \ll 1\), the following approximation can be used :

\[(1 + x)^{n} \approx 1 + nx\]

Ex Find \((1.03)^{1/3}\)
Sol. \(\ (1 + 0.03)^{1/3}\left( x = 0.03\& n = \frac{1}{3} \right)\)
\[\simeq 1 + \frac{0.03}{3} = 1.01\]

Q. 1 The angle subtended by the moon's diameter at a point on the earth is about \({0.50}^{\circ}\). Use this and the fact that the moon is about 384000 km away to find the approximate diameter of the moon.

Sol. \(tan\theta \simeq \theta = \frac{D}{r_{m}}\theta\) is radians
\(180^{\circ} = \pi\)-radians

\[{\Rightarrow \ {0.5}^{\circ} = \frac{\pi}{180} \times \frac{1}{2}rad = \frac{\pi}{360}rad }{\Rightarrow \ D = \theta \cdot r_{m} }{= \frac{\pi}{360} \times 384000\text{ }km = 3350\text{ }km }\]Q. 2 Find \(\frac{dx}{dt}\) (derivation of \(x\) with respect to \(t\) )
(i) \(x = \left( t^{2} + 1 \right)^{3}\)
(ii) \(x = sin2t\)

Sol. (i) \(\frac{dx}{dt} = 3\left( t^{2} + 1 \right)^{2}(2t)\)
\[= 6t\left( t^{2} + 1 \right)^{2} \](ii) \(\frac{dy}{dt} = 2cos2t\)
Q. 3 Find the derivative of \(y(x) = x^{3}/(x + 1)^{2}\) with respect to \(x\).

Sol. We can rewrite this function as \(y(x) = x^{3}(x + 1)^{- 2}\) and apply Equation ()
\[{\frac{dy}{dx} = (x + 1)^{- 2}\frac{d}{dx}\left( x^{3} \right) + x^{3}\frac{d}{dx}(x + 1)^{- 2} }{= (x + 1)^{- 2}3x^{2} + x^{3}( - 2)(x + 1)^{- 3} }{\frac{dy}{dx} = \frac{3x^{2}}{(x + 1)^{2}} - \frac{2x^{3}}{(x + 1)^{3}} }\]Q. 4 The velocity of particle is given by \(v = \sqrt{gx}\). Find its acceleration.

Sol. \(\ \frac{dv}{dt} = \frac{1}{2}(gx)^{- 1/2}\left( \text{ }g\frac{dx}{dt} \right)\)
\[{= \frac{1}{2}(gx)^{- 1/2}gv }{= \frac{1}{2}\text{ }g }\]Q. 5 If \(\overrightarrow{r} = \lbrack ucos\theta(\widehat{i}) + usin\theta(\widehat{j})\rbrack t + \frac{1}{2}\text{ }g( - \widehat{j})t^{2}\) then calculate equation of trajectory.

Sol. \(\ x = ucos\theta t\& y = usin\theta t - \frac{1}{2}gt^{2}\)

\[\begin{matrix} & y = u\left( \frac{x}{ucos\theta} \right) - \frac{1}{2}\text{ }g\left( \frac{x}{ucos\theta} \right)^{2} \\ & y = xtan\theta - \frac{{gx}^{2}}{2u^{2}\cos^{2}\theta} \end{matrix}\]

Q. 6 Find the value of definite integral: \(\int_{0}^{\pi}\mspace{2mu}\left( \frac{\pi t}{2} - \frac{t^{2}}{2} \right)dt\)

Sol. \(\ \int_{0}^{\pi}\mspace{2mu}\left( \frac{\pi t}{2} - \frac{t^{2}}{2} \right)dt\)

\[\begin{matrix} & \ = \left( \frac{\pi}{2}\left( \frac{t^{2}}{2} \right) - \frac{t^{3}}{6} \right)_{0}^{\pi} \\ & \ = \frac{\pi^{3}}{4} - \frac{\pi^{3}}{6} = \frac{\pi^{3}}{12} \end{matrix}\]

Q. 7 The velocity of a body moving in a straight line is given by \(v = \left( 3x^{2} + x \right)m/s\). Find acceleration at \(x = 2\text{ }m\).

Sol. \(\ v = \left( 3x^{2} + x \right)\)

\[\begin{matrix} & \frac{dv}{dx} = 6x + 1 \\ & a = v\frac{dv}{dx} = \left( 3x^{2} + x \right)(6x + 1) \\ & \text{~}\text{at}\text{~}x = 2m \\ & a = \left( 3 \times 2^{2} + 2 \right)(6 \times 2 + 1) = 182\text{ }m/s^{2} \end{matrix}\]

Q. 8 Find \((104)^{1/2}\)

Sol. \((100 + 4)^{1/2}\)

\[\begin{matrix} & \ = 10\lbrack 1 + 0.04\rbrack^{1/2} \\ & \ \simeq 10\lbrack 1 + 0.02\rbrack = 10.2 \end{matrix}\]