4.7 Motion in a plane

In this section, we shall see how to describe motion in two dimensions using vectors. Before we study motion in two dimensions using vectors, let’s refresh the topics from Integral calculus.

ELEMENTS OF CALCULUS

1. Differential Calculus
2. Integral Calculus

Differential Calculus: \(\frac{d y}{d x}\)

Using the concept of ‘differentlal coefflcient’ or ‘derivative’, we can easily define velocity and acceleration. Suppose we have a quantity \(y\) whose value depends upon a single variable \(x\), and is expressed by an equation defining \(y\) as some specific function of \(x\). This is represented as:
\(
y=f(x)
\)
This relationship can be visualised by drawing a graph of function \(y=f(x)\) regarding \(y\) and \(x\) as Cartesian coordinates, as shown in Figure 4k.

 

Figure 4k

Consider the point \(\mathrm{P}\) on the curve \(y=f(x)\) whose coordinates are \((x, y)\) and another point \(\mathrm{Q}\) where coordinates are \((x+\Delta x, y+\Delta y)\). The slope of the line joining \(\mathrm{P}\) and \(\mathrm{Q}\) is given by:

\(
\tan \theta=\frac{\Delta y}{\Delta x}=\frac{(y+\Delta y)-y}{\Delta x}
\)

Suppose now that the point \(Q\) moves along the curve towards \(P\). In this process, \(\Delta y\) and \(\Delta x\) decrease and approach zero; though their ratio \(\frac{\Delta y}{\Delta x}\) will not necessarily vanish. What happens to the line PQ as \(\Delta y \rightarrow 0, \Delta x \rightarrow 0\). You can see that this line becomes a tangent to the curve at point P. This means that \(\tan \theta\) approaches the slope of the tangent at \(\mathrm{P}\), denoted by \(m\):

\(
m=\lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{(y+\Delta y)-y}{\Delta x}
\)

The limit of the ratio \(\Delta y / \Delta x\) as \(\Delta x\) approaches zero is called the derivative of \(y\) with respect to \(x\) and is written as \(\mathrm{d} y / \mathrm{d} x\). It represents the slope of the tangent line to the curve \(y=f(x)\) at the point \((x, y)\).
Since \(y=f(x)\) and \(y+\Delta y=f(x+\Delta x)\), we can write the definition of the derivative as:

\(
\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d} f(x)}{\mathrm{dx}}=\lim _{\Delta \mathrm{x} \rightarrow 0} \frac{\Delta y}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left[\frac{f(x+\Delta x)-f(x)}{\Delta x}\right]
\)

VELOCITY AND ACCELERATION

The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero:

\(v=\lim _{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t}=\frac{\mathrm{d} x}{\mathrm{~d} t}\)

The acceleration (instantaneous acceleration) is the limiting value of the average acceleration as the time interval approaches zero:

\(a=\lim _{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}=\frac{\mathrm{d} v}{\mathrm{~d} t}=\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}\)

POINTS TO REMEMBER:

\(\frac{d y}{d x}\) for some common functions

y	\(\frac{d y}{d x}\)
\(x^{n}\)	\(n x^{n-1}\)
\(\ln x\)	\(\frac{1}{x}\)
\(e^{x}\)	\(e^{x}\)
\(\sin x\)	\(\cos x\)
\(\cos x\)	\(-\sin x\)
\(\tan x\)	\(\sec ^{2} x\)
\(\cot x\)	\(-\operatorname{cosec}^{2} x\)
\(\sec x\)	\(\sec x \tan x\)
\(\operatorname{cosec} x\)	\(-\operatorname{cosec} x \cot x\)
\(\frac{d}{d x}(a y)\)	\(a \frac{d y}{d x}\)
\(\frac{d}{d x}(u+v)\)	\(\frac{d u}{d x}+\frac{d v}{d x}\)
\(\frac{d}{d x}(u v)\)	\(u \frac{d v}{d x}+v \frac{d u}{d x}\)
\(\frac{d}{d x}\left(\frac{u}{v}\right)\)	\(\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}\)
\(\frac{d y}{d x}\)	\(\frac{d y}{d u} \cdot \frac{d u}{d x}\)

Example 1:

Find \(\frac{d y}{d x}\) if \(y=e^{x} \sin x\)

Solution: \(y=e^{x} \sin x\).
\(
\begin{aligned}
\frac{d y}{d x} &=\frac{d}{d x}\left(e^{x} \sin x\right)=e^{x} \frac{d}{d x}(\sin x)+\sin x \frac{d}{d x}\left(e^{x}\right) \\
&=e^{x} \cos x+e^{x} \sin x=e^{x}(\cos x+\sin x)
\end{aligned}
\)

MAXIMA AND MINIMA

One of the great powers of calculus is in the determination of the maximum or minimum value of a function. 

Figure 4l

Assume a quantity \(y\) depends on another quantity \(x\) in a manner shown in figure 4l. It becomes maximum at \(x_{1}\) and minimum at \(x_{2}\).

Just before the maximum, the slope is positive, at the maximum, it is zero and just after the maximum, it is negative. Thus, \(\frac{d y}{d x}\) decreases at a maximum, and hence the rate of change of \(\frac{d y}{d x}\) is negative at a maximum i.e.
\(
\frac{d}{d x}\left(\frac{d y}{d x}\right)<0 \text { at a maximum. }
\)
The quantity \(\frac{d}{d x}\left(\frac{d y}{d x}\right)\) is the rate of change of the slope. It is written as \(\frac{d^{2} y}{d x^{2}}\).

Thus, the condition of a maximum is
\(\begin{aligned}
&\frac{d y}{d x}=0 \\
&\frac{d^{2} y}{d x^{2}}<0
\end{aligned}\)

Similarly, at a minimum the slope changes from negative to positive. The slope increases at such a point and hence \(\frac{d}{d x}\left(\frac{d y}{d x}\right)>0\).

The condition of a minimum is 

\(\begin{aligned}
&\frac{d y}{d x}=0 \\
&\frac{d^{2} y}{d x^{2}}>0
\end{aligned}\)

Example 2:

The height reached in time \(t\) by a particle thrown upward with a speed \(u\) is given by
\(
h=u t-\frac{1}{2} g t^{2}
\)
where \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) is a constant. Find the time taken in reaching the maximum height.

Solution: The height \(h\) is a function of time. Thus, \(h\) will be maximum when \(\frac{d h}{d t}=0\). We have,
\(
\begin{aligned}
h &=u t-\frac{1}{2} g t^{2} \\
\frac{d h}{d t} &=\frac{d}{d t}(u t)-\frac{d}{d t}\left(\frac{1}{2} g t^{2}\right)
\end{aligned}
\)

\(\begin{aligned}
&=u \frac{d t}{d t}-\frac{1}{2} g \frac{d}{d t}\left(t^{2}\right) \\
&=u-\frac{1}{2} g(2 t)=u-g t
\end{aligned}
\)
For maximum \(h\),
\(
\frac{d h}{d t}=0
\)
or, \(u-g t=0 \quad\) or, \(\quad t=\frac{u}{g} .\)

INTEGRAL CALCULUS

Integral is the representation of the area of a region under a curve. Let us take the example shown in Figure 4m. Suppose a variable force \(F(x)\) acts on a particle in its motion along \(x\) – axis from \(x=\) a to \(x=b\). The problem is to determine the work done \((W)\) by the force on the particle during the motion. 

Figure 4m shows the variation of \(F(x)\) with \(x\). If the force were constant, work would be simply the area \(F(b-a)\) as shown in Figure 4m. But in the general case, force is varying.

Figure 4m

To calculate the area under this curve(Figure 4m), let us employ the following trick. Divide the interval on \(x\)-axis from \(a\) to \(b\) into a large number \((N)\) of small intervals: \(x_{0}(=a)\) to \(x_{1}, x_{1}\) to \(x_{2} , x_{2}\) to \(x_{3}\), \(x_{N-1}\) to \(x_{N}(=b)\). The area under the curve is thus divided into \(N\) strips. Each strip is approximately a rectangle, since the varlation of \(F(x)\) over a strip is negligible. The area of the \(i^{\text {th }}\) strip shown [Figure 4m] is then approximately
\(
\Delta A_{i}=F\left(x_{i}\right)\left(x_{i}-x_{i-1}\right)=F\left(x_{i}\right) \Delta x
\)
where \(\Delta x\) is the width of the strip which we have taken to be the same for all the strips. You may wonder whether we should put \(F\left(x_{i-1}\right)\) or the mean of \(F\left(x_{i}\right)\) and \(F\left(x_{i-1}\right)\) in the above expression. If we take \(N\) to be very very large \((N \rightarrow \infty)\), it does not really matter, since then the strip will be so thin that the difference between \(F\left(x_{i}\right)\) and \(F\left(x_{i-1}\right)\) is vanishingly small. The total area under the curve then is:
\(
A=\sum_{i=1}^{N} \Delta A_{i}=\sum_{i=1}^{N} F\left(x_{i}\right) \Delta x
\)
The limit of this sum as \(N \rightarrow \infty\) is known as the integral of \(F(x)\) over \(x\) from a to b. It is given a special symbol as shown below:
\(
A=\int_{a}^{b} F(x) d x
\)

Indefinite Integral

An indefinite integral is a function that takes the antiderivative of another function. A most significant mathematical fact is that integration is, in a sense, an inverse of differentiation.
Suppose we have a function \(g(x)\) whose derivative is \(f(x)\), i.e. \(f(x)=\frac{d g(x)}{d x}\)
The function \(g(x)\) is known as the indefinite integral of \(f(x)\) and is denoted as:
\(
g(x)=\int f(x) d x
\)

Definite IntegralAn integral with lower and upper limits is known as a definite integral. It is a number. Indefinite integral has no limits; it is a function. A fundamental theorem of mathematics states that
\(
\int_{a}^{b} f(x) d x=\left.g(x)\right|_{a} ^{b} \equiv g(b)-g(a)
\)

Integration Formulae:

Some useful rules for integration are as follows:
(1) \(\int c f(x) d x=c \int f(x) d x\) where \(c\) is a constant

(2) \(Let \int f(x) d x=F(x)\) then \(\int f(c x) d x=\frac{1}{c} F(c x) .\)

(3) \(\int[f(x)+g(x)] d x=\int f(x) d x+\int g(x) d x\).

Example 3: The position of a particle is given by
\(
\mathbf{r}=3.0 t \hat{\mathbf{i}}+2.0 t^2 \overline{\mathbf{j}}+5.0 \hat{\mathbf{k}}
\)
where \(t\) is in seconds and the coefficients have the proper units for \(\mathbf{r}\) to be in metres. (a) Find \(\mathbf{v}(t)\) and \(\mathbf{a}(t)\) of the particle. (b) Find the magnitude and direction of \(\mathbf{v}(t)\) at \(t=1.0 \mathrm{~s}\).

Answer:
\(
\begin{aligned}
\mathbf{v}(t) & =\frac{\mathrm{d} \mathbf{r}}{\mathrm{d} t}=\frac{\mathrm{d}}{\mathrm{d} t}\left(3.0 t \hat{\mathbf{i}}+2.0 t^2 \hat{\mathbf{j}}+5.0 \hat{\mathbf{k}}\right) \\
& =3.0 \hat{\mathbf{i}}+4.0 t \hat{\mathbf{j}}
\end{aligned}
\)
\(\mathbf{a}(t)=\frac{\mathrm{d} \mathbf{v}}{\mathrm{d} t}=+4.0 \hat{\mathbf{j}}\)
\(a=4.0 \mathrm{~m} \mathrm{~s}^{-2}\) along \(y\)-direction
At \(t=1.0 \mathrm{~s}, \quad \mathbf{v}=3.0 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}}\)
It’s magnitude is \(v=\sqrt{3^2+4^2}=5.0 \mathrm{~m} \mathrm{~s}^{-1}\) and direction is
\(
\theta=\tan ^{-1}\left(\frac{v_y}{v_x}\right)=\tan ^{-1}\left(\frac{4}{3}\right) \cong 53^{\circ} \text { with } x \text {-axls. }
\)

Example 4: From the curve given in the figure below find \(\frac{d y}{d x}\) at \(x=2\), 6 and 10.

Answer: The tangent to the curve at \(x=2\) is \(A C\). Its slope is \(\tan \theta_1=\frac{A B}{B C}=\frac{5}{4}\)
Thus, \(\quad \frac{d y}{d x}=\frac{5}{4}\) at \(x=2\).
The tangent to the curve at \(x=6\) is parallel to the \(X\)-axis.
Thus, \(\quad \frac{d y}{d x}=\tan \theta=0\) at \(x=6\).
The tangent to the curve at \(x=10\) is \(D F\). Its slope is
\(
\tan \theta_2=\frac{D E}{E F}=-\frac{5}{4} .
\)
Thus, \(\quad \frac{d y}{d x}=-\frac{5}{4}\) at \(x=10\).

Example 5: \(\text { Evaluate } \int_3^6\left(2 x^2+3 x+5\right) d x \text {. }\)

Answer:

\(
\begin{aligned}
\int\left(2 x^2+\right. & 3 x+5) d x \\
& =\int 2 x^2 d x+\int 3 x d x+\int 5 d x \\
& =2 \int x^2 d x+3 \int x d x+5 \int x^0 d x
\end{aligned}
\)
\(
\begin{aligned}
& =2 \frac{x^3}{3}+3 \frac{x^2}{2}+5 \frac{x^1}{1} \\
& =\frac{2}{3} x^2+\frac{3}{2} x^2+5 x
\end{aligned}
\)
Thus,
\(
\begin{aligned}
& \int_3^6\left(2 x^2+3 x+5\right) d x=\left[\frac{2}{3} x^3+\frac{3}{2} x^2+5 x\right]_3^6 \\
& \quad=\frac{2}{3}(216-27)+\frac{3}{2}(36-9)+5(6-3) \\
& \quad=126+40 \cdot 5+15=181 \cdot 5 .
\end{aligned}
\)