Differential equation: A special case ($f'(x) - f(x) =...$)

Question:
Solve a differential equation.
\[
f'(x) - f(x) = -2\sin x
\]
The initial condition is $f(0) = 1$.

Answer:
When you find the form, $f'(x) - f(x) =...$, we can utilize the following relation:
\[
[f(x)e^{-x}]' = e^{-x}[f'(x) - f(x)]
\]
If we apply this formula to the above differential equation, we get
\[
[f(x)e^{-x}]' = e^{-x}[-2\sin x]
\]
For the right hand side, we used $f'(x) - f(x) = -2\sin x$. Then, integrate both sides of above equation.
\begin{equation}
f(x)e^{-x} = -2\int e^{-x}\sin x dx
\end{equation}
For the right hand side, we just consider integration by parts of the following factor:
\begin{eqnarray*}
& & \int e^{-x}\sin x dx = -e^{-x}\sin x - \int (-1)e^{-x}\cos x dx  \\
\rightarrow & & \int e^{-x}\sin x dx = -e^{-x}\sin x - e^{-x}\cos x - \int (-1)e^{-x}(-1)\sin x dx  \\
\rightarrow & & \int e^{-x}\sin x dx + \int e^{-x}\sin x dx = -e^{-x}\sin x - e^{-x}\cos x   \\
\rightarrow & & 2\int e^{-x}\sin x dx = -e^{-x}(\sin x + \cos x)   \\
\rightarrow & & \int e^{-x}\sin x dx = -\frac{1}{2}e^{-x}(\sin x + \cos x) + C
\end{eqnarray*}
Plug this into (1).
\begin{eqnarray}
f(x)e^{-x} &=& e^{-x}(\sin x + \cos x) + C  \\
\end{eqnarray}
Use the initial condition, $f(0) = 1$.
\[
f(0) = \sin 0 + \cos 0 + C = 1 + C = 1
\]
Thus, $C = 0$. The final result is
\[
f(x) = \sin x + \cos x
\]

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A wheeled inclined plane with an object: Newton's equation of motion

Question:
Consider an inclined cart with motorized wheels as shown. Assume that the coefficients of static and kinetic frictions are equal on the surface of the cart.
(1) After an object is put on the cart gently, it starts falling with acceleration of $a$. The cart is fixed, so it does not move at all. Suppose the inclined angle is the angle where the object just starts falling. Find the coefficient of static friction.
(2) The cart moves ahead and begins accelerated. The, the object becomes stopped falling. Find the acceleration of the cart.

Answer:
(1) From the diagram, we can set up two equations of motion for the moving axis and the axis perpendicular to it.
\begin{eqnarray}
mg\sin\theta -\mu N &=& ma  \\
N - mg\cos\theta &=& 0
\end{eqnarray}
where $\mu$, $N$, and $\mu N$ are the coefficient of static friction, the normal force and the frictional force, $f$. From (2), we obtain the normal force, $N = mg\cos\theta$. Then, equation (1) becomes
\begin{equation}
mg\sin\theta - \mu mg\cos\theta = ma
\end{equation}
All $m$'s are cancelled out. Solve for $\mu$.
\begin{eqnarray}
& & g\sin\theta - \mu g\cos\theta = a  \\
& & \rightarrow \mu g\cos\theta = g\sin\theta - a  \\
& & \rightarrow \mu = \frac{g\sin\theta - a}{g\cos\theta}
\end{eqnarray}

(2) As shown in the figure, the acceleration of the cart works against the gravitational falling force. The equation of motion in the moving direction becomes
\begin{equation}
mg\sin\theta - \mu N -ma\cos\theta = 0
\end{equation}
The left hand side is the net forces in that direction. The right hand side is zero because there is no motion. Now, set up the other equation of motion for the axis perpendicular to the moving direction.
\begin{equation}
N - ma\sin\theta - mg\cos\theta = 0
\end{equation}
From (8), solve for $N$.
\begin{equation}
N  = ma\sin\theta + mg\cos\theta
\end{equation}
Plug this into (7).
\begin{eqnarray}
& &mg\sin\theta - \mu (ma\sin\theta + mg\cos\theta) -ma\cos\theta = 0  \\
& & \rightarrow g(\sin\theta - \mu\cos\theta) - a(\mu\sin\theta + \cos\theta) = 0  \\
& & \rightarrow a = \frac{\sin\theta - \mu\cos\theta}{\mu\sin\theta + \cos\theta}g
\end{eqnarray}

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Inscribed and circumscribed circles

Question:
Two circles are given
\begin{eqnarray}
& & x^2+y^2-6ax+2ay+20a-10 = 0  \\
& & x^2+y^2 = 4
\end{eqnarray}
Find $a$ so these circles are inscribed or circumscribed.

Answer:
Equation (1) can be arranged as follows:
\begin{eqnarray}
& & x^2 -6ax + (3a)^2 -(3a)^2 + y^2 +2ay + (a)^2 -(a)^2 = 10 - 20a  \nonumber \\
& & \rightarrow x^2 -6ax + (3a)^2 + y^2 +2ay + (a)^2  = 10 - 20a + (3a)^2 + (a)^2 \nonumber  \\
& & \rightarrow (x - 3a)^2 + (y + a)^2 = 10a^2 -20a +10  \nonumber \\
& & \rightarrow (x - 3a)^2 + (y + a)^2 = 10(a-1)^2
\end{eqnarray}
Therefore, (2) has the center $(0, 0)$ and the radius $2$. Circle (3) has the center $(3a, -a)$ and radius $\sqrt{10}|a-1|$. The distance between those centers is
\[
d = \sqrt{(3a-0)^2+(-a-0)^2}=\sqrt{10a^2}=\sqrt{10}|a|
\]
If $a=1$, the radius becomes zero. Thus, $a \neq 1$. Let $r_1$ and $r_2$ be radii of two circles. In general, when $d=r_1 + r_2$, they are circumscribed. When $d=|r_1 - r_2|$, they are inscribed. Then, these conditions can be combined as $d = |r_1 \pm r_2|$. Now consider the case in $a>1$. Use the definition above:
\begin{equation}
\sqrt{10}|a| = | \sqrt{10}(a-1) \pm 2 |
\end{equation}
Square both sides and solve for $a$.
\begin{equation}
a = \frac{5 \pm \sqrt{10}}{10} < 1
\end{equation}
The result is contradicted. Consider when $a<1$.
\begin{equation}
\sqrt{10}|a| = | \sqrt{10}(1-a) \pm 2 |
\end{equation}
Likewise, solve for $a$.
\begin{equation}
a = \frac{5 \pm \sqrt{10}}{10}
\end{equation}
which is the proper result.

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Linear transformation: Two dimensional rotation matrix

Question:
There is an ellipse:
\[
x^2 + \frac{y^2}{4} = 1
\]
If it is on a new coordinate that is rotated by 45$^o$, find the equation of the ellipse with the new coordinate, ($X, Y$).

Answer:
When the coordinate axes are rotated, the old coordinate ($x, y$) will be expressed by the new coordinate ($X, Y$).
\[
\left( \begin{array}{c} x \\ y \end{array} \right)
 =
\left( \begin{array}{cc} \cos\theta & -\sin\theta \\
                                      \sin\theta & \cos\theta
\end{array} \right)
\left( \begin{array}{c} X \\ Y  \end{array} \right)
\]
Therefore, we have
\[
\left\{
\begin{array}{l}
x = X\cos\theta -Y\sin\theta = X\cos 45^o -Y\sin 45^o =\frac{X-Y}{\sqrt{2}} \\
y = X\sin\theta + Y\cos\theta = X\sin 45^o + Y\cos 45^o = \frac{X+Y}{\sqrt{2}}
\end{array}
\right.
\]
Then, they are plugged into the original equation of ellipse.
\[
\frac{1}{2}(X-Y)^2 + \frac{1}{8}(X+Y)^2 = 1
\]
Or you can also express it as
\[
\frac{1}{8}(5X^2 -6XY +5Y^2) = 1
\]

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A sliding block on a large cart: Conservation of energy and momentum

Question:
A block B is placed on the cart A that has the frictionless surface and going to be sliding the height of $h$ until the frictional bottom (the shaded part). The cart has frictionless wheels and it moves due to block's falling on the slope. The masses of the block and the cart are $m$ and $M$, respectively. Find each final velocity.

Answer:
The momentum in the horizontal direction should be conserved. Let velocities be $V_A$ and $V_B$. The initial total momentum is zero since they are initially at rest. We assume that the final total momentum is the sum of these momenta. From conservation of momentum, we have
\begin{equation}
0 = MV_A+mV_B
\end{equation}
The total energies in initial and final states must be conserved. The LHS and RHS of the following equation are initial total and final total energies, respectively.
\begin{eqnarray}
 (0+mgh) + (0 + 0) &=& \left(\frac{1}{2}mV_B^2+0\right) + \left(\frac{1}{2}MV_A^2 + 0 \right) \nonumber  \\
\rightarrow \ \frac{1}{2}MV_A^2 &+& \frac{1}{2}mV_B^2 = mgh
\end{eqnarray}
Now, derive $V_B$ and $V_A$ from these equations. From (1), we have
\begin{equation}
V_A = -\frac{m}{M}V_B
\end{equation}
Plug it in (2).
\begin{eqnarray}
M \left(\frac{m}{M}\right)^2 V_B^2 + mV_B^2 &=& 2mgh  \\
\left(\frac{m}{M}+1\right)V_B^2 &=& 2gh  \\
V_B = \sqrt{\frac{2Mgh}{m+M}}
\end{eqnarray}
Then, plug (6) in (3).
\begin{equation}
V_A = -m\sqrt{\frac{2gh}{M(m+M)}}
\end{equation}

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Value of an infinite series I

Question:
Consider the following infinite series:
\[S_n = 2-\frac{3}{2}+\frac{3}{2}-\frac{4}{3}+\frac{4}{3}-\frac{5}{4}+ \cdots\]
Find the value(s) when $n \rightarrow \infty$.

Answer:
This series can be generalized as follows:
\[
S_n = \sum^{\infty}_{n=1}\left(\frac{n+1}{n}-\frac{n+2}{n+1}\right)
\]
The value depends on the number of terms. When $n$ is even, the last term should be $-\frac{n+2}{n+1}$. Namely,
\begin{eqnarray}
S_n &=& 2-\frac{3}{2}+\frac{3}{2}-\frac{4}{3}+\frac{4}{3}-\frac{5}{4}+ \cdots +\frac{n+1}{n}-\frac{n+2}{n+1}  \\
   &=& 2 - \frac{n+2}{n+1}
\end{eqnarray}
Except the first and last terms, all of them are cennceled out. Therefore, the value of the infinite series is
\[
\lim_{n \rightarrow \infty}S_n = \lim_{n \rightarrow \infty}\left(2-\frac{1+\frac{2}{n}}{1+\frac{1}{n}}\right)=1
\]
When $n$ is odd, the series ends up with $+\frac{n+1}{n}$. Namely,
\begin{eqnarray}
S_n &=& 2-\frac{3}{2}+\frac{3}{2}-\frac{4}{3}+\frac{4}{3}- \cdots -\frac{n+1}{n}+\frac{n+1}{n}  \\
   &=& 2
\end{eqnarray}
Thus we have the value.
\[
\lim_{n \rightarrow \infty}S_n = 2
\]
The value oscillates between 1 and 2.

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Kinematic diagram: Time vs velocity

Question:
The time vs velocity diagrams of a 10-kg object for $x$ and $y$ directions are given as shown. Describe each motion for each time interval.
(1) 0 < $t$ < 2 seconds
(2) 2 < $t$ < 3 seconds
(3) 3 < $t$ < 4 seconds
(4) at 2 seconds

Answer:
(1) From the diagram, the object moves at the constant velocity, 1 m/s, in $x$ direction, but it gets acceleration of $\frac{3}{2}$ m/s$^2$ in $y$ direction. The slope indicates the acceleration. Thus, it is exerted by 15 N in that direction because $F=ma$.

(2) For both directions, during this time interval, the object moves at the constant velocity. The diagrams show the flat slopes; namely, no acceleration and no force on the object.

(3) The accelerations from 3 s to 4 s are
\begin{equation*}
a_x = a_y = \frac{0-3}{4-3} = -3 \ \mathrm{m/s^2}
\end{equation*}
The magnitude of the negative acceleration in both directions is equal, so the force is directed in opposite to 45 degrees on $x-y$ plane. The magnitude of force is given by
\[
|F| = 10 \times \sqrt{3^2+3^2} = 42 \ \mathrm{N}
\]

(4) At 2 seconds, the velocity in $x$ direction is suddenly changed from 1 m/s to 3 m/s. This indicates the change in momenta; namely, there is an impulse in positive $x$ direction. The impulse is calculated as
\[
I = mv_f - mv_i = 10 \cdot 3 - 10 \cdot 1 = 20 \ \mathrm{N\cdot s}
\]

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