The figure below shows a rod of resistive material. Its resistance per unit length increases in the positive direction of the x axis. At any position $x$ along the rod, the resistance $dR$ of a narrow section of width $dx$ is given by $dR=5x dx$, where $dR$ is in ohms and $x$ is in meters. I want to slice off a length of the rod between $x=0$ and some position $x=L$ and then connect that length to a battery with a potential difference $V=5V$. I want the current in the lenght to transfer energy to thermal energy at the rate of $200 W$. I am going to show how to find the position $x=L$ such that these conditions are met.

Since I am dealing with a resistor, I can relate the given rate of energy transfer to thermal energy, $P$, to the Resistance, $R$, using the equation $P=\frac{V^2}{R}$, which is the resistive dissipation of the resistive battery. Essentially electric potential energy is converted to internal thermal energy via collisions between charge carriers and atoms. After plugging in the given voltage for the battery and the rate of thermal energy transfer, and then solving for $R$, I obtain $R=1.25\times10^{-1}\Omega$, which is the resistance of the battery. Next, I can integrate the given differential, $dR=5x dx$, which will equal $R$ and set it equal to $R=1.25\times10^{-1}\Omega$. I am doing this because the rate of thermal energy transfer depends upon the Resistance, which in turn depends on the length of the rod. After applying the Fundamental Theorem of Calculus to the integral and solving for $R$, I obtain $L=2.236\times10^{-1}m$

In the figure below a box (total mass $m_1$ = 1.65 kg) and another box (total mass $m_2$ =3.30 kg) slide down an inclined plane while attached b a massless rod parallel to the plane. The angle if incline is $\Theta$ = 30.0°. The coefficient of kinetic friction between $m_1$ and the incline is $\Theta$ =30.0°. The coefficient of kinetic friction between $\mu_1$ and the incline is $\mu_1$ = .226; that between $m_2$ and the incline is $\mu_2$ = .113. Compute the tension in the rod and the magnitude of the common acceleration of the two boxes.

Once the force vectors are labeled and the axis components are resolved we can set up Force Systems to compute the tension and magnitude of common acceleration. The systems are as follows:

System 1a: m1, perpendicular, a = 0

Force balance, $|C_1| = |m_1gcos \theta|$.

System 1b: m1, parallel, a

$+|m_1gsin \theta|+|T|-|F_{f1}|=m_1a$

System 2a: m2, perpendicular, a = 0

Force balance, $|C_2| = |m_{2}gcos \theta|$

System 2b: m2, parallel, a

$+|m_2gsin \theta| - |T| - |F_{f2}|=m_2a$

$F_{f1}$ Model

$|F_{f1}| = .226C_1$

$F_{f2}$ Model

$|F_{f2}| = .113C_2$

Since there is no acceleration in the vertical dimension we only add the horizontal systems for both masses to obtain $\sum F$ of the system. This gives us:

$gsin \theta(m_1+m_2) - F_{f1} - F_{f2} = (m_1 + m_2)a$

$a = g[sin 30(m_1+m_2) - (\mu m_1 + \mu m_2)cos 30$

$a = 3.62m/s^2$

$|T| = |m_1a| + |F_f|-|m_1gsin \theta |$

$|T| = |(1.65)(3.62)| + |.726m_1gcos30|-|(1.65)(9.8)sin30|$

$|T| = 1.05 N$

In this post I am going to introduce the formal definition of a limit. I am still trying to figure out the symbol formatting so bear with me. If $f(x)$ becomes arbitrarily close to a single number $L$ as $x$ approaches $c$ from either side, then the limit of $f(x)$ as $x$ approaches $c$ is $L$, which can be written as $\displaystyle\lim_{x\to\ c} f(x) = L$. This definition may seem formal but isn’t because exact meanings have not yet been given to the two phrases  “$f(x)$ becomes arbitrarily close to $L$” and “$x$ approaches $c$“. In the figure below we can let $\epsilon$ represent a (small) positive number. Then the phrase $"f(x)$ becomes arbitrarily close to $L"$ means that $f(x)$ lies in the interval $(L - \epsilon, L + \epsilon)$. Using absolute value, we can write this as $|f(x) - L| < \epsilon$. Similarly, the phrase “$x$ approaches $c$” means that there exists a positive number $\delta$ such that $x$ lies in either the interval $(c - \delta, c)$ or the interval $(c, c + \delta)$. This fact can be concisely expressed by the double inequality $0 < |x -c| < \delta$. The first inequality $0 < |x -c|$ states that the distance between $x$ and $c$ is more than 0, and expresses the fact that $x \neq c$. The second inequality $|x - c| < \delta$ says that $x$ is within $\delta$ units of $c$.

This brings us to a formal definition which can be stated as following: Let $f$ be a function defined on an open interval containing $c$ (except possibly at $c$) and let $L$ be a real number. The statement $\displaystyle\lim_{x\to\ c} f(x) = L$ means that for each $\epsilon > 0$ there exists a $\delta > 0$ such that if $0 < |x-c| < \delta$, then $|f(x) - L| < \epsilon$. This can be expressed symbolically as $\forall \epsilon > 0 \exists \delta > 0 : \forall x (0 < |x - c| < \delta \Rightarrow |f(x) - L| < \epsilon)$  .