# Solve Things Symbolically

You might noticed that most of the physics problem I’ve posted are solved symbolically. This is because solving problems symbolically can have many advantages, especially in physics. From my experience, they are :

1. Faster. We spend less time picking up our calculator.
2. More Accurate. Rounding up the answer throughout the calculations may lead to a major error at the end. By solving symbolically, we will only need to perform the round-up once when we are finished with the final substitution.
3. Less Mistake. It’s very easy to overlook 6 as a 0, especially when you are nervous in your exam. By solving symbolically, we are not likely to make mistakes for substitutions and arithmetic operations.
4. Stronger Understanding of Problem. Does acceleration matters in this problem? If the time increased, will it travel further? Since symbolic answers normally looks nice, all these questions can be answered from it! Other than that, we can determine other special cases in the problem.
5. Reusable. When you calculated $g$ as $9.8$ instead of the problem requirement $g=10$, you don’t have to solve the problem all over again but substitute the correct value.

Although it’s better solving problems symbolically, it really depends on the type of problem. For example, you won’t want to solve a system of 3 equations with only symbols. Now, let’s look at a typical problem regarding heat transfer, and solve it symbolically.

Two metals $A$ and $B$ with length $50\text{ cm}$ and $70\text{ cm}$ is wielded together. If we constantly heat up the end of metal $A$ and $B$ to $473\text{ K}$ and $323\text{ K}$, what is the temperature of the wielding section of the rod?

( $k$ is the thermal conductivity of metal, with unit $Wm^{-1}k^{-1}$ )

Solution :

The wielding section will reaches thermal equilibrium when the heat transferred from metal $A$ and $B$ is the same. Hence,

\begin{aligned} \displaystyle Q_A &= Q_B \\ \\ k_AA\frac{T_A-T}{l_A} &= k_BA\frac{T-T_B}{l_B} \\ \\ T_Ak_Al_B - Tk_Al_B &= Tk_Bl_A - T_Bk_Bl_A \\ \\ T(k_Al_B+k_Bl_A) &= T_Ak_Al_B+T_Bk_Bl_A \\ \\ T\bigg(\frac{k_A}{l_A}+\frac{k_B}{l_B}\bigg) &= T_A\frac{k_A}{l_A}+T_B\frac{k_B}{l_B} \\ \\ T &= \frac{T_A\frac{k_A}{l_A}+T_B\frac{k_B}{l_B}}{\frac{k_A}{l_A}+\frac{k_B}{l_B}}\end{aligned}

With that formula in hand, we only have to substitute the values into their responding variable :

\begin{aligned} \displaystyle T &= \frac{T_A\frac{k_A}{l_A}+T_B\frac{k_B}{l_B}}{\frac{k_A}{l_A}+\frac{k_B}{l_B}} \\ &= \frac{473\times\frac{200}{50}+323\times\frac{100}{70}}{\frac{200}{50}+\frac{100}{70}} \\ &= 433.5\text{ K}\end{aligned}

Since this problem often appears as a sub-problem in the UEC exam, it’s worth to memorize it. Next, let’s solve a similar problem, but UEC-leveled.

What is the value of $T_1$ and $T_2$?

Consider the heat transfer between metal $P$ and $Q$,

\begin{aligned} \displaystyle T_1 &= \frac{473\frac{200}{0.5}+T_2\frac{100}{0.5}}{\frac{200}{0.5}+\frac{100}{0.5}} \\ 600T_1 &= 189200 + 200T_2 \\ 3T_1 - T_2 &= 926 \end{aligned}

Consider the heat transfer between $Q$ and $R$,

\begin{aligned} \displaystyle T_2 &= \frac{T_1\frac{100}{0.5}+323\frac{400}{0.5}}{\frac{100}{0.5}+\frac{400}{0.5}} \\ 1000T_2 &= 200T_1 + 258400 \\ 5T_2 - T_1 &= 1292\end{aligned}

Solve the two system of equation :

$\displaystyle \left\{ \begin{array}{l l} 3T_1-T_2 &= 866\\ 5T_2-T_1 &= 1292 \end{array} \right.$

$T_1 = 430.14\text{ K} \\ T_2 = 344.43\text{ K}$

It is true that the only way to be good at problem solving is through solving problems. However, there’s no point doing the same thing over and over. Fully understand the problem is a much better approach.