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Economics

Why is the supply curve upward-sloping?

The law of supply states that as price increases, so does quantity supplied and vice versa, as seen in the supply schedule below: |Price|Quantity supplied| |:-:|:-:| |$1|5| |$2|10| |$3|15| |$4|20| |$5|25| Because it reflects this relationship, the supply curve is positively sloped. To better understand the law of supply, keep in mind that suppliers can make choices about what they produce. For example, a restaurant can produce a wide range of meals. If the price consumers are willing to pay for one type of meal increases relative to the others, it makes sense that restaurants would begin making more of that type of meal. This **profit motive** is one reason the supply curve is upward sloping. Alongside this idea is the idea of **opportunity cost**. Every time a restaurant makes one type of meal (e.g., pasta), they give up the ability to make a different meal (e.g., pizza). When the price of pasta goes up the opportunity cost of making pizza increases, and restaurateurs reallocate their resources to producing more pasta. This is especially true if other firms, seeing high prices in the industry, switch their production to seek these higher prices. In the example above, restaurants increase production by reallocating their existing resources. But what if restaurants want to produce more meals than they can with their given resources? To do so, the restaurant must buy new machinery or hire more workers, meaning the restaurant will face additional costs. This is a third reason price has a positive relationship with quantity supplied—production *may* only be able to increase if prices rise enough to cover **increased production costs**.

Physics

How to calculate percentage uncertainty

In science, all measurements have an associated uncertainty. This uncertainty communicates the precision to which the measurement was taken. If a scientific result is the product of a series of calculations of measured values, then the result itself will have an associated uncertainty to show the degree of confidence in the result. The percent uncertainty of a value is the same as the fractional or relative uncertainty of the value expressed as a percentage. It can be found from the following formula: $\hspace{3em}$ % uncertainty = $\dfrac{\textrm{absolute uncertainty}}{\textrm{measured value}} \times 100 \% $ For example, if the time for a cart to roll down a ramp is measured to be 3.6 s ± 0.2 s we can see the absolute uncertainty on the measurement is 0.2 s. We want to express this as a percentage of the measured value: $\hspace{3em}$ % uncertainty = $\dfrac{\textrm{absolute uncertainty}}{\textrm{measured value}} \times 100 \% = \dfrac{0.2}{3.6}=5.6 \%$ The method used to find the percentage uncertainty of a calculated value depends on the mathematical operation being performed. If the values involved are being added or subtracted, then the absolute uncertainties need to be added to find the absolute uncertainty on the result, and then the percentage uncertainty can be found. For example, if three length measurements are being added together: $\hspace{3em}$ (2.0 ± 0.1) cm + (8.2 ± 0.2) cm + (2.5 ± 0.5) cm The result will be the sum of the lengths and the absolute uncertainty will be the sum of the individual uncertainties: $\hspace{3em}$ Length = 2.0 + 8.2 + 2.5 = 12.7 cm $\hspace{3em}$ Absolute uncertainty = 0.1 + 0.2 + 0.5 = 0.8 cm $\hspace{3em}$ The result is expressed as 12.7 cm ± 0.8 cm. We can now find the percentage uncertainty on the answer as follows: $\hspace{3em}$ % uncertainty = $\dfrac{\textrm{absolute uncertainty}}{\textrm{measured value}} \times 100 \% = \dfrac{0.8}{12.7}=6.3 \%$ If two values are being divided or multiplied together, their individual percentage uncertainties must be found. The final result's percentage uncertainty will be the sum of the individual percentage uncertainties. For example, consider the following calculation of the density of an object. The mass $m$ and volume $V$ of the object are measured to be $\hspace{3em}$ $m$ = 4.5 g ± 0.2 g $\hspace{3em}$ $V$ = 2.5 cm$^3$ ± 0.5 cm$^3$ The density can be found from the formula $\hspace{3em}$ $\rho= \dfrac{m}{V}$ Giving $\hspace{3em}$ $\rho=\dfrac{4.5}{2.5}=1.8$ g cm$^{-3}$ Now we need to find the percentage uncertainty of the result. First, we find the percentage uncertainties on the mass and the volume: $\hspace{3em} \dfrac{\Delta m}{m} \times 100 \% = \dfrac{0.2}{4.5} \times 100 \% =4.4 \% \hspace{2em} \hspace{2em}$ and $\hspace{3em}\dfrac{\Delta V}{V} \times 100 \%= \dfrac{0.5}{2.5} \times 100 \%=20 \%$ Because we are dividing the terms, we add the percentage uncertainties: $\hspace{3em}$ density % uncertainty = mass % uncertainty + volume % uncertainty = 4.4% +20% $\hspace{10em}$ = 24%

Mathematics

What is the domain of a function?

The domain of a function is all of the allowed input values of a function. The domain can be defined in a number of different ways. Here are some examples. - Using set notation - $x=\{2,3,4,5\}$, this tells us that $x$, the input to the function, can be any of the numbers in the curly braces. - $x \in \mathbb{R}, x \neq -2 \hspace{1em}$ this means $x$, the input to the function, can be any real number except $-2$. - Using inequalities - $-1 < x \leq 4$ where $x \in \mathbb{R}\hspace{1em}$ this means $x$, the input, can be any real number larger than $-1$ and less than or equal to $4$. - $x > 5$ this means $x$, can be any real number larger than $5$. - Using words - $x$ can be any integer less than or equal to $-2$. Sometimes we need to identify the domain from given information. For example, if we are given the graph of a function, we can identify the domain by looking at the $x$-axis. The end-points of the function can either be a - Filled circle $\rightarrow$ indicating that the end-point is included in the domain - Open circle $\rightarrow$ indicating that the end-point is __not__ included in the domain - Arrow $\rightarrow$ indicating that the domain tends to infinity If we are dealing with a rational (fractional) function then any $x$-values that make the denominator of the function equal zero must be excluded from the domain. For example, if $f(x)=\dfrac{1}{x-3}$ then the domain could be $x \in \mathbb{R}, x \neq 3$, the domain restriction of $x \neq 3$ would be represented by a vertical asymptote with equation $x=3$.

Physics

How to find total mechanical energy?

Mechanical energy is defined as the sum of an object’s energies due to its motion and position. The amount of mechanical energy an object possesses determines the amount of work it can do on other objects, or in other words, the amount of energy it is capable of transferring to other objects. Given the above definition, we can be more specific about the types of energy that contribute to the total mechanical energy of an object: - Kinetic energy $E_k$ is the energy associated with an object’s motion and is given by the formula $E_k = \dfrac{1}{2}mv^2$ where $m$ is the mass of the object and $v$ is its speed. Note that this is the macroscopic energy of motion of the particle, and does not include the microscopic energy of the movement of its particles which contributes to the object's internal energy. The energy associated with an object’s position is called its potential energy. Two main types of potential energy will contribute to mechanical energy: - Gravitational potential energy $E_p$ is the energy of an object due to its position in a gravitational field. For an object at a height $h$ above a reference point, its gravitational potential energy is calculated with the formula $E_p = mgh$, where $m$ is the mass of the object and $g$ is the gravitational field strength. At the Earth’s surface, the value for $g$ is 9.8 N kg$^{-1}$. - Elastic potential energy $E_H$ is is the energy stored due to the deformation of an elastic object. Work can be done in changing the shape of an object, for example stretching or compressing a string, and energy is stored in the object as a result. This stored energy is released when the object returns to its original shape. The standard formula for elastic potential energy is derived from Hooke's Law: $E_H = \dfrac{1}{2}kx^2$. Where $k$ is the spring constant and $x$ is the displacement from the equilibrium position. Another potential energy that can be considered to contribute to total mechanical energy is electric potential energy. Like gravitational potential energy, a charged object will have stored energy due to its position in an electric field. Because the idea of mechanical energy is normally applied to larger objects and not small charged particles, we will ignore it here. Having discussed the different types of energy that contribute to the mechanical energy of an object, we can create a formula for total mechanical energy. In words, the formula is $\hspace{3em} $ Mechanical Energy = Kinetic Energy + Potential Energy or more specifically $\hspace{3em} $ Mechanical Energy = Kinetic Energy + Gravitational Potential energy + Elastic Potential Energy In the form of an equation, the total mechanical energy can be expressed as $\hspace{3em} E_{tot}=E_k + E_p + E_H$ or $\hspace{3em} E_{tot}=\dfrac{1}{2}mv^2+ mgh+ \dfrac{1}{2}kx^2$

Mathematics

What is the change of base formula?

The change of base formula, shown below, is used to change the base of a logarithm from one value to another. $$ \begin{align*} \log_a{x}=\frac{\log_b{x}}{\log_b{a}} \end{align*} $$ Where $a,b \neq 1$ and $x>0$. We can transform a logarithm with base $a$ into the expression on the R.H.S. with base $b$. The formula is particularly useful as we can choose any value we like for $b$. We can use the formula for a range of different things, such as - simplifying expressions with logarithms - solving equations with logarithms - comparing the size/value of logarithms with different bases For example, if we want to simplify $\log_2{x}+\log_4{y}$ we need to change one of the terms in the expression so that they both have the same base. This then allows us to use other logarithm laws. Let's use the formula to change the logarithm in the second term to base $2$. So that means we choose $b=2$. Using other well-known logarithm laws, we get. $$ \begin{align*} \log_2{x}+\log_4{y}&=\log_2{x}+\frac{\log_2{y}}{\log_2{4}}\hspace{2em}(\text{recall }\rightarrow 2^2=4)\\[12pt] &=\log_2{x}+\frac{\log_2{y}}{2}\\[12pt] &=\log_2{x}+\log_2{\sqrt{y}}\\[12pt] &=\log_2{x\sqrt{y}} \end{align*} $$

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