Problem 29 - Algebra 2

Question

The optimal height \(h\) of the letters of a message printed on pavement is given by the formula \(h=\frac{0.0052 d^{227}}{e} .\) Here \(d\) is the distance of the driver from the letters and \(e\) is the height of the driver's eye above the pavement. All of the distances are in meters. Find \(h\) for the given values of \(d\) and \(e .\) $$d=25 \mathrm{m}, e=2.3 \mathrm{m}$$

Step-by-Step Solution

Answer

The optimal height \(h\) of the letters is approximately 0.64 meters.

Step 1: Identify the Given Values

Identify the given values from the problem. Here, the distance of the driver from the letters (\(d\)) is given as 25m, and the height of the driver's eye above the pavement (\(e\)) is given as 2.3m.

Step 2: Substitution

Substitute values of \(d\) and \(e\) into the formula. Hence, \(h = \frac{0.0052 \times 25^{2.27}}{2.3}\).

Step 3: Calculation

Calculate the value of \(h\) using a calculator to ensure accuracy. After calculation, we get \(h\) approximately equal to 0.6391m. Considering the physical significance of \(h\), which represents the optimal height of the letters, we can round off the obtained result to two decimal places. Hence, \(h = 0.64\)m.

Key Concepts

Understanding the Substitution Method

The substitution method is like plugging in values to find specific solutions from general equations. For instance, with our problem of letter height on pavement, we have a formula that uses different variables. When you understand the given formula, \[ h = \frac{0.0052 \times d^{2.27}}{e} \], it means the letter height depends on both the distance of the driver (\(d\)) and the eye level height (\(e\)).

Substitution involves inserting the given values into this equation. It’s like solving a puzzle where you know most of the pieces and just need to find one missing part. You replace \(d\) with 25 and \(e\) with 2.3. Replacing these known values helps simplify the expression, making it ready for solving. It's simple! Just put the numbers where they belong into the original formula.

This method is useful when formulas or equations involve multiple variables, as it lets us see the specific outcome for a particular scenario.

Grasping the Concept of Exponents

Exponents are a way of expressing repeated multiplication of a number by itself. When we see something like \(d^{2.27}\), it means that the distance \(d\) is multiplied by itself 2.27 times.

This use of decimals in exponents might seem unusual, but it's common in scientific and precise measurements. Exponents allow for specifying large or small number scales in a compact form.

In our context, the exponent \(2.27\) plays a crucial role in determining how dramatically the distance affects the letter height. It can be easily calculated using a scientific calculator by typing in the base, \(d\), and then using the exponent function to input \(2.27\). This computation is crucial for finding the precise height measurement when combined with other elements of the formula.

Exploring Distance and Measurement

Distance and measurement concepts are fundamental in real-world applications, like in our exercise, where they determine how effectively a driver can see a message on the pavement. The distance variable \(d\) is how far the driver is from the letter, while \(e\) refers to how high above the pavement the driver's eye is.

Accurate measurement ensures correct data is inputted into the formula, leading to a reliable outcome. Here, both \(d\) and \(e\) need to be precise, as small errors could significantly affect the letter visibility calculation.

Understanding these measurements helps us apply the mathematical concepts correctly, ensuring any signs or messages are legible for drivers from the calculated optimal distance and angle.

Problem-Solving in Algebra

Problem-solving in algebra involves breaking down a problem into manageable parts, just like solving our exercise. First, identify what's given: the formula and specific values. Next, break down the process step-by-step.

We use multiplication, division, and powers to find the answer. This exercise showcases how algebra helps model real-world problems. Using step-by-step problem-solving, you can approach each part methodically, substituting, calculating, and rounding as needed.

Algebra trains us to approach problems logically and systematically. Each step leads to the next, ensuring no part is left out. With practice and a clear method, solving such exercises becomes intuitive.