Question

Find a formula for \(g\) by scaling the input and/or output of \(f\). Let \(f(s)\) give the volume of water in liters in a reservoir when the depth measures \(s\) meters, and \(g(d)\) give the volume of water in liters when the depth measures \(d\) centimeters. Use the fact that \(1 \mathrm{~m}\) equals \(100 \mathrm{~cm}\).

Short answer

Question: Given a function \(f(s)\) that calculates the volume of water in liters based on the depth given in meters, find the formula for \(g(d)\) where the depth is given in centimeters. Answer: The formula for \(g(d)\) is given by: $$ g(d) = f\left(\frac{d}{100}\right) $$

Step-by-step solution

Convert depth from centimeters to meters

Since \(1 \ \text{m} = 100 \ \text{cm}\), we can convert the depth \(d\) from centimeters to meters by dividing \(d\) by 100. In other words, \(\text{depth in meters} = \frac{d}{100}\).

Plug the converted depth into \(f(s)\)

Now that we have the depth in meters, we can plug it into the function \(f(s)\) to get the volume in liters. Hence, we have \(f\left(\frac{d}{100}\right)\).

Define the function \(g(d)\)

Finally, we define the function \(g(d)\) as the result of step 2. Therefore, we have the formula for \(g(d)\): $$ g(d) = f\left(\frac{d}{100}\right) $$

Key concepts

Conversion of Units

Converting units is a crucial skill in math and science. It's all about translating one measurement into another compatible form. In the given exercise, we are converting depth measurements from centimeters to meters. This is important because the function \( f(s) \) uses meters as its unit of measurement for depth. The conversion factor between meters and centimeters is 100, meaning 1 meter is equivalent to 100 centimeters. For example, if you have a depth \( d \) in centimeters, you convert it to meters by using the formula: \[ \text{depth in meters} = \frac{d}{100} \] This step ensures consistency in units, allowing us to use the function \( f(s) \) correctly. Consistent units are vital for accurately calculating and comparing quantities.

Scaling of Input

Scaling is all about adjusting the input to fit the required model or formula. In this exercise, once the depth is converted from centimeters to meters, we scale the input by taking this new depth value and using it in our function. This step is crucial because \( f(s) \) operates based on depth values in meters. So, after converting \( d \) to \( \frac{d}{100} \), we use it as a scaled input to compute the volume. The formula becomes: \[ g(d) = f\left(\frac{d}{100}\right) \] Scaling is often necessary when dealing with different measurement units, requiring conversions to maintain function compatibility and predict accurate outcomes. It deals with adjusting inputs so that they are suitable inputs for the given function.

Volume Calculation

Volume calculation in this context involves determining the volume of water in a reservoir. The function \( f(s) \) returns this volume given the depth \( s \) in meters. When we want the function \( g(d) \), which uses the depth in centimeters, we ensure the units are converted, and the appropriate input is scaled. The function \( g(d) = f\left(\frac{d}{100}\right) \) helps calculate the same volume but now references centimeter measurements. Understanding how to calculate volume is essential in many fields such as engineering and environmental science, where accurate measurements are vital for planning and development. By using the correct formulas and conversions, we can ensure these calculations are precise and valuable for real-world applications.