Problem 58 - Calculus: Early Transcendentals

Question

If the units for \( x \) are feet and the units for \( a(x) \) are pounds per foot, what are the units for \( da/dx \)? What units does \( \displaystyle \int^8_2 a(x) \, dx \) have?

Step-by-Step Solution

Answer

Units for \( \frac{da}{dx} \) are pounds per foot squared, and for \( \int_2^8 a(x) \, dx \), it is pounds.

Step 1: Understanding the Function

We are given a function \( a(x) \) with units of pounds per foot (\( \frac{\text{pounds}}{\text{foot}} \)). This function depends on \( x \), which is measured in feet.

Step 2: Find Units of the Derivative

The derivative \( \frac{da}{dx} \) represents the rate of change of the function \( a(x) \) with respect to the change in \( x \). It has units of \( \frac{\text{pounds}}{\text{foot}} \) divided by \( \text{foot} \), which simplifies to \( \frac{\text{pounds}}{\text{foot}^2} \).

Step 3: Understanding the Integral Expression

The integral \( \int_2^8 a(x) \, dx \) represents the area under the curve of the function \( a(x) \) from \( x = 2 \) to \( x = 8 \).

Step 4: Determine Units of the Integral

The units of an integral are the units of the integrand \( a(x) \) times the units of \( dx \). Given \( a(x) \) has units of \( \frac{\text{pounds}}{\text{foot}} \) and \( dx \) has units of feet, the resulting units for the integral are \( \text{pounds} \).

Key Concepts

Derivative Units

Understanding the units of a derivative involves recognizing how a function changes with respect to the independent variable. The derivative of a function like \( a(x) \), denoted \( \frac{da}{dx} \), tells us how quickly or slowly the function's values change as \( x \) changes. In our example, \( a(x) \) describes a relationship where the dependent variable is in pounds per foot, and \( x \) is measured in feet.

• The derivative \( \frac{da}{dx} \) represents "change in pounds per foot" divided by the "change in feet." So its units become \( \frac{\text{pounds/foot}}{\text{foot}} \).
• This simplifies to \( \frac{\text{pounds}}{\text{foot}^2} \). This means for every square foot, there is a pound unit change.

By understanding derivative units, one can better grasp how different quantities related to real-world scenarios change under the influence of variables.

Integral Units

The units of an integral, like \( \int^8_2 a(x) \, dx \), describe a total accumulation of quantities across an interval. Integrals provide us with a way to sum up or accumulate a quantity over an interval, in this case from \( x = 2 \) to \( x = 8 \). The units calculation for the integral depends on both the integrand and the differential element.

• The integrand \( a(x) \) is in pounds per foot (\( \frac{\text{pounds}}{\text{foot}} \)).
• The differential element \( dx \) has units of feet, as it is the element of length over which integration occurs.
• Thus, when multiplying the two: \( \frac{\text{pounds}}{\text{foot}} \times \text{foot} = \text{pounds} \), we find that the units of the integral result in pounds.

Think of it as summing up small slices of weight per foot, over the range of feet specified by the integral. This gives a total weight that encompasses the interval concerned.

Unit Analysis

Unit analysis is a crucial method in calculus and science for checking the correctness of calculations and understanding the relationship between different physical quantities. When conducting mathematical operations involving different components, analyzing units helps ensure dimensional consistency.

• For derivatives, unit analysis involves differentiating the output and input units, providing insight into the nature of rate change in a physical context.
• With integrals, unit analysis confirms the resulting unit from adding up an accumulation over an interval, ensuring our expressions rightfully convey the concept of total accumulation.
• Applying unit analysis helps in confirming that expressions make physical sense. For instance, pounds per foot when multiplied by feet yields pounds, as expected in our integral example.

In essence, unit analysis serves as a key tool in problem-solving, verifying computations, and facilitating a deeper understanding of calculus problems set in real-world contexts.