Problem 77 - Applied Calculus

Question

An average young female in the United States gains weight at the rate of \(14(x-10)^{-1 / 2}\) pounds per year, where \(x\) is her age \((10 \leq x \leq 19)\). Find the total weight gain from age 11 to \(19 .\)

Step-by-Step Solution

Answer

The total weight gain from age 11 to 19 is 56 pounds.

Step 1: Understand the Problem

We need to find the total weight gain of a young female from age 11 to 19. The rate of weight gain is given as a function of age, i.e., \( 14(x-10)^{-1/2} \). Therefore, we must integrate this function from 11 to 19 to find the total weight gain.

Step 2: Establish the Integral

The total weight gain can be represented as the integral of the rate function from age 11 to 19. Therefore, we set up the integral: \[ \int_{11}^{19} 14(x-10)^{-1/2} \, dx \].

Step 3: Solve the Integral

To solve the integral, we need to find the antiderivative of the integrand. The antiderivative of \( (x-10)^{-1/2} \) is \( 2(x-10)^{1/2} \). Thus, the integral becomes: \[ \int_{11}^{19} 14(x-10)^{-1/2} \, dx = 14 \cdot 2 (x-10)^{1/2} + C \],where \(C\) is the constant of integration.

Step 4: Evaluate the Definite Integral

Now evaluate the definite integral from 11 to 19: \[ 14 \cdot 2 \left[ (x-10)^{1/2} \right]_{11}^{19} = 28 \left[ (19-10)^{1/2} - (11-10)^{1/2} \right] \]. Calculate the values inside the brackets:

Step 5: Calculate the Values

First, compute the value at the upper limit: \( (19-10)^{1/2} = 3 \). Next, compute the value at the lower limit: \( (11-10)^{1/2} = 1 \). Insert these into the integral expression.

Step 6: Final Calculation

Substitute the values back into the evaluated integral: \[ 28 \left[ 3 - 1 \right] = 28 \times 2 = 56 \]. Therefore, the total weight gain from age 11 to 19 is 56 pounds.

Key Concepts

Understanding Definite Integrals

Definite integrals are extremely useful in calculating the accumulation of quantities, like distance, area, or in this case, weight gain over a certain range.
The symbol \(\int_{a}^{b} f(x) \, dx\) denotes the definite integral, where \(a\) is the starting point and \(b\) is the end point.
For this exercise, "a" is 11 and "b" is 19, making the integral show how much weight is gained between these ages.

• The function \(f(x) = 14(x-10)^{-1/2}\) represents the rate of weight gain.
• The purpose of integrating this function is to sum up all the tiny amounts of weight gained each year from 11 to 19 years.

After finding the integral, we evaluate it at the boundaries \(x=11\) and \(x=19\) to reveal the total weight gained.
This evaluation gives us a specific numerical amount, essentially summing the changes over that time frame.

Discovering Antiderivatives

Antiderivatives, sometimes known as indefinite integrals, are the reverse operation of differentiation.
When solving the integral \(\int_{11}^{19} 14(x-10)^{-1/2} \, dx\), we first need to determine the antiderivative of \((x-10)^{-1/2}\).

• The antiderivative of \((x-10)^{-1/2}\) is found to be \(2(x-10)^{1/2}\).
• When multiplying by the constant 14, we adjust the antiderivative to \(28(x-10)^{1/2}\).

This step is crucial, as it prepares the function for evaluation between the two limits (11 and 19).
Finding antiderivatives involves recognizing patterns from derivatives, effectively "reversing" known derivative rules.

Interpreting Rate of Change

The rate of change is a fundamental concept in calculus that measures how one quantity changes with respect to another.
In this problem, \(14(x-10)^{-1/2}\) expresses how much weight increases per year in pounds, known as the rate of weight gain.

• This function can vary depending on different ages, which is captured by the variable \(x\).
• The negative power indicates that as age goes on (and \(x\) increases), the rate of weight gain slows down.

Understanding rates of change helps students comprehend how dynamic processes operate over time.
By integrating this "rate" function, we discover not just the instantaneous change at any given age but the total accumulated change over ages 11 to 19, which sums to a total weight gain of 56 pounds.