Problem 78
Question
An average child of age \(x\) years grows at the rate of \(6 x^{-1 / 2}\) inches per year (for \(2 \leq x \leq 16\) ). Find the total height gain from age 4 to age \(9 .\)
Step-by-Step Solution
Verified Answer
The total height gain from age 4 to age 9 is 12 inches.
1Step 1: Understand the Problem
We need to calculate the total height gain of a child from age 4 to age 9 years. The rate of growth is given as a function of age: \(6x^{-1/2}\) inches per year.
2Step 2: Set Up the Integral for Total Height Gain
The total height gain can be determined by integrating the growth rate function over the interval from 4 to 9 years. The definite integral \(\int_{4}^{9} 6x^{-1/2} \, dx\) will give us the total height gain over this period.
3Step 3: Integrate the Growth Function
Integrate the function \(6x^{-1/2}\). The antiderivative of \(x^{-1/2}\) is \(2x^{1/2}\). Therefore, the integral becomes:\[\int 6x^{-1/2} \, dx = 6 \cdot 2x^{1/2} + C = 12x^{1/2} + C\]
4Step 4: Evaluate the Definite Integral
Substitute the limits 4 and 9 into the antiderivative to find the total height gain:\[12\left[ x^{1/2} \right]_{4}^{9} = 12\left(9^{1/2} - 4^{1/2}\right) = 12(3 - 2) = 12(1) = 12\]
5Step 5: Interpret the Result
The result \(12\) means that from age 4 to age 9, the child has a total height gain of 12 inches.
Key Concepts
Growth Rate FunctionDefinite IntegralAntiderivative
Growth Rate Function
In the context of calculus, a growth rate function is a mathematical expression representing how a quantity increases over time. In this exercise, the growth rate function is given as \(6x^{-1/2}\), where \(x\) represents the child's age in years. This function describes how fast the child's height is increasing at any given age.
Key features of a growth rate function include:
Key features of a growth rate function include:
- It depends on the independent variable \(x\), which, in our example, is the age of the child.
- A higher value of the function indicates a faster growth rate at that point in time.
- The rate can change over different values of \(x\), illustrating the dynamic nature of growth during different ages.
Definite Integral
A definite integral is a powerful calculus tool that allows us to calculate the total change in a quantity across an interval. In this specific exercise, we use it to determine the total height gain of the child from ages 4 to 9.
The notation for a definite integral is \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the lower and upper bounds of the interval, respectively. This represents the accumulation of growth over the interval from \(x = a\) to \(x = b\).
Benefits of using definite integrals include:
The notation for a definite integral is \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the lower and upper bounds of the interval, respectively. This represents the accumulation of growth over the interval from \(x = a\) to \(x = b\).
Benefits of using definite integrals include:
- Computation of total quantities from rates, such as height from growth rate functions.
- Ability to evaluate functions over specific intervals, providing precise numerical results.
- Once evaluated, provides a singular value representing the total accumulation, like the 12-inch height gain in our example.
Antiderivative
An antiderivative, or indefinite integral, is essentially the reverse process of differentiation. It helps us find a function whose derivative gives us the original function. In this exercise, finding the antiderivative of the growth rate function \(6x^{-1/2}\) leads us to an expression for total height gain.
Finding the antiderivative involves:
Finding the antiderivative involves:
- Recognizing common patterns in derivatives to apply antiderivative rules, like \(x^{-1/2}\) whose antiderivative is \(2x^{1/2}\).
- Including a constant \(C\), which becomes obsolete in definite integrals as limits negate its effect.
- Applying known integration techniques across various functions to simplify linked computations.
Other exercises in this chapter
Problem 78
If the values of a function on an interval are always positive, can the average value of the function over that interval be negative?
View solution Problem 78
An average young male in the United States gains weight at the rate of \(18(x-10)^{-1 / 2}\) pounds per year, where \(x\) is his age \((11 \leq x \leq 20)\). Fi
View solution Problem 79
If the values of a function on an interval are always greater than 7 , what can you say about the average value of the function on that interval?
View solution Problem 79
A friend says that finding differentials is as easy as finding derivatives-you just multiply the derivative by \(d x\). Is your friend right?
View solution