Problem 76 - College Algebra and Calculus: An Applied Approach

Question

Tree Growth An evergreen nursery usually sells a certain shrub after 6 years of growth and shaping. The growth rate during those 6 years is approximated by \(d h / d t=1.5 t+5\), where \(t\) is the time in years and \(h\) is the height in centimeters. The seedlings are 12 centimeters tall when planted \((t=0)\). (a) Find the height after \(t\) years. (b) How tall are the shrubs when they are sold?

Step-by-Step Solution

Answer

a) After t years, the height of the shrub is given by \(h(t) = 0.75t^2 + 5t + 12\). b) The shrubs are 51 cm tall when they are sold.

Step 1: Integrate the Growth Rate Function

First, integrate the given growth rate function \(dh/dt = 1.5t + 5\). This yields the function \(h(t) = 0.75t^2 + 5t + C\), where C is the constant of integration.

Step 2: Find the Constant of Integration

Using the initial condition which says that the seedlings are 12 centimeters tall when planted \(t=0\), we can find the constant of integration. Substituting \(h(0) = 12\), we get the equation \(12 = 0.75*0^2 + 5*0 + C\). Solving this for C, we find that \(C = 12\).

Step 3: Compute the Height Function h(t)

Plugging the found value of C back into our integrated function, we now have the function \(h(t) = 0.75t^2 + 5t + 12\). This function will give us the height of the shrub after t years.

Step 4: Determine the Height of the Shrubs When They are Sold

Now, we substitute t=6 into the function \(h(t) = 0.75t^2 + 5t + 12\) to get the height when the shrubs are sold. Doing this we get \(h(6) = 0.75*6^2 + 5*6 + 12 = 51\).

Key Concepts

Integration

Integration is a fundamental concept in calculus used to find functions from their derivatives. In this exercise, we are given the growth rate of a shrub as a derivative, expressed as \( \frac{dh}{dt} = 1.5t + 5 \), representing how the shrub's height changes over time. To determine the actual height as a function of time, we perform integration.

Integration involves finding the antiderivative of a given function. In this case, integrating the growth rate equation results in the height function \( h(t) = 0.75t^2 + 5t + C \). Here, \( C \) represents the constant of integration, a crucial component because integration can yield infinitely many solutions differing by a constant.

• The antiderivative of \( 1.5t \) is \( 0.75t^2 \) because the power rule states that the integral of \( t^n \) is \( \frac{t^{n+1}}{n+1} \).
• The integral of \( 5 \) is \( 5t \) because it involves integrating a constant.

Integration helps us to derive the original function, giving us insight into physical quantities, in this case, the shrub height over time.

Growth Rate

The growth rate of a plant, or any living organism, describes how quickly it increases in size over a period of time. In this scenario, the growth rate is mathematically described by the equation \( \frac{dh}{dt} = 1.5t + 5 \). This represents a linear growth rate, which varies with time \( t \).

Linear growth rate means the height of the shrub increases in a predictable way as time progresses. Here are some key points about the growth rate:

• The term \( 1.5t \) implies the rate of growth accelerates as time increases, contributing more to the height as \( t \) becomes larger.
• The constant \( 5 \) represents the initial growth rate when \( t = 0 \), meaning that even at the start, the shrub grows by 5 units per year due to environmental or genetic factors.

The awareness of the growth rate facilitates understanding how growth patterns might change and helps in predicting future growth at any given time.

Initial Conditions

Initial conditions provide known values at the onset of a scenario, allowing one to find specific solutions to differential equations. In this exercise, the initial condition states that the seedling is 12 centimeters tall when \( t = 0 \).

This specific information is crucial for solving the constant of integration, \( C \). By using the initial condition \( h(0) = 12 \), we substitute \( t = 0 \) into the integrated function to find the exact height. This substitution yields the equation \( 12 = 0.75 \times 0^2 + 5 \times 0 + C \), simplifying to \( C = 12 \).

These initial conditions allow us to refine the general solution obtained from integration into a precise function, \( h(t) = 0.75t^2 + 5t + 12 \). This is particularly pivotal because, without \( C \), the function would not correctly fit the initial condition or describe the actual growth. Thus, initial conditions are foundational in generating tailored solutions that accurately describe real-world phenomena.