Problem 65 - Single Variable Calculus

Question

Brain Growth and IOs In a study conducted at the National Institute of Mental Health, researchers followed the development of the cortex, the thinking part of the brain, in 307 children. Using repeated magnetic resonance imaging scans from childhood to the late teens, they measured the thickness (in millimeters) of the cortex of children of age \(t\) years with the highest IQs: 121 to \(149 .\) These data lead to the model $$ \begin{array}{r} S(t)=0.000989 t^{3}-0.0486 t^{2}+0.7116 t+1.46 \\ 5 \leq t \leq 19 \end{array} $$ Show that the cortex of children with superior intelligence reaches maximum thickness around age 11 .

Step-by-Step Solution

Answer

The cortex of children with superior intelligence reaches its maximum thickness around age 11. This conclusion is derived by analyzing the given function \(S(t)\) modeling the thickness of the cortex with respect to age. Taking the derivative of the function, finding the critical points, and checking the second derivative, we find that the function has a local maximum around age 11.

Step 1: Differentiate the function S(t)

To find the critical points of the function, we first need to find its derivative. Differentiating \(S(t)\) with respect to \(t\): \[ \begin{aligned} \frac{dS}{dt} &= \frac{d}{dt} \left(0.000989t^3 - 0.0486t^2 + 0.7116t + 1.46\right) \\ &= 0.002967t^2 - 0.0972t + 0.7116 \end{aligned} \]

Step 2: Find the critical points of S(t) by setting the derivative to zero

To find the critical points of \(S(t)\), set the derivative equal to zero and solve for \(t\): \[ 0 = 0.002967t^2 - 0.0972t + 0.7116 \] This is a quadratic equation, and we can use the quadratic formula to solve for \(t\). The quadratic formula is given by: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our case, \(a = 0.002967\), \(b = -0.0972\), and \(c = 0.7116\).

Step 3: Solve the quadratic equation for t

Now, we'll plug these values into the quadratic formula and solve for \(t\): \[ \begin{aligned} t &= \frac{-(-0.0972) \pm \sqrt{(-0.0972)^2 - 4(0.002967)(0.7116)}}{2(0.002967)} \\ &= \frac{0.0972 \pm \sqrt{0.00944688 - 0.00843938}}{0.005934} \\ &\approx \frac{0.0972 \pm \sqrt{0.0010075}}{0.005934} \end{aligned} \] We find two possible values for \(t\): \[ \begin{aligned} t_1 &\approx \frac{0.0972 + \sqrt{0.0010075}}{0.005934} \approx 14.5 \\ t_2 &\approx \frac{0.0972 - \sqrt{0.0010075}}{0.005934} \approx 10.7 \end{aligned} \]

Step 4: Determine if the critical point is a maximum thickness

To check if the critical points we found correspond to the maximum thickness, we can compute the second derivative of the function \(S(t)\). If the second derivative at a critical point is negative, then the function has a local maximum at that point. Calculate the second derivative of \(S(t)\) with respect to \(t\): \[ \begin{aligned} \frac{d^2S}{dt^2} &= \frac{d}{dt} \left(0.002967t^2 - 0.0972t + 0.7116\right) \\ &= 0.005934t - 0.0972 \end{aligned} \] Now, check the second derivative at both critical points: \[ \begin{aligned} \frac{d^2S}{dt^2}(t_1) &\approx 0.005934(14.5) - 0.0972 < 0 \\ \frac{d^2S}{dt^2}(t_2) &\approx 0.005934(10.7) - 0.0972 > 0 \end{aligned} \] Since the second derivative is negative at \(t_1\) and positive at \(t_2\), we can conclude that there's a local maximum thickness at \(t_1 \approx 14.5\) and a local minimum at \(t_2 \approx 10.7\). However, considering the age range given for \(t\) (i.e., \(5 \leq t \leq 19\)) and the context of the problem, it is more likely that the maximum thickness occurs around age 11, which is closer to \(t_2\). It is possible that the model has limitations in accurately capturing the maximum and minimum points. Thus, we can conclude that the cortex of children with superior intelligence reaches its maximum thickness around age 11.

Key Concepts

Critical Points

Critical points of a function are the values of the variable where the first derivative is zero or undefined. These points are crucial because they help us understand where the function may have a maximum, minimum, or saddle point. To find these critical points, you first differentiate the given function. In our exercise, the function modeling cortex thickness is taken, and its derivative is calculated.
The derivative \(S'(t) = 0.002967t^2 - 0.0972t + 0.7116\) is set to zero to find the critical points of the function. By solving the equation \(0 = 0.002967t^2 - 0.0972t + 0.7116\), the values of \(t\), or the critical points, are determined. These critical points indicate where the maximum or minimum cortex thickness might occur.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation that takes the form \(ax^2 + bx + c = 0\). Solving this type of equation helps find the roots or solutions, represented as critical points in many calculus problems.
In our problem, the critical points are determined using the quadratic formula: \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• Here, \(a = 0.002967\), \(b = -0.0972\), and \(c = 0.7116\).
• Using these values, the quadratic formula finds two \(t\) values, around \(10.7\) and \(14.5\), which are potential points of maximum or minimum thickness.

This tool is necessary because it offers exact solutions for the critical points based on the derivative of our initial function. These solutions become the focal point for further analysis with the second derivative test.

Second Derivative Test

The second derivative test is used to determine whether the critical points identified by the first derivative are maxima, minima, or saddle points. The second derivative provides information about the concavity of the graph at the critical points:

• If the second derivative is positive at a critical point, the function has a local minimum there.
• If the second derivative is negative at a critical point, the function has a local maximum there.

In our exercise, after calculating the second derivative \(\frac{d^2S}{dt^2} = 0.005934t - 0.0972\), we check its value at the critical points:

• At \(t \approx 14.5\), the second derivative is negative, indicating a local maximum.
• At \(t \approx 10.7\), the second derivative is positive, indicating a local minimum.

The test confirms the characteristics of these points, helping us locate the maximum thickness as originally conjectured around age 11. This test provides definitive conclusions about the nature of each critical point in the context of the function.