Problem 66 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, consider a college student who works from 7 P.M. to 11 P.M. assembling mechanical components. The number \(N\) of components assembled after \(t\) hours is given by the function. At what time is the student assembling components at the greatest rate? $$ N=\frac{20 t^{2}}{4+t^{2}}, \quad 0 \leq t \leq 4 $$

Step-by-Step Solution

Answer

The student assembles components at the greatest rate at 9 P.M.

Step 1: Finding Derivative

Differentiate the function \(N(t) =\frac{20 t^{2}}{4+t^{2}}\). Utilize the formula for the derivative of a quotient: \( (u/v)' = (u'v - uv') / v^{2} \). With \(u = 20t^{2}\) and \(v = 4 + t^{2}\) you find the derivative \(N'(t) = \frac{4 t (4 - t^{2})}{(4 + t^{2})^{2}}\).

Step 2: Find critical points

Set derivative equal to zero, and solve for \(t\). So, we solve the equation \(\frac{4 t (4 - t^{2})}{(4 + t^{2})^{2}} = 0\). This gives \(t = 0\) and \(t = 2\) as critical points.

Step 3: Second Derivative Test

Compute the second derivative \(N''(t)\) to apply the second derivative test for local maxima and minima. The second derivative \(N''(t)\) is \(\frac{32 t^{3} - 48 t}{(4 + t^{2})^{3}}\). Substituting \(t = 0\) and \(t = 2\) into \(N''(t)\), we get \(N''(0) = 0\) and \(N''(2) < 0\). Since \(N''(2)\) is negative, \(t = 2\) corresponds to a local maximum.

Step 4: Considering the Domain

The critical points \(0\) and \(2\) are within the given domain of \(0 \leq t \leq 4\). We evaluate \(N'(t)\) at the endpoints of the domain and at \(t = 2\). We find that \(N'(2) > N'(0)\), and \(N'(2) > N'(4)\), so the student assembles components at the greatest rate at \(t = 2\) hours after 7 P.M.

Key Concepts

Derivative of a Quotient

When analyzing the rate at which a student assembles mechanical components, understanding the derivative of a quotient is crucial. The derivative of a quotient follows a specific rule, typically called the quotient rule, and is given by \( (u/v)' = (u'v - uv') / v^{2} \). It is used when you have a function defined as one expression divided by another.
Consider the function \( N(t) =\frac{20 t^{2}}{4+t^{2}} \) from our exercise. By defining \(u = 20t^{2}\) and \(v = 4 + t^{2}\), students can apply the quotient rule to find the derivative \(N'(t)\), which gives the rate of change of components assembled over time. Through this process, students learn how to handle complex functions that involve division and are better equipped to solve similar problems in calculus.

Critical Points Calculus

Finding the moments when the rate of assembling components is greatest involves identifying critical points. In calculus, a critical point occurs where a function's derivative is either zero or undefined. These points are essential because they indicate where a function's graph has a horizontal tangent line and could potentially be a maximum or minimum.
To find these, take the derivative found previously \(N'(t)\) and set it equal to zero: \(\frac{4 t (4 - t^{2})}{(4 + t^{2})^{2}} = 0\). Solving this gives us the critical values \(t=0\) and \(t=2\). By understanding how critical points are determined students can apply this knowledge to a variety of problems, analyzing where functions peak or bottom out.

Second Derivative Test

Once critical points are found, the second derivative test helps determine whether these points are local maxima, minima, or points of inflection. To perform this test, we compute the second derivative \(N''(t)\) of our assembly function, given by \(\frac{32 t^{3} - 48 t}{(4 + t^{2})^{3}}\).
By substituting our critical points—\(t=0\) and \(t=2\)—into the second derivative, we analyze their concavity. \(N''(0) = 0\) does not give us conclusive information, but \(N''(2) < 0\) indicates a concave down graph at \(t=2\), thus confirming a local maximum. The second derivative test is a powerful tool students can use to solidify their findings and draw reliable conclusions about a function's behavior at critical points.

Calculus Optimization Problems

Optimization is a fundamental concept in calculus that enables students to find the best, most efficient outcome based on given constraints. In our scenario, we sought to maximize the rate of component assembly for a college student.
After finding the critical points and applying the second derivative test, students should also consider the domain of the function to make sure they're finding the absolute maximum. This is done by comparing the derivative—or rate of assembly—at the critical points and the domain's endpoints. After computing that \(N'(2) \)>\( N'(0)\) and \(N'(2) \)>\( N'(4)\), we confidently conclude that the student assembles components at the greatest rate two hours into the shift. Optimization problems like these not only have practical applications but also help students develop an intuition for maximizing or minimizing functions in a range of contexts.