Problem 65 - 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? $$ N=-0.12 t^{3}+0.54 t^{2}+8.22 t, \quad 0 \leq t \leq 4 $$

Step-by-Step Solution

Answer

The student is assembling components at the greatest rate at approximately 7:40 P.M.

Step 1: Differentiate the Function

To determine the point in time at which the rate of assembly is fastest, the first step is to differentiate the function with respect to \(t\). This gives us \(\frac{dN}{dt}= -0.36t^2 + 1.08t + 8.22\).

Step 2: Find Possible Critical Points

We equate the first derivative to zero to find possible critical points for which the rate might be a maximum. Solving \(-0.36t^2 + 1.08t + 8.22 = 0\) we find \(t\) approximately equal to 0.67 and 3.49.

Step 3: Identify Maximum Point

We then perform the second derivative test to identify if these points are maximums, minimums or inflection points. The second derivative is obtained by differentiating the first derivative \(-0.72t + 1.08\), which simplifies as \( -0.72\). The second derivative is negative for all values of \(t\), hence both points are maximums. However, considering the interval \(0 \leq t \leq 4\), only \(t = 0.67\) fits the constraint.

Step 4: Convert to Real Time

The last step is to convert the \(t\)-value back into real time. The student begins work at 7 P.M., therefore, \(t = 0\) corresponds to 7 P.M. With each \(t\) representing an hour, \(t = 0.67\) approximately equals to 40 minutes. This translates to 7:40 P.M., that's when the student builds components at the maximum rate.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that helps us understand how functions change. When we differentiate a function, we are essentially calculating its derivative, which represents the rate of change.

In the context of this exercise, we have a function that describes the number of components, \(N\), a student assembles over time, \(t\). To find out how quickly the student is assembling these components at any given moment, we need to differentiate the function.

The function given is \(N = -0.12 t^{3} + 0.54 t^{2} + 8.22 t\). By differentiating it with respect to \(t\), you can find the first derivative, \(\frac{dN}{dt} = -0.36t^2 + 1.08t + 8.22\), which tells us the rate of change of \(N\) with respect to \(t\).

This new function, \(\frac{dN}{dt}\), describes how fast the number of components increases per hour. It is the key function we use to determine when the assembly rate is highest.

Critical Points

Critical points are values of \(t\) where the derivative of a function is zero or doesn't exist. Finding these points helps us understand where the function's rate of change may peak, drop, or remain constant.

In this problem, after obtaining the first derivative of the function, which is \(-0.36t^2 + 1.08t + 8.22\), the next step is to identify the critical points by setting the derivative equal to zero and solving for \(t\).

• We solve the equation \(-0.36t^2 + 1.08t + 8.22 = 0\) using the quadratic formula or by factoring to find possible values of \(t\).
• This yields the critical points at approximately \(t = 0.67\) and \(t = 3.49\).

These points indicate where the rate of assembly could be at a maximum or minimum during the student's work shift. However, not all critical points are necessarily maximums; that's where further analysis, like the second derivative test, comes in.

Second Derivative Test

The second derivative test is a method used in calculus to classify critical points obtained from the first derivative. It helps to determine whether these points are maxima (peaks), minima (valleys), or points of inflection.

For our function, once we have identified potential critical points from the first derivative, we can apply the second derivative test.

• We first find the second derivative by differentiating the first derivative, resulting in \(-0.72t + 1.08\).
• Evaluating this second derivative at our critical points provides insight into the nature of these points.
• If the second derivative is negative at a critical point, it indicates a local maximum; if positive, a local minimum.

In this scenario, the second derivative \(-0.72\) is constant and negative, indicating that all critical points are maxima.

Thus, between our found critical points and within the interval \(0 \leq t \leq 4\), the point \(t = 0.67\) not only satisfies the problem's constraints but is also where the student assembles components at the fastest rate, making it the point of interest.