Question

The spread of a virus can be modeled by \(N=-t^{3}+12 t^{2}, \quad 0 \leq t \leq 12\) where \(N\) is the number of people infected (in hundreds), and \(t\) is the time (in weeks). (a) What is the maximum number of people projected to be infected? (b) When will the virus be spreading most rapidly? (c) Use a graphing utility to graph the model and to verify your results.

Short answer

The maximum number of people projected to be infected is 51200 after 8 weeks and the virus will be spreading most rapidly after 4 weeks.

Step-by-step solution

Find the Derivative of the given function

The first derivative of the function \(N(t)=-t^{3} + 12t^{2}\) is determined by the power rule. This states that the derivative of \(t^n\) is \(nt^{n−1}\). Thus, if we take derivative of \(N(t)\), we get \(N'(t)=-3t^{2} + 24t\).

Solve the first derivative for \(t\)

To find critical points we set \(N'(t)\) equal to zero, \(0=-3t^{2} + 24t\). Solving for \(t\) we get that \(t=0,8\).

Evaluate the function at those points

The values of function \(N\) for the critical points \(t=0\) and \(t=8\) and the end points \(t=0\) and \(t=12\) must be evaluated because we are asked to find the maximum of a function on a closed interval [\0,12]. Substituting the values of \(t\) into the function \(N(t)\), we obtain \(N(0)=0, N(8)=512, N(12)=0\). The maximum value among these is 512, which means the virus will be projected to infect maximum 51200 people at 8 weeks.

Find the maximum of the first derivative

To know when the virus will be spreading most rapidly, take the second derivative of the function \(N(t)\). The derivative of \(N'(t)\) will be \(N''(t)=-6t + 24\). The critical point is found by setting this equal to zero, which gives \(t=4\). This means the virus will be spreading most rapidly at 4 weeks.

Graph the Function

To verify the results, graph the function \(N(t)\). The shape of the curve, the maximum point at \(t=8\), and the most rapid increase reach at \(t=4\) will confirm our calculations.

Key concepts

Derivative

The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. In simple terms, it provides the rate at which one quantity changes with respect to another.
For our virus model, the function given is:
\( N=-t^3 + 12t^2 \), where \( N \) is the number of people infected, and \( t \) is the time in weeks.
To analyze how quickly the infection spreads over time, we need the first derivative of this function.

• The derivative is calculated using the power rule: the derivative of \( t^n \) is \( nt^{n-1} \).
• Applying this to our function, we find \( N'(t) = -3t^2 + 24t \).

This derivative tells us how the number of infected people changes with each passing week.

Critical Points

Critical points are specific values of \( t \) where the derivative is zero or undefined. These points are crucial because they can indicate where a function's graph changes direction, thus identifying local maximums, minimums, or points of inflection.
To find these points for our virus function, we set the first derivative to zero:
\(-3t^2 + 24t = 0\).
By factoring, we find:

• \( t(-3t + 24) = 0 \)
• This gives us solutions \( t = 0 \) and \( t = 8 \).

Evaluating the function at these critical points and the endpoints of the interval helps determine where the maximum number of people will be infected. Here, we find that the function reaches its maximum value at \( t = 8 \).

Graphing Utility

A graphing utility is a handy tool that simplifies visualizing mathematical functions and their behavior. It allows us to see the shape of the graph, understand how the function behaves over an interval, and verify analytical methods visually.
For the virus spreading model, graphing the function \( N(t) = -t^3 + 12t^2 \) on a graphing utility confirms our analytical findings:

• The maximum number of people infected is at \( t = 8 \).
• The shape of the graph demonstrates this peak clearly.

Using a graphing utility helps solidify your understanding by showing the function's behavior over the entire interval \([0, 12]\). It's a powerful way to confirm calculations and gain intuition about the problem.

Second Derivative

The second derivative provides information about the concavity of a function and identifies points where the function changes its rate of growth or decay most rapidly.
For our virus model, the second derivative is the derivative of the first:
\( N''(t) = -6t + 24 \).
This derivative helps find when the infection is spreading at its quickest pace.

• Setting \( N''(t) = 0 \), we solve for \( t \), leading to \( t = 4 \).
• Therefore, at \( t = 4 \), the number of infections is increasing most rapidly.

Understanding the second derivative gives us deeper insights into the inflection points of the function, further illustrating changes in the function's growth rate.