Question

The flow of a small stream is monitored for 90 days between May 1 and August 1. The total water that flows past a gauging station is given by $$V(t)=\left\\{\begin{array}{ll} \frac{4}{5} t^{2} & \text { if } 0 \leq t<45 \\ -\frac{4}{5}\left(t^{2}-180 t+4050\right) & \text { if } 45 \leq t<90 \end{array}\right.$$ where \(V\) is measured in cubic feet and \(t\) is measured in days, with \(t=0\) corresponding to May 1 a. Graph the volume function. b. Find the flow rate function \(V^{\prime}(t)\) and graph it. What are the units of the flow rate? c. Describe the flow of the stream over the 3 -month period. Specifically, when is the flow rate a maximum?

Short answer

Question: Graph the volume function of a stream and its flow rate function, given in piecewise form, describe the flow of the stream, and find the maximum flow rate. Answer: The maximum flow rate of the stream occurs at t = 45 days, with a value of 9 cubic feet per day.

Step-by-step solution

Determine points for both parts of the function

To graph the volume function, first find the points for the two parts of the piecewise function. For the first part of the function, calculate the volume at \(t = 0\) and \(t = 45\): $$V(0)=\frac{4}{5} 0^{2} = 0$$ $$V(45)=\frac{4}{5} 45^{2} = 729$$ For the second part of the function, calculate the volume at \(t = 45\) and \(t = 90\): $$V(45)=-\frac{4}{5}(45^{2}-180 \cdot 45+4050) = 729$$ $$V(90)=-\frac{4}{5}(90^{2}-180 \cdot 90+4050) = 0$$ Now, create the graph using these points. #b. Find the flow rate function V'(t) and graph it. What are the units of the flow rate?#

Differentiate the volume function

To find the flow rate function, differentiate the volume function with respect to time: For the first part, when \(0 \leq t < 45\): $$V^{\prime}(t) = \frac{d}{dt} \biggl(\frac{4}{5} t^{2}\biggr) = \frac{8}{5}t$$ For the second part, when \(45 \leq t < 90\): $$V^{\prime}(t) = \frac{d}{dt} \biggl(-\frac{4}{5}(t^{2}-180t+4050)\biggr) = -\frac{8}{5}t + 144$$

Graph the flow rate function

Calculate the flow rate at \(t = 0\), \(t = 45\), and \(t = 90\) to help graph the function: $$V^{\prime}(0) = \frac{8}{5}(0) = 0$$ $$V^{\prime}(45) = -\frac{8}{5}(45) + 144 = 9$$ $$V^{\prime}(90) = -\frac{8}{5}(90) + 144 = 0$$ Now, create the graph using these points.

Determine the units of flow rate

The units of the flow rate will be the units of volume (cubic feet) per unit time (days). So, the flow rate is measured in cubic feet per day. #c. Describe the flow of the stream over the 3-month period. Specifically, when is the flow rate a maximum?#

Analyze the flow rate function

From the graph of the flow rate function, we can see that the flow rate is increasing in the first part and decreasing in the second part. Also, the flow rate's maximum occurs at the transition point between the two function parts, which is at \(t = 45\).

Conclusion

Throughout the 3-month period: - The volume of the stream increases in a quadratic manner before the midpoint (\(t = 45\)) and decreases in a quadratic manner after the midpoint (\(t = 45\)). - The flow rate is a piecewise function, increasing linearly before the midpoint and decreasing linearly after the midpoint. - The maximum flow rate occurs at the midpoint (\(t = 45\)) with a value of 9 cubic feet per day.

Key concepts

Understanding Piecewise Functions

A piecewise function is a function composed of multiple sub-functions, each of which applies to a certain interval of the domain. In calculus, representing functions in a piecewise manner is particularly useful when the behavior of a function changes across different intervals.

For example, let's look at the volume function for our exercise: \[ V(t)=\left\{\begin{array}{ll} \frac{4}{5} t^{2} & \text { if } 0 \leq t<45 \-\frac{4}{5}\left(t^{2}-180 t+4050\right) & \text { if } 45 \leq t<90 \end{array}\right. \]This function represents the total volume of water that has passed a gauging station over a period of 90 days. Here, we have two quadratic expressions, each valid for a different range of days. The sudden change in expressions at the 45-day mark likely represents a physical change in the flow dynamics of the stream.

Graphing the Volume Function

Graphing the volume function from our exercise involves analyzing how the volume of the stream changes over time. In this case, the volume V is a function of time t. To graph this function, we find key points and observe that the volume grows in a quadratic manner for the first segment and follows a different quadratic equation for the second segment.

The value of the function at transition points and end points, such as \( t = 0, 45 \), and \( 90 \) days, helps us sketch the function accurately. By connecting these points with the appropriate curves—a parabola opening upwards for the first interval and another opening downwards for the second interval—we can visualize the change in volume over time.

Such a visual representation aids students in understanding the concept of accumulation and how flow rates affect the total volume over periods of time.

Differentiation and Flow Rate Analysis

Differentiation plays a crucial role in understanding changes in functions over time. In the flow rate's context, we differentiate the volume function to determine the rate at which the volume is changing with respect to time. This rate of change is called the flow rate.

The differentiated volume function, denoted as \( V'(t) \), gives us a new piecewise function representing the flow rate. The units for this flow rate are 'cubic feet per day,' clearly indicating how much volume of water flows past a point each day.

By analyzing the flow rate function's graph, we can discern that the maximum flow rate occurs at the point where the first function shifts to the second, which is at \( t = 45 \) days. At this moment, the change in flow rate from increasing to decreasing indicates a peak, which has practical implications for predicting and managing water behavior in the stream.