Problem 1 - Calculus Early Transcendentals

Question

A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume \(V\) of water remaining in the tank (in gallons) after t minutes. $$\begin{array}{|c|c|c|c|c|c|c|}\hline t(\min ) & {5} & {10} & {15} & {20} & {25} & {30} \\ \hline V(g a l) & {694} & {444} & {250} & {111} & {28} & {0} \\\ \hline\end{array}$$ $$\begin{array}{l}{\text { (a) If } P \text { is the point }(15,250) \text { on the graph of } V \text { , find the slopes }} \\ {\text { of the secant lines PQ when } Q \text { is the point on the graph }} \\ {\text { with } t=5,10,20,25, \text { and } 30 \text { . }}\end{array} $$ $$\begin{array}{l}{\text { (b) Estimate the slope of the tangent line at } P \text { by averaging the }} \\ {\text { slopes of two secant lines. }}\end{array}$$ $$\begin{array}{l}{\text { (c) Use a graph of the function to estimate the slope of the }} \\ {\text { tangent line at } P . \text { (This slope represents the rate at which the }} \\ {\text { water is flowing from the tank after } 15 \text { minutes.) }}\end{array}$$

Step-by-Step Solution

Answer

The estimated slope of the tangent line at \( t = 15 \) is about \(-33\) gallons per minute.

Step 1: Calculate Slope of Secant Line for t=5

We start by calculating the slope of the secant line using the points \((15, 250)\) and \((5, 694)\). The formula for the slope \(m\) of a line between two points \((t_1, V_1)\) and \((t_2, V_2)\) is \(m = \frac{V_2 - V_1}{t_2 - t_1}\). Substitute \(t_1 = 15\), \(V_1 = 250\), \(t_2 = 5\), and \(V_2 = 694\):\[ m = \frac{694 - 250}{5 - 15} = \frac{444}{-10} = -44.4 \].

Step 2: Calculate Slope of Secant Line for t=10

Calculate the slope of the secant line using the points \((15, 250)\) and \((10, 444)\). Substitute \(t_1 = 15\), \(V_1 = 250\), \(t_2 = 10\), and \(V_2 = 444\):\[ m = \frac{444 - 250}{10 - 15} = \frac{194}{-5} = -38.8 \].

Step 3: Calculate Slope of Secant Line for t=20

Calculate the slope of the secant line using the points \((15, 250)\) and \((20, 111)\). Substitute \(t_1 = 15\), \(V_1 = 250\), \(t_2 = 20\), and \(V_2 = 111\):\[ m = \frac{111 - 250}{20 - 15} = \frac{-139}{5} = -27.8 \].

Step 4: Calculate Slope of Secant Line for t=25

Calculate the slope of the secant line using the points \((15, 250)\) and \((25, 28)\). Substitute \(t_1 = 15\), \(V_1 = 250\), \(t_2 = 25\), and \(V_2 = 28\):\[ m = \frac{28 - 250}{25 - 15} = \frac{-222}{10} = -22.2 \].

Step 5: Calculate Slope of Secant Line for t=30

Calculate the slope of the secant line using the points \((15, 250)\) and \((30, 0)\). Substitute \(t_1 = 15\), \(V_1 = 250\), \(t_2 = 30\), and \(V_2 = 0\):\[ m = \frac{0 - 250}{30 - 15} = \frac{-250}{15} \approx -16.67 \].

Step 6: Estimate Slope of the Tangent Line at P using Two Secants

To estimate the slope of the tangent line at \((15, 250)\), we can average the slopes of two secant lines. We choose the closest points: the secant slopes with \(t = 10\) and \(t = 20\). Use slopes \(-38.8\) and \(-27.8\):\[ \text{Average slope} = \frac{-38.8 + (-27.8)}{2} = \frac{-66.6}{2} = -33.3 \].

Step 7: Estimate Slope of the Tangent Line Using the Graph

If you plot the values of \(t\) and \(V\), you can visually inspect the graph and determine that the tangent line at the point \((15, 250)\) has a slope approximately consistent with past calculations, possibly confirming the estimate of around \(-33 \) gallons per minute.

Key Concepts

Secant and Tangent Lines

In calculus, understanding the concepts of secant and tangent lines is essential for grasping how functions behave and change. A secant line intersects a curve at two or more points, showing the average rate of change between those points. Think of a secant line as a snapshot of how quickly a car travels between two points on a road. On the other hand, a tangent line touches the curve at exactly one point, indicating the immediate rate of change at that point. It's similar to looking at the car's speedometer at a specific moment. These lines provide useful insights into the behavior of curves, making them crucial tools in analyzing real-world applications like water flow in a tank.

Slope of a Secant Line

The slope of a secant line between two points on a curve calculates the average rate of change of the function over an interval. The slope formula is given by \( m = \frac{V_2 - V_1}{t_2 - t_1} \). Here, \( (t_1, V_1) \) and \( (t_2, V_2) \) are the coordinates of the two points on the curve. For instance, in this exercise, to find the slope of the secant line from \((15, 250)\) to \((5, 694)\), the calculation \( m = \frac{694 - 250}{5 - 15} = -44.4 \) is performed. This slope indicates how much water volume decreases per minute between those two times. By calculating slopes at different intervals, we can see how the rate at which water leaves the tank changes over time.

Graphical Analysis

By graphing a function or a data set, we can gather visual insights that help us understand how it behaves. When plotting the values from the table, each point on the graph shows the volume of water left in the tank at a certain time. By drawing a secant line, we visualize the average rate of water leaving the tank between two moments, but more importantly, the tangent line at a point gives us an instant rate of flow. Graphical analysis complements numerical calculations, confirming estimates like the slope of the tangent line at a specific point. Looking at the graph, if a line just brushes a curve at one point, that's where the tangent line provides the slope signifying the leak rate.

Rate of Change

The rate of change is how we describe how a function changes concerning another variable. In the context of this exercise, it's about how quickly the volume of water in the tank decreases over time. For example, when we average slopes of secant lines close to a specific point, we get an estimate of the tangent's slope at that point - the instantaneous rate of change.The slope calculated for the tangent line at a given point, like \((15, 250)\), helps us understand how fast water is draining at that exact moment. This rate of change is essential for predicting real-world outcomes and is foundational in calculus studies.