Problem 27 - University Calculus: Early Transcendentals

Question

It takes 12 hours to drain a storage tank by opening the valve at the bottom. The depth \(y\) of fluid in the tank \(t\) hours after the valve is opened is given by the formula $$ y=6\left(1-\frac{t}{12}\right)^{2} \mathrm{m} $$ a. Find the rate \(d y / d t(\mathrm{m} / \mathrm{h})\) at which the tank is draining at time \(t\) b. When is the fluid level in the tank falling fastest? Slowest? What are the values of \(d y / d t\) at these times? c. Graph \(y\) and \(d y / d t\) together and discuss the behavior of \(y\) in relation to the signs and values of \(d y / d t\)

Step-by-Step Solution

Answer

Rate of draining is \(-\left(1-\frac{t}{12}\right)\) m/hour. Fastest at \(t = 12\), slowest at \(t = 0\).

Step 1: Differentiate to find the rate

Given the formula for the depth of the fluid \(y = 6\left(1-\frac{t}{12}\right)^{2}\), we need to differentiate it with respect to \(t\) to find the rate \(\frac{dy}{dt}\). Using the chain rule, differentiate \(y\) as follows:\[ y = 6u^{2} \text{ where } u = 1 - \frac{t}{12} \]The derivative \(\frac{dy}{dt}\) can be found by applying the chain rule:\[ \frac{dy}{dt} = \frac{d}{du}(6u^{2})\cdot\frac{du}{dt} = 12u\cdot\left(-\frac{1}{12}\right) = -u = -\left(1-\frac{t}{12}\right) \]

Step 2: Find rate at a specific time

The rate \(\frac{dy}{dt}\) when the tank is draining fastest occurs when \(|\frac{dy}{dt}|\) is maximized. Since \(\frac{dy}{dt} = -\left(1-\frac{t}{12}\right)\), its absolute value is maximum when the expression \(1 - \frac{t}{12}\) is minimized around its possible values: 0 to 1. Thus, \(\frac{dy}{dt}\) is fastest and has a value of -1 at \(t = 12\). This is when the fluid is almost fully drained.

Step 3: Determine slowest draining rate

From the function \(\frac{dy}{dt} = -\left(1-\frac{t}{12}\right)\), it's slowest when \(t = 0\). At this time, \(\frac{dy}{dt} = -(1-0) = -1\). When the valve is first opened, the rate of change in depth is greatest since the pressure is highest.

Step 4: Graphing and discussing

First, plot the depth \(y = 6(1-\frac{t}{12})^{2}\) and its rate of change \(\frac{dy}{dt} = -\left(1-\frac{t}{12}\right)\) against time \(t\). The graph of \(y\) should be a quadratic curve opening upwards starting at maximum depth at \(t=0\) decreasing to \(t=12\) where \(y=0\). The derivative \(\frac{dy}{dt}\) is a linear function starting at \(-1\) and increasing linearly to \(0\). When \(\frac{dy}{dt}\) crosses from negative to zero, \(y\) transitions from decreasing at a slow rate to stopping. The behavior of the graph confirms the depth decreases faster initially and slower as \(t\) approaches 12.

Key Concepts

Rate of Change

In differential calculus, the concept of rate of change is fundamental. It measures how a quantity, such as the depth of fluid in a tank, changes over time. In this example, we're interested in the rate at which the fluid level decreases as described by the function:\[ y = 6\left(1-\frac{t}{12}\right)^{2} \]To find the rate of change, we differentiate this equation with respect to time \(t\). The derivative, denoted as \( \frac{dy}{dt} \), gives us the rate of change of the fluid level, which tells us how quickly the fluid level is dropping. By applying the chain rule, we derived:\[ \frac{dy}{dt} = -\left(1-\frac{t}{12}\right) \]This equation shows that the rate of change is dependent on \(t\). Initially, when the valve is first opened \((t=0)\), the rate is highest at \(-1\) m/h, due to high-pressure differential, which rapidly decreases as \(t\) moves towards 12 hours, reflecting how the tank drains slower over time.

Chain Rule

The chain rule is a key technique in differential calculus used to differentiate compositions of functions. Understanding and applying this rule allows us to find the rate of change for complex functions by breaking them into simpler parts. In our exercise, the depth function of the fluid depends on a nested function structure:\[ y = 6u^{2} \text{ where } u = 1 - \frac{t}{12} \]To differentiate \(y\) with respect to \(t\), we use the chain rule, which states that:\[ \frac{dy}{dt} = \frac{d}{du}(6u^{2}) \cdot \frac{du}{dt} \]Calculating these components separately:

• The derivative of \(6u^2\) with respect to \(u\) is \(12u\).
• The derivative of \(u = 1 - \frac{t}{12}\) with respect to \(t\) is \(-\frac{1}{12}\).

Multiplying these together, we find:\[ \frac{dy}{dt} = 12u \cdot \left(-\frac{1}{12}\right) = -u = -\left(1-\frac{t}{12}\right) \]The chain rule efficiently confirms how changes in time directly alter the rate at which the fluid drains.

Graphing Functions

Graphing the function of depth in the tank and its rate of change over time provides a visual understanding of how these quantities interact. Starting with our depth function:\[ y = 6\left(1-\frac{t}{12}\right)^{2} \]This function represents a quadratic curve. Initially, at time \(t = 0\), the fluid level is at its maximum because \(y = 6\times1^{2} = 6\). As \(t\) increases towards 12, the fluid level decreases, forming a downward-opening parabola.The rate of change is represented by:\[ \frac{dy}{dt} = -\left(1-\frac{t}{12}\right) \]This is a linear function starting at \(-1\) when \(t=0\) and increasing to 0 at \(t=12\). The negative values indicate the depth is decreasing, showing how the fluid drains faster initially and then slows down. When plotting these two functions on a graph, we observe:

• The depth \(y\) decreases from 6 meters to zero over 12 hours.
• The slope \(\frac{dy}{dt}\) linearly progresses from \(-1\) m/h, slowing to zero.

The intersection of these plots visualizes the rate dynamically decreasing, indicating the physics of draining—a powerful tool for understanding calculus in real-world scenarios.