Question

Draining a Tank 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 water level is changing at time. (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 \(d y / d t .\)

Short answer

The rate at which the water level in the tank changes is given by the derivative of y, \(dy/dt = -(t-6)/3\). The fastest rate is observed at t=0 and is -2 m/h while the slowest rate is at t=12 and is -1 m/h. The graphs of y and dy/dt illustrate these findings.

Step-by-step solution

Find the derivative of y with respect to t

We first need to find the derivative of y with respect to t, which will represent the rate of change of the water level in the tank. Using the power rule and the chain rule, we have \(dy/dt = 2\times6(1- t/12)\times (-1/12) = - (t-6)/3\).

Find values for fastest and slowest rates of change

At fastest rate, derivative of y would reach its maximum. This will occur at the end point since the derivative is decreasing function of t, the fastest draining rate is at t=0, i.e. \(dy/dt = -2\) m/h. The slowest draining rate happens when the tank is empty, i.e., at t=12, given \(dy/dt = -1\) m/h.

Graph y and dy/dt and discuss

The graph of y will show a curve that begins at (0, 6) and ends at \(12, 0), precisely a concave quirve reflecting the water level dropping over time. Its slope decreases as t increases, reflecting the fact that the water drains slower as the tank gets less and less full. The graph of dy/dt will show a straight line with a negative slope, representing a constant decrease in the rate at which the tank is draining.

Key concepts

Related Rates

Related rates problems in calculus involve finding the rate at which one quantity changes with respect to another. This requires understanding how two or more quantities are connected, usually in a physical or geometric scenario.

In our case, the rate at which the water level, represented by the depth y, changes with respect to time t in the draining tank is a classic related rates problem. Here, as the valve opens, the level of fluid decreases over time, and we explore the relationship between these two variables. By differentiating y with respect to t, we obtain a rate of change that is 'related' to both the volume of the fluid and the time elapsed.

Differentiation

Differentiation is the process of finding the derivative of a function. It is a fundamental tool in calculus used to determine the instantaneous rate of change of a quantity. The derivative gives us a mathematical way to predict and understand motion, growth, and decay patterns.

In the draining tank exercise, differentiation allows us to calculate the rate at which the tank's water level decreases over time, signifying the water's flow rate out of the tank. The derivative is expressed as \(dy/dt\), which represents the change in depth (y) with respect to time (t). This approach allows us to explore how quickly the tank is draining at any given moment.

Power Rule

The power rule is an efficient way to differentiate functions of the form \(x^n\), where n is any real number. According to this rule, the derivative of such a function is given by \(nx^{n-1}\).

Applying the power rule to our problem, where the depth y is a function of time t in the power of 2, we differentiate the expression to find the tank's draining rate. Since we have a chain of functions, we also use the chain rule, which states that the derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function. The power rule simplifies the differentiation process, making it easier to solve related rates problems like the one presented.

Real-world Applications of Calculus

Calculus has vast applications in science, engineering, economics, and beyond. It helps us understand and model the physical world by allowing us to predict future states of dynamic systems.

For instance, the draining tank example is practical in various engineering disciplines where fluid dynamics and changes in volume over time are crucial. To design systems and structures that handle fluids efficiently, engineers use calculus to model and predict rates of flow and volume changes. The ability to find rates of change through differentiation is particularly powerful in forecasting and optimizing real-world scenarios.

Graphing Derivatives

Graphing derivatives aids in visualizing the behavior of rates of change. The graph of a derivative provides insights into the increasing or decreasing nature of the original function and indicates points of maximum and minimum rates.

By graphing \(dy/dt\) alongside the original function for the tank's water depth, we can see how the rate at which the tank drains changes over time. Initially, the rate is steeper, meaning that the tank drains faster. As time passes, this rate decreases, depicted by the less steep slope of the \(dy/dt\) graph. The derivative's graph is a powerful tool to analyze and understand the dynamic nature of the water level over time, giving a clear picture of the tank's draining behavior.