Problem 64 - A Graphical Approach to Precalculus with Limits

Question

Solve each problem. W(t)$ represents the gallons of water in a tank after I minutes. Complete the following. (a) Find the initial amount of water in the tank. (b) Find the amount of water in the rank after 12 minutes. (c) Is the rate of change in the amount of water in the rank constant? Explain. (d) Find the rate of change in the amount of water at 12 minutes. $$W(t)=6000-20 t^{2}, \text { for } 0 \leq t \leq 15$$

Step-by-Step Solution

Answer

(a) 6000 gallons. (b) 3120 gallons. (c) No, it's not constant. (d) -480 gallons/min.

Step 1: Understand the Function

The function \(W(t) = 6000 - 20t^2\) represents the amount of water in a tank as a function of time \(t\), where \(0 \leq t \leq 15\). We need to find initial amounts and rates of change associated with this function.

Step 2: Finding Initial Amount of Water

To find the initial amount of water, evaluate \(W(t)\) at \(t=0\). So: \(W(0) = 6000 - 20(0)^2 = 6000\). The initial amount of water in the tank is 6000 gallons.

Step 3: Calculating Water After 12 Minutes

To find the amount of water after 12 minutes, substitute \(t=12\) into the function: \(W(12) = 6000 - 20(12)^2 = 6000 - 20(144) = 6000 - 2880 = 3120\). Thus, there are 3120 gallons after 12 minutes.

Step 4: Checking if the Rate of Change is Constant

The function \(W(t) = 6000 - 20t^2\) is quadratic, not linear, as indicated by the \(t^2\) term. Therefore, the rate of change of \(W(t)\) with respect to \(t\) is not constant.

Step 5: Calculating Rate of Change at 12 Minutes

The rate of change of water in the tank is represented by the derivative \(W'(t)\). We find the derivative: \(W'(t) = \frac{d}{dt}(6000 - 20t^2) = -40t\). Evaluate \(W'(12)\): \(W'(12) = -40(12) = -480\). The rate of change at 12 minutes is -480 gallons per minute.

Key Concepts

Quadratic Functions

Quadratic functions are mathematical expressions of the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The general shape of a quadratic is a parabola, which can open upwards or downwards depending on the sign of \( a \). In our exercise, the function \( W(t) = 6000 - 20t^2 \) is a quadratic function where \( a = -20 \), \( b = 0 \), and \( c = 6000 \).

Key characteristics of this function include:

• Vertex: The peak point of the curve, which, in downward-opening parabolas like this one, is the maximum point.
• Axis of symmetry: A vertical line that passes through the vertex, dividing the parabola into two symmetrical halves.

Quadratics are used in various contexts, from modeling projectile motion to calculating areas. In our context, understanding the quadratic nature helps to assess how the water level changes over time.

Rate of Change

The rate of change in a function measures how one quantity changes in relation to another. In the context of our quadratic function \( W(t) = 6000 - 20t^2 \), it indicates how the amount of water in a tank changes over time. For quadratic functions:

• The rate of change is nonlinear, meaning it varies at different points along the curve.
• The change is determined by the function's slope, which is influenced by the \( t^2 \) term.

The term "rate of change" is critical when analyzing real-world situations, helping us understand speed, growth, or decline in various scenarios. It is a way to phrase how quickly or slowly things are happening, and for quadratic functions, this rate is not constant as shown by the decreasing slope when moving along the function.

Derivatives

Derivatives are a fundamental concept in calculus. For a function, the derivative represents the rate at which the function value changes as its input changes. It can be thought of as the slope of the function at any given point. In our problem, the derivative of \( W(t) \), denoted \( W'(t) \), helps us determine how fast the amount of water is changing at any point \( t \).

To find the derivative of \( W(t) = 6000 - 20t^2 \), we apply the power rule, a basic derivative calculation technique:

• Subtract 1 from the exponent
• Multiply by the original exponent

This results in \( W'(t) = -40t \). Evaluating at \( t = 12 \) gives \( W'(12) = -480 \), indicating that the water is decreasing at a rate of 480 gallons per minute at that moment.

Understanding derivatives allows us to make precise calculations about rates of change, which is essential in fields such as physics, engineering, and economics.

Initial Value Problem

An initial value problem (IVP) in mathematics involves determining the value of a function given the initial conditions. These problems set the stage for finding specific solutions from a broad range of possibilities and often include understanding the starting point, which dictates the solution trajectory.

In our exercise, finding the initial amount of water in the tank can be seen as solving an initial value problem. We determine this by evaluating the function \( W(t) \) at \( t=0 \). This calculation provides a baseline from which changes in the system can be observed.

The importance of initial value problems in calculus cannot be overstated, as they provide the necessary conditions to solve differential equations—equations that describe many natural phenomena, from population growth to heat distribution. By solving the IVP, we ensure the calculated models comport with reality from the beginning, allowing for predictive analysis.