Problem 68 - Algebra and Trigonometry

Question

Volume of Water Between \(0^{\circ} \mathrm{C}\) and \(30^{\circ} \mathrm{C},\) the volume \(V\) (in cubic centimeters) of 1 \(\mathrm{kg}\) of water at a temperature \(T\) is given by the formula $$ V=999.87-0.06426 T+0.0085043 T^{2}-0.0000679 T^{3} $$ Find the temperature at which the volume of 1 \(\mathrm{kg}\) of water is a minimum.

Step-by-Step Solution

Answer

The volume is minimum at approximately \( T = 4^{\circ} \mathrm{C} \).

Step 1: Understand the problem

We need to find the temperature \( T \) between \(0^{\circ} \mathrm{C}\) and \(30^{\circ} \mathrm{C}\) at which the volume \( V \) of 1 \( \mathrm{kg} \) of water is minimum, according to the given cubic polynomial equation.

Step 2: Identify the function to be minimized

The volume of water is represented by the function:\[V(T) = 999.87 - 0.06426T + 0.0085043T^2 - 0.0000679T^3\]We need to find the minimum of this function over the interval from \(0^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\).

Step 3: Find the derivative

Calculate the first derivative \( V'(T) \) to find the critical points:\[V'(T) = -0.06426 + 2 \times 0.0085043 \times T - 3 \times 0.0000679 \times T^2\]Simplifying:\[V'(T) = -0.06426 + 0.0170086T - 0.0002037T^2\]

Step 4: Set the derivative to zero

To find critical points, set \( V'(T) = 0 \):\[-0.06426 + 0.0170086T - 0.0002037T^2 = 0\]Re-arrange the terms and solve this quadratic equation for \( T \).

Step 5: Solve the quadratic equation

Use the quadratic formula \( T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = -0.0002037, b = 0.0170086, \text{ and } c = -0.06426 \). Calculate the discriminant:\[b^2 - 4ac = (0.0170086)^2 - 4(-0.0002037)(-0.06426)\]Find the roots to determine \( T \).

Step 6: Evaluate the solutions

Compute the roots from the quadratic formula. Verify which fall within the interval \([0, 30]\). Determine the corresponding volumes if needed to confirm the minimum between critical points and endpoints.

Step 7: Compare critical points and endpoints

Calculate \( V(T) \) at the critical points found and at the endpoints \( T = 0 \) and \( T = 30 \). Choose the smallest volume value to determine at what \( T \) the volume is minimum.

Key Concepts

Critical Points

Critical points are key in optimization problems as they help identify where a function may reach its local extrema, such as minima or maxima. In our problem, we need to find the critical points of the volume function, which is a cubic polynomial, to determine the temperature where the volume of water is minimized.

To find the critical points, we locate where the derivative of the function equals zero or does not exist. In this exercise, the derivative is continuous, so we only need to find the zeros. By setting the first derivative of the volume function to zero, we derive a quadratic equation, whose solutions indicate potential critical points.

Once critical points are found, they are tested along with the function's endpoints to determine which gives the minimum volume within the given interval. This ensures that we account for the possible minimum values occurring at the boundaries of the temperature range or at one of the critical points found.

Quadratic Formula

The quadratic formula is a fundamental tool in algebra used to find the roots of a quadratic equation, which is an equation of the form \( ax^2 + bx + c = 0 \). In the context of our problem, solving a quadratic equation is a crucial step for finding critical points.

Once the derivative of the cubic volume function is set to zero, we obtain a quadratic equation in terms of temperature \( T \):

• \( a = -0.0002037 \)
• \( b = 0.0170086 \)
• \( c = -0.06426 \)

The quadratic formula \( T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is applied here.

Calculating the discriminant \( b^2 - 4ac \) helps determine the type and number of solutions. A positive discriminant indicates two distinct real roots, which we need to explore further to see if they lie within our interval of interest.

Cubic Polynomial

Cubic polynomials include terms up to the third degree, represented as \( ax^3 + bx^2 + cx + d \). In this exercise, the given polynomial represents the volume \( V(T) \) of water as a function of temperature \( T \).

Such functions are interesting because they can have various shapes, including having maxima and minima, which is why they are often analyzed in optimization tasks. The real-world behavior described by cubic polynomials, like changes in water volume with temperature, showcases their application.

Unlike linear and quadratic functions, cubic functions can have an inflection point and more than one critical value, complicating the process of finding extreme values. Nevertheless, derivatives help us deal with these intricacies by reducing the problem to simpler quadratic equations that we can solve to learn about the trends and critical points of the original function.

Derivative Calculation

The derivative of a function is the mathematical procedure for determining the rate at which the function's value changes. Calculating the derivative is fundamental in optimization exercises as it helps identify where the slope of the function is zero, indicating potential minima or maxima.

Let's break down the derivative calculation for the given cubic polynomial for volume, \( V(T) \):

• The constant term disappears upon differentiation.
• The term \( -0.06426T \) becomes \( -0.06426 \).
• The term \( 0.0085043T^2 \) becomes \( 2 \times 0.0085043 \times T = 0.0170086T \).
• The term \( -0.0000679T^3 \) becomes \( 3 \times -0.0000679 \times T^2 = -0.0002037T^2 \).

Thus, the derivative \( V'(T) = -0.06426 + 0.0170086T - 0.0002037T^2 \). Setting this to zero lets us solve for important temperatures where the volume might be minimized, allowing us to assess the critical points.