Problem 57 - Essential Calculus Early Transcendentals

Question

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 approximately by the formula $$V=999.87-0.06426 T+0.0085043 T^{2}-0.0000679 T^{3}$$ Find the temperature at which water has its maximum density.

Step-by-Step Solution

Answer

Water has maximum density at \(4^{\circ} \text{C}\).

Step 1: Understand the formula

The given formula \( V = 999.87 - 0.06426 T + 0.0085043 T^{2} - 0.0000679 T^{3} \) models the volume \(V\) of water at a temperature \(T\). To find the temperature at which water has maximum density, we need to find the temperature \(T\) at which the volume \(V\) is minimized, since density is inversely proportional to volume.

Step 2: Differentiate the volume formula

To find the minimum volume, take the derivative of \(V\) with respect to \(T\). This gives: \[ \frac{dV}{dT} = -0.06426 + 2 \times 0.0085043 T - 3 \times 0.0000679 T^{2} \] Which simplifies to: \[ \frac{dV}{dT} = -0.06426 + 0.0170086 T - 0.0002037 T^{2} \] .

Step 3: Find critical points

Set the first derivative equal to zero and solve for \(T\): \[ -0.06426 + 0.0170086 T - 0.0002037 T^{2} = 0 \] This is a quadratic equation in the form of \(aT^{2} + bT + c = 0\) where \(a = -0.0002037\), \(b = 0.0170086\), and \(c = -0.06426\). Use the quadratic formula \(T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find \(T\).

Step 4: Apply the quadratic formula

Substitute the values into the quadratic formula: \[ T = \frac{-0.0170086 \pm \sqrt{(0.0170086)^2 - 4(-0.0002037)(-0.06426)}}{2(-0.0002037)} \] Calculate the discriminant \(b^2 - 4ac\), then find the two possible values for \(T\).

Step 5: Verify the solution

Evaluate the calculated values of \(T\) within the given range, \([0^{\circ} \text{C}, 30^{\circ} \text{C}]\), and check if they are valid. The one that yields the minimum value of \(V\) corresponds to the maximum density.

Step 6: Choose the correct temperature

Based on calculations, the temperature \(T\) that minimizes \(V\) within \(0^{\circ} \text{C}\) and \(30^{\circ} \text{C}\) and therefore maximizes density is \(T = 4^{\circ} \text{C}\).

Key Concepts

Volume-Temperature Relationship

The volume-temperature relationship is crucial in understanding the behavior of water as it is heated or cooled. Generally, substances expand when heated and contract when cooled. However, water exhibits an exception, especially near its freezing point. As the temperature of water rises from 0°C to 4°C, it actually contracts rather than expands. This contraction continues until approximately 4°C, where water reaches its maximum density. Beyond this temperature, water starts expanding again as typical behavior. - This unique behavior is due to the hydrogen bonding in water that causes a special dip in volume. - As water warms from 0°C to 4°C, molecules rearrange into a more efficient packing, leading to decreased volume. - Understanding this relationship helps in calculating when water reaches maximum density, aiding numerous scientific and real-world applications.

Derivative of Polynomial Functions

Derivatives are essential tools in calculus for determining the behavior of functions. Taking the derivative of a polynomial function involves using power rules to find the rate of change at a particular point. The given function representing the volume of water is a cubic polynomial: \[ V = 999.87 - 0.06426 T + 0.0085043 T^2 - 0.0000679 T^3 \] To find where the volume is minimized (thus maximizing density), we must differentiate this function with respect to temperature, T. The derivative is calculated by: - Applying power rules: the derivative of \(aT^n\) is \(naT^{n-1}\).- For this function, differentiate each term to get: \[ \frac{dV}{dT} = -0.06426 + 0.0170086 T - 0.0002037 T^2 \] Once the derivative is found, critical points can be sought where the derivative is zero, indicating potential minima, maxima, or inflection points.

Quadratic Formula

To find the critical points where the derivative is zero, set \[ \frac{dV}{dT} = -0.06426 + 0.0170086 T - 0.0002037 T^2 = 0 \] This equation is a quadratic equation of the form \[ aT^2 + bT + c = 0 \] where \( a = -0.0002037 \), \( b = 0.0170086 \), and \( c = -0.06426 \). The solution to this equation can be found using the quadratic formula:\[ T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] - The discriminant \(b^2 - 4ac\) determines the nature of the roots: whether there are real roots and if they are distinct.- Substituting the coefficients into the formula gives potential temperatures that either minimize or maximize volume, and hence provide information on density.

Critical Points in Calculus

Critical points in calculus are where the derivative of a function is zero or undefined. At these points, a function's behavior can change, such as moving from increasing to decreasing, indicating local maxima or minima. - For the volume formula, finding where the derivative equals zero helps identify the temperature at which volume is minimized. - Determining whether these points are minima or maxima involves further investigation, such as the second derivative test or comparing values. In this exercise, one of the critical points within the given temperature range will correspond to a minimum volume, which conversely represents maximum density. By finding and verifying these critical points, we can conclude that the temperature at which water has maximum density is around 4°C, confirmed by checking the effectiveness of these points in reducing the overall water volume.