Question

scalcet9 2.7.053. the quantity of oxygen that can dissolve in water depends on the temperature of the water. (so thermal pollution influences the oxygen content of water.) the graph shows how oxygen solubility s varies as a function of the water temperature t. (a) what is the meaning of the derivative s’(t)? what are its units? (\bigcirc) (s(t)) is the rate at which oxygen solubility changes with respect to the time. the units are (\bigcirc) (s(t)) is the rate at which temperature changes with respect to time. the units are (^circ)c/s. (\bigcirc) (s(t)) is the current level of oxygen solubility for a given temperature. the units are mg/l. (\bigcirc) (s(t)) is the rate at which temperature changes with respect to the oxygen solubility. the u (\bigcirc) (s(t)) is the rate at which oxygen solubility changes with respect to the water temperature. (b) estimate the value of (s(8)) and interpret it. (round your answer to three decimal places.) (s(8) approx square)

Explanation:

Part (a)

To determine the meaning of \( S'(T) \), recall that the derivative of a function \( y = f(x) \) represents the rate of change of \( y \) with respect to \( x \). Here, \( S \) is oxygen solubility (in mg/L) and \( T \) is water temperature (in \(^\circ\text{C}\)). So \( S'(T) \) is the rate at which oxygen solubility changes with respect to water temperature.

Now let's analyze the options:

• Option 1: Talks about change with respect to time, but \( T \) is temperature, not time. Eliminate.
• Option 2: Talks about temperature change with respect to time, but \( S(T) \) is about solubility vs temperature. Eliminate.
• Option 3: Describes the level of solubility (which is \( S(T) \), not its derivative). Eliminate.
• Option 4: Talks about temperature change with respect to solubility (inverse of what we need). Eliminate.
• Option 5: Correctly states \( S'(T) \) is the rate of solubility change with respect to temperature.

For units, the derivative \( S'(T)=\frac{dS}{dT} \), so units are \( \frac{\text{units of } S}{\text{units of } T}=\frac{\text{mg/L}}{^\circ\text{C}} \) (or mg/(L·°C)).

Step 1: Identify points for estimation

To estimate \( S'(8) \), we use the slope of the secant line between two points around \( T = 8 \). From the graph, at \( T = 0 \), \( S(0)=15 \) (approx, since the first point is at (0,15) or (0,14 - 16, let's take (0,15) and (8,12) as visible points). Wait, looking at the graph: when \( T = 0 \), \( S \approx 15 \) mg/L (the first dot is at (0,15) maybe? Wait the y-axis: at T=0, S is 14 - 16, let's take (0,15) and (8,12). Wait, at T=8, S=12. At T=0, S=15.

Step 2: Calculate the slope

The derivative \( S'(T) \) at \( T = 8 \) can be estimated by the slope between \( T = 0 \) and \( T = 8 \) (or nearby points). Let's use points \( (T_1, S_1)=(0, 15) \) and \( (T_2, S_2)=(8, 12) \).

The slope formula is \( \frac{\Delta S}{\Delta T}=\frac{S_2 - S_1}{T_2 - T_1} \).

Substitute values: \( \frac{12 - 15}{8 - 0}=\frac{-3}{8}=-0.375 \). Wait, but maybe check another interval. Wait, maybe the first point is (0,14) or (0,15)? Wait the graph: at T=0, S is 14 - 16, let's see the grid. The y-axis has 12, 16, etc. At T=0, the dot is at 14 - 16, maybe 15. At T=8, it's at 12. At T=16, maybe 10? Wait, let's re-examine.

Wait, the graph: T=0, S≈15; T=8, S=12; T=16, S=10; T=24, S=9; T=32, S=8; T=40, S=7? Wait no, the points: let's list the coordinates. From the graph, the first point (T=0, S=15), T=8, S=12; T=16, S=10; T=24, S=9; T=32, S=8; T=40, S=7? Wait maybe my initial points were wrong. Wait, let's take T=0, S=15; T=8, S=12. Then the slope between T=0 and T=8 is (12 - 15)/(8 - 0)= -3/8 = -0.375. Alternatively, between T=8 and T=16: S(16)=10, so (10 - 12)/(16 - 8)= -2/8 = -0.25. But since we need S’(8), the best is to use the points adjacent to T=8, i.e., T=0 and T=8, or T=8 and T=16. Wait, the problem says "estimate", so using the two closest points: T=0 (S=15) and T=8 (S=12). So the slope is (12 - 15)/(8 - 0)= -3/8 = -0.375. Wait, but maybe the first point is (0,14) or (0,16)? Wait the graph shows at T=0, S is 14 - 16, let's check the y-axis: 20,16,12,8,4. So the first dot is at 14 - 16, maybe 15. At T=8, it's at 12. So the change in S is 12 - 15 = -3, change in T is 8 - 0 = 8. So slope is -3/8 = -0.375.

Wait, but let's confirm. The derivative at a point is the slope of the tangent line, but we estimate it using the secant line between two points. So using T=0 (S=15) and T=8 (S=12), the slope is (12 - 15)/(8 - 0) = -3/8 = -0.375. So \( S'(8) \approx -0.375 \) mg/(L·°C). This means that at 8°C, the oxygen solubility is decreasing at a rate of approximately 0.375 mg/L per °C increase in temperature.

Answer:

The correct option is: \( S'(T) \) is the rate at which oxygen solubility changes with respect to the water temperature. The units are mg/(L·°C) (or similar, but the option's description about the meaning is what matters here; the last option in the list as per the problem's options is the correct one for meaning, and units follow from \( \frac{dS}{dT} \)).

Part (b)