A tabular cooling-coffee temperature function is given. Students estimate a derivative via an average rate of change, approximate a definite integral with a left Riemann sum and interpret its average value, use a supplied rate model to recover a later temperature, and analyze whether the rate of change is increasing or decreasing via the second derivative.
A tabular cooling-coffee temperature function is given. Students estimate a derivative via an average rate of change, approximate a definite integral with a left Riemann sum and interpret its average value, use a supplied rate model to recover a later temperature, and analyze whether the rate of change is increasing or decreasing via the second derivative.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
Approximate the instantaneous rate of change $C'(5)$ using the symmetric difference quotient over the table interval $[3,7]$ that brackets $t=5$: $$C'(5)\approx\frac{C(7)-C(3)}{7-3}=\frac{69-85}{4}=\frac{-16}{4}=-4.$$ So $C'(5)\approx -4$ degrees Celsius per minute. The negative sign reflects the beverage cooling at $t=5$.
1 point for the estimate from a correct difference quotient using the table values bracketing t=5, i.e. (69-85)/(7-3); 1 point for correct units (degrees Celsius per minute). Equivalent unsimplified forms such as (69-85)/4 also earn the estimate point. A response missing the units forfeits the units point.
Approximate $\int_0^{12} C(t)\,dt$ with a **left Riemann sum** using the three subintervals from the table, $[0,3]$, $[3,7]$, $[7,12]$ (widths $3,4,5$), and left-endpoint values $C(0)=100$, $C(3)=85$, $C(7)=69$: $$\int_0^{12} C(t)\,dt \approx 3\cdot C(0)+4\cdot C(3)+5\cdot C(7)=3(100)+4(85)+5(69)=300+340+345=985.$$ The expression $\frac{1}{12}\int_0^{12} C(t)\,dt$ is the **average temperature** of the beverage, in degrees Celsius, over the time interval from $t=0$ to $t=12$ minutes (its approximate value is $\frac{985}{12}\approx 82.083^{\circ}\mathrm{C}$).
1 point for the correct form of the left Riemann sum with subinterval widths 3,4,5; 1 point for the numerical estimate 985 with table values substituted; 1 point for interpreting (1/12) times the integral as the average temperature of the beverage over 0 <= t <= 12 (with units). Using right endpoints would give the alternate value 3(85)+4(69)+5(55)=806 and is not the requested left sum.
On the interval $12\le t\le 20$ the rate is given by a closed-form model for $C'(t)$. By the Fundamental Theorem of Calculus, the later temperature is the earlier temperature plus the accumulated change in temperature: $$C(20)=C(12)+\int_{12}^{20} C'(t)\,dt.$$ With $C(12)=55$ from the table and the modeled rate integrated by calculator, $\int_{12}^{20} C'(t)\,dt\approx -14.671$ (the beverage continues to cool), so $$C(20)=55+(-14.670812)=40.329188\approx 40.329.$$ The temperature of the beverage at $t=20$ minutes is approximately $40.329^{\circ}\mathrm{C}$.
1 point for the correct integral setup applying the FTC, C(20) = C(12) + integral_{12}^{20} C'(t) dt; 1 point for using the initial condition C(12)=55 from the table; 1 point for the answer 40.329 (degrees Celsius). The calculator must be in radian mode if the model contains trig/exponential terms; degree-mode evaluation would forfeit the relevant point.
On $12\le t\le 20$, the given model makes the second derivative $C''(t)>0$ throughout the interval. Since $C''(t)$ is the rate of change of the rate $C'(t)$, a positive $C''$ means $C'(t)$ is **increasing** on $12\le t\le 20$. In context: although the beverage is still cooling, the rate of change of its temperature is increasing over this interval (the temperature is changing at an increasing rate).
1 point for the correct answer with a supporting reason tied to the sign of C''(t): because C''(t) > 0 on 12 <= t <= 20, the rate of change of temperature C'(t) is increasing there. A bare 'increasing' with no reference to C'' (or to C' increasing) does not earn the point.
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