A tabular reading-rate function is given. Students estimate a derivative using an average rate of change with units, justify the existence of an intermediate value via the Intermediate Value Theorem, approximate an accumulated total with a trapezoidal sum, and integrate a supplied polynomial rate model for a second reader.
A tabular reading-rate function is given. Students estimate a derivative using an average rate of change with units, justify the existence of an intermediate value via the Intermediate Value Theorem, approximate an accumulated total with a trapezoidal sum, and integrate a supplied polynomial rate model for a second reader.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
Approximate $R'(1)$ by the average rate of change of $R$ on $[0,2]$ using the table values $R(0)=90$ and $R(2)=100$: $$R'(1)\approx \frac{R(2)-R(0)}{2-0} = \frac{100-90}{2-0} = \frac{10}{2} = 5.$$ The units are words per minute per minute (the rate of change of a reading rate), i.e. $R'(1)\approx 5\ \text{words/min}^2$.
2 points. P1: the answer 5 supported by the difference quotient (100-90)/(2-0) using table values; the symbolic form (R(2)-R(0))/(2-0) by itself, without the numeric work, is not enough. P2: correct units 'words per minute per minute' (equivalently words/minute^2), whether or not attached to the value.
$R$ is differentiable on $[0,10]$, and differentiability implies continuity, so $R$ is continuous on $[0,10]$. From the table, $R(0)=90$ and $R(10)=162$, so $$R(0)=90 < 155 < 162 = R(10).$$ Since $155$ lies strictly between $R(0)$ and $R(10)$, the Intermediate Value Theorem guarantees a value $c$ with $0<c<10$ such that $R(c)=155$. Yes, such a $c$ must exist.
2 points. P3: states that R is continuous because it is differentiable (a bare 'R is continuous' with no reason does not earn P3). P4: notes R(0)<155 (or R(2)<155 or R(8)<155) and R(10)>155, invokes continuity, and answers 'yes.' The theorem need not be named, but if named it must be the IVT. P4 does not require P3.
Use a trapezoid on each of the three subintervals $[0,2]$, $[2,8]$, $[8,10]$ (widths $2$, $6$, $2$): $$\int_0^{10} R(t)\,dt \approx \frac{R(0)+R(2)}{2}(2) + \frac{R(2)+R(8)}{2}(6) + \frac{R(8)+R(10)}{2}(2).$$ Substituting, $$=\frac{90+100}{2}(2) + \frac{100+150}{2}(6) + \frac{150+162}{2}(2) = 190 + 750 + 312 = 1252.$$ The trapezoidal estimate of the total words read is $1252$.
2 points. P5: the correct trapezoidal form — three terms, each a product including the 1/2 factor and the correct subinterval width (2, 6, 2); at least five of the six factors must be correct. P6: the value 1252, eligible only after P5 and only if all six factors are correct. A correct left (1080) or right (1424) Riemann sum earns P5 but not P6.
The total words read by the teacher in $10$ minutes is $\displaystyle\int_0^{10} W(t)\,dt$ with $W(t)=-\tfrac{3}{10}t^2+8t+100$: $$\int_0^{10}\Big(-\tfrac{3}{10}t^2+8t+100\Big)\,dt = \Big[-\tfrac{1}{10}t^3 + 4t^2 + 100t\Big]_0^{10}.$$ Evaluating at the bounds: $$=\Big(-\tfrac{1}{10}(1000) + 4(100) + 100(10)\Big) - 0 = -100 + 400 + 1000 = 1300.$$ The teacher reads $1300$ words by the end of the $10$ minutes.
3 points. P7: a definite (or indefinite) integral with integrand W(t), with or without dt. P8: the correct antiderivative -(1/10)t^3 + 4t^2 + 100t (constant of integration optional). P9: the answer 1300, eligible only after P8; once the correct bound-substitution is shown it banks P9, so later arithmetic slips are not penalized.
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