A Mixed Free-Response Routine
A calculator can help find numerical values, but a strong free response also explains what those values mean and why they answer the question. In this practice set, the tasks are deliberately different: find an instantaneous rate from a model, calculate accumulated change with a definite integral, and identify an absolute extremum from a graph.
For each task, distinguish the mathematical setup, the calculator evidence, and the written conclusion. Keep units attached to quantities, use enough calculator digits before rounding, and make a check that fits the calculation.
Worked Example: Instantaneous Rate of Water in a Tank
Worked Example: Instantaneous Rate of Water in a Tank
A tank contains \(V(t)=80+24t-1.5t^2\) liters of water \(t\) hours after a measurement begins. Find the rate at which the amount of water is changing when \(t=4\), and interpret the sign of the rate.
1. Set up the rate. The requested quantity is the instantaneous rate of change of volume, \(V'(4)\), measured in liters per hour. Differentiate the model:
Therefore,
2. Record calculator evidence. A numerical derivative command applied to \(V(t)\) at \(t=4\) gives \(12\) L/h. A nearby-value check can also be made with the central difference quotient using \(h=0.001\):
3. State the conclusion. To three significant figures, the volume is increasing at \(12.0\) liters per hour when \(t=4\) hours. The positive sign means that the amount of water in the tank is increasing at that instant.
Check. Substituting \(t=4\) into the derivative gives \(V'(4)=12\), agreeing with the numerical derivative and the nearby-value calculation. The units are liters per hour because a change in liters is divided by a change in hours.
Worked Example: Total Amount from a Numerical Integral
Worked Example: Total Water Delivered
Water flows into a reservoir at the rate \(r(t)=5+2\sin(t/2)\) liters per minute, where \(t\) is in minutes. Find the amount delivered during the first \(6\) minutes. Report the result to three decimal places.
1. Set up the accumulation. The rate is in liters per minute. Integrating it over a time interval in minutes gives the total amount in liters:
2. Record calculator evidence. Enter the definite integral with lower bound \(0\), upper bound \(6\), and the complete rate expression as the integrand. The numerical integral is approximately \(37.959970\) liters. Retain the displayed digits during the calculation; round only when reporting the answer.
3. Check the numerical value. An antiderivative of the rate is \(5t-4\cos(t/2)\), since
Evaluating this antiderivative at the bounds gives
Using \(\cos 3\approx-0.989992497\),
This agrees with the numerical integral. Rounded to three decimal places, the reservoir receives \(37.960\) liters during the first \(6\) minutes.
Check the interpretation. The integrand is a rate, and the limits describe the requested time interval. The integral therefore represents the total delivered amount, not the flow rate at the end of the interval. Also, \(r(t)\) is positive on this interval, so the accumulated amount is positive.
Worked Example: Absolute Extrema from a Graph
Worked Example: Absolute Maximum and Minimum on a Closed Interval
Use a calculator graph of \(g(x)=x^3-3x^2-9x+5\) on the closed interval \([-2,4]\) to find the absolute maximum and absolute minimum values, including where they occur.
1. Graph and record the window. Enter \(g(x)\) and use an \(x\)-window from \(-2\) to \(4\). A \(y\)-window from \(-25\) to \(15\) shows the relevant range. The graph suggests an interior high point near \(x=-1\) and an interior low point near \(x=3\). A calculator’s maximum and minimum commands, with starting guesses near those locations, return \((-1,10)\) and \((3,-22)\), respectively. These are numerical observations; check them against the endpoints and the model.
2. Identify candidates on the interval. The function is a polynomial, so it is continuous on \([-2,4]\) and differentiable in its interior. Its derivative is
The interior critical numbers satisfy \(g'(x)=0\), giving \(x=-1\) and \(x=3\). For an absolute-extremum check on a closed interval, compare the function values at these critical numbers and at both endpoints:
3. Compare and conclude. The largest candidate value is \(10\), at \(x=-1\), and the smallest is \(-22\), at \(x=3\). Thus \(g\) has an absolute maximum value of \(10\) at \(x=-1\), and an absolute minimum value of \(-22\) at \(x=3\), on \([-2,4]\). The graph’s calculator results agree with the checked candidate values.
Check. Both reported \(x\)-values lie in the specified interval, and direct substitution gives \(g(-1)=10\) and \(g(3)=-22\). The endpoint values, \(3\) and \(-15\), do not exceed the maximum or fall below the minimum.
Why the Extremum Check Is Necessary
A graph helps locate likely high and low points, but it can be misleading if the viewing window is poor or a feature is hard to see. The candidate check has a mathematical reason: a continuous function on a closed interval attains both an absolute maximum and an absolute minimum. If such an extremum occurs at an interior point where the function is differentiable, the derivative there must be zero.
For the interior condition, suppose a differentiable function has a local maximum at \(c\). For small positive \(h\), \(f(c+h)-f(c)\leq 0\), so the right-hand difference quotient is nonpositive. For small negative \(h\), \(f(c+h)-f(c)\leq 0\), but dividing by negative \(h\) makes the quotient nonnegative. If the derivative exists, both one-sided limits equal \(f'(c)\). It must therefore be both nonpositive and nonnegative, so \(f'(c)=0\). The same reasoning applies at an interior local minimum.
Common Pitfall: Reporting the Wrong Quantity
In the tank example, a tempting but incorrect response is to calculate \(V(4)=152\) and report “the rate is \(152\) liters per hour.” That value is wrong for two reasons: \(V(4)\) is a volume, not a rate, and its units are liters, not liters per hour.
The correction is to identify the requested quantity before using the calculator. The instantaneous rate is \(V'(4)\), which is \(12\) L/h. In an accumulation question, by contrast, the requested total is an integral of the rate. In an extremum question, the requested result includes both the extreme value and the input where it occurs. Matching the operation and units to the question prevents a plausible-looking calculator output from becoming a contextual error.
Writing the Final Response
A concise free response can still show the reasoning the reader needs. For a numerical derivative or integral, state the expression and bounds or input, report the calculator result with units and suitable precision, and finish with a sentence interpreting it. For a graph-based absolute extremum, report the viewing interval, identify the graphical candidates, and compare the endpoint and interior candidate values.
Decide whether the question asks for a rate, an accumulated amount, or an extreme value.
Enter the correct expression and interval, and note the numerical result and relevant graph window.
Verify the result in a suitable way, round only at the end, include units, and answer in context.
Check Your Understanding
For each question, identify the setup and the check that would make the final response convincing.
- A model gives the volume of a container in liters as a function of time in minutes. What quantity and units should you report for the instantaneous rate at a specified time?
- A flow rate is given in gallons per minute. What does its definite integral over a time interval represent, and what units should the result have?
- A graph suggests an absolute maximum inside a closed interval. Which values should you compare before claiming it is the absolute maximum?
- Why is a calculator’s numerical maximum command not, by itself, enough to establish an absolute maximum on the full interval?
- When should you round a calculator result, and what information should accompany the rounded number in a contextual answer?
