- What quantity does \(Q(t)\) measure, and what does \(t\) measure? - A derivative compares the output quantity with the input quantity, so combine their units as “output units per input unit.” - Use the sign of the derivative to decide whether the amount is increasing or decreasing.

1. Because \(Q\) is measured in gallons and \(t\) in minutes, \(Q'(t)\) is measured in gallons per minute. 2. The positive value \(Q'(6)=4.2\) means that exactly \(6\) minutes after the pump starts, the amount of water in the tank is increasing at an instantaneous rate of \(4.2\) gallons per minute.

At \(t=6\) minutes, the amount of water in the tank is increasing at an instantaneous rate of \(4.2\,\text{gal/min}\).

- A negative derivative describes the direction of change, not whether the original quantity is negative. - What does a zero derivative say about change at one instant? - Keep the units of energy divided by the units of time.

1. \(E'(3)=-7\,\text{MJ/h}\) means that at \(t=3\) hours, the stored energy is decreasing at an instantaneous rate of \(7\) megajoules per hour. 2. \(E'(8)=0\,\text{MJ/h}\) means that at \(t=8\) hours, the stored energy is momentarily not changing. 3. Neither derivative value tells the amount of energy stored at those times; each describes only its instantaneous rate of change.

At \(t=3\) hours, the stored energy is decreasing at \(7\,\text{MJ/h}\). At \(t=8\) hours, the stored energy is momentarily not changing. These statements describe rates, not the amount of stored energy.

- Which notation describes the rate at one instant rather than the temperature itself? - What sign should the rate have if temperature is falling?

1. A dropping temperature means its instantaneous rate of change is negative. 2. The rate after four hours is \(T'(4)=-2\,{}^\circ\text{F}/\text{h}\).

\(T'(4)=-2\,{}^\circ\text{F}/\text{h}\).

- Read two points on the straight tangent, not two points on the curved graph. - Use rise over run and keep the sign of the slope.

1. On the tangent, the change from \((0,25)\) to \((4,17)\) gives slope \(\frac{17-25}{4-0}=-2\,\text{L/min}\). 2. Therefore, \(Q'(2)\approx-2\,\text{L/min}\). The water amount is falling at that instant.

\(Q'(2)\approx-2\,\text{L/min}\); the amount is falling.

- What geometric quantity is represented by \(x^2\)? What about \(x^3\)? - Differentiate each function with respect to \(x\). - Compare the units of each geometric quantity with the units of \(x\).

1. By the power rule, \(A'(x)=2x\) and \(V'(x)=3x^2\). 2. \(A(x)=x^2\) is the area of a square, so \(A'(x)=2x\) describes how its area changes with side length, in square units per unit of length. \(V(x)=x^3\) is the volume of a cube, so \(V'(x)=3x^2\) describes how its volume changes with edge length, in cubic units per unit of length.

1. \(A'(x)=2x\) and \(V'(x)=3x^2\) 2. \(A\) is a square's area, and \(A'\) is its area-change rate per unit increase in side length. \(V\) is a cube's volume, and \(V'\) is its volume-change rate per unit increase in edge length. The units are square units per length unit and cubic units per length unit, respectively.

- Identify the variable of differentiation and treat the other letters as constants. - Which circle measurement is represented by \(2\pi r\)? - Think about how a cylinder's lateral area changes when only its height changes. - For the sphere, interpret the derivative as a rate for an extremely small change in radius.

1. For part a), treat \(r\) as constant: \(M'(h)=2\pi r\). This derivative equals the circumference of the circular base and gives the exact increase in lateral surface area per unit increase in height. 2. For part b), \(S'(r)=8\pi r\). This is the instantaneous rate of change of surface area with respect to radius, in square units per unit of radius. It is also four times the circumference \(2\pi r\) of a great circle.

a) \(M'(h)=2\pi r\); it is the cylinder base's circumference and the rate at which lateral area changes with height. b) \(S'(r)=8\pi r\); it is the instantaneous rate at which a sphere's surface area changes with radius.

- Decide whether each statement describes a value at one time or a change over an interval. - Recall the difference between an instantaneous rate of change and an average rate of change. - Which notation represents the instantaneous rate of change? - Pay attention to whether the statement gives a height, a total change, or a rate.

1. The height at the planting time is represented by \(h(0)\), so \(h(0)=1.2\). 2. An instantaneous growth rate is represented by the derivative, so \(h'(10)=0.8\). 3. The total growth from year \(5\) to year \(10\) is the difference in heights, so \(h(10)-h(5)=3.5\). 4. Stopping growth after year \(150\) means the instantaneous growth rate remains zero, so \(h'(t)=0\) for all \(t\geq150\). 5. The average growth rate over \([0,20]\) is the difference quotient, so \(\frac{h(20)-h(0)}{20}=0.6\).

a) \(h(0)=1.2\) b) \(h'(10)=0.8\) c) \(h(10)-h(5)=3.5\) d) \(h'(t)=0\) for all \(t\geq150\) e) \(\frac{h(20)-h(0)}{20}=0.6\)

- Treat the cylinder's height as a constant when differentiating the volume. - Substitute that same fixed height into the lateral area formula. - To compare the two radii, evaluate the derivative at each one, not the total volumes.

1. Differentiate the volume function: \(V'(r)=20\pi r\). 2. With \(h=10\), the lateral area is \(M(r)=20\pi r\), so \(V'(r)=M(r)\). 3. \(V'(3)=60\pi\,\text{cm}^2\) and \(V'(6)=120\pi\,\text{cm}^2\). The volume changes twice as fast per centimeter of radius at \(r=6\) as at \(r=3\).

a) \(V'(r)=20\pi r\) b) \(V'(r)=M(r)\) c) At \(r=6\,\text{cm}\), the rate is \(120\pi\,\text{cm}^2\), twice its value of \(60\pi\,\text{cm}^2\) at \(r=3\,\text{cm}\).

- Use the derivative column, not the amount column, to compare instantaneous rates. - For decreasing quantities, compare the negative derivative values by how negative they are. - Interpret \(D'(5)\) using milligrams per minute.

1. A negative derivative means dissolved oxygen is decreasing. Among the listed rates, \(-4\,\text{mg/min}\) has the greatest decrease magnitude, so the fastest listed decrease occurs at \(t=9\) minutes. 2. Since \(D'(5)=-1\,\text{mg/min}\), at \(t=5\) minutes the amount of dissolved oxygen is decreasing at an instantaneous rate of \(1\) milligram per minute.

a) \(t=9\) minutes b) At \(t=5\) minutes, the dissolved oxygen is decreasing at \(1\,\text{mg/min}\).

- Use the straight tangent, not a line connecting two points on the curve. - Look for two readable points on the tangent and calculate rise over run. - Give the slope's units and explain what its sign says about the liquid.

1. Two points on the tangent line are approximately \((1,25.5)\) and \((5,29.5)\). 2. Its slope is \(\frac{29.5-25.5}{5-1}=1\,\text{gal/min}\). 3. Thus, \(Q'(3)\approx1\,\text{gal/min}\): after \(3\) minutes, the amount of liquid is increasing at about \(1\) gallon per minute.

\(Q'(3)\approx1\,\text{gal/min}\). At \(t=3\), the tank's liquid amount is increasing at about \(1\) gallon per minute.

- First identify the units of \(P'(t)\), then differentiate the units once more with respect to time. - Interpret the first and second derivatives separately before combining them. - A negative second derivative does not by itself mean that the original power output is decreasing.

1. Since \(P'(t)\) is measured in kilowatts per minute, \(P''(t)\) is measured in kilowatts per minute squared. 2. \(P''(5)=-3\,\text{kW/min}^2\) means that at \(t=5\) minutes, the rate at which power output is changing is itself decreasing by \(3\,\text{kW/min}\) per minute. 3. Because \(P'(5)=12\,\text{kW/min}>0\), the power output is increasing at that instant. Because \(P''(5)<0\), that positive increase is becoming slower.

\(P''(5)=-3\,\text{kW/min}^2\) means the power-change rate is decreasing by \(3\,\text{kW/min}\) per minute at \(t=5\). Since \(P'(5)=12\,\text{kW/min}>0\), the power output is increasing, but its rate of increase is decreasing.

- Decide first whether \(R'(4)\) is a quantity value or a rate value. - Check which statement uses units of “thousand gallons per hour.” - Distinguish a derivative at one instant from a difference quotient over an interval.

1. Interpretation B is correct. The derivative value \(R'(4)=-6\) describes the instantaneous rate of change at \(t=4\), with units of thousand gallons per hour. 2. Interpretation A is incorrect because \(R'(4)\) is a rate, not the reservoir volume \(R(4)\). 3. Interpretation C is incorrect because \(R'(4)\) is an instantaneous rate at one time, not an average rate over the interval \([0,4]\).

B is correct. \(R'(4)=-6\) means that at \(t=4\) hours, the reservoir's volume is decreasing at \(6\) thousand gallons per hour. A confuses \(R'\) with \(R\), and C confuses an instantaneous rate with an average rate.

- Separate what happens to the temperature from what happens to its rate. - Decide the sign of each change before writing the units. - The second derivative is measured per unit time twice.

1. A falling temperature means \(T'(3)=-4\,{}^\circ\text{F}/\text{min}\). 2. The temperature-change rate \(T'\) is increasing, so \(T''(3)=1.5\,{}^\circ\text{F}/\text{min}^2\). 3. Temperature is still falling, but its rate of decrease is becoming less negative.

\(T'(3)=-4\,{}^\circ\text{F}/\text{min}\) and \(T''(3)=1.5\,{}^\circ\text{F}/\text{min}^2\). Temperature is falling more slowly.

- Each straight segment shows the tangent slope at a different time. - Read the rise and run of each segment separately. - Connect positive and negative rates to increasing and decreasing volume.

1. The tangent at \(t=1\) rises \(8\,\text{L}\) over \(2\,\text{min}\), so \(V'(1)\approx4\,\text{L/min}\). 2. The tangent at \(t=4\) falls \(4\,\text{L}\) over \(2\,\text{min}\), so \(V'(4)\approx-2\,\text{L/min}\). 3. The volume is increasing at \(t=1\) and decreasing at \(t=4\).

\(V'(1)\approx4\,\text{L/min}\), filling; \(V'(4)\approx-2\,\text{L/min}\), emptying.

- If the first derivative is a rate, what does the derivative of that rate represent? - What does the sign of the second derivative tell you about how the first derivative changes? - What must be true of \(W'(t)\) while the water volume is increasing? - Examine the value of the first derivative at a time beyond the model's interval.

1. The first derivative \(W'(t)\) is the net rate of change of the water volume. Therefore, \(W''(t)\) is the rate at which that net inflow rate is changing, measured in \(\frac{\text{m}^3}{\text{min}^2}\). 2. Differentiate: \(W'(t) = -0.6t^2 + 6t + 20\) and \(W''(t) = -1.2t + 6\). Then \(W''(2)=3.6\,\frac{\text{m}^3}{\text{min}^2}\), so the net inflow rate is increasing at \(t=2\). Also, \(W''(8)=-3.6\,\frac{\text{m}^3}{\text{min}^2}\), so the net inflow rate is decreasing at \(t=8\). 3. Extending the model to \(t=15\) gives \(W'(15)=-0.6 \cdot 15^2 + 6 \cdot 15 + 20 = -25\,\frac{\text{m}^3}{\text{min}}\). This negative rate would mean the water volume is decreasing, so the extended model would no longer describe a continuously filling basin.

a) \(W''(t)\) is the rate of change of the basin's net inflow rate. b) \(W''(2)=3.6\,\frac{\text{m}^3}{\text{min}^2}\), so the net inflow rate is increasing. \(W''(8)=-3.6\,\frac{\text{m}^3}{\text{min}^2}\), so the net inflow rate is decreasing. c) \(W'(15)=-25\,\frac{\text{m}^3}{\text{min}}\), which would make the modeled volume decrease.

- What quantity does the derivative compare with time in this model? - After differentiating, evaluate the same rate formula at each requested time and use its sign in your interpretation. - Average rate uses two function values over an interval, while instantaneous rate describes one time.

1. The derivative \(c'(t)\) gives the instantaneous rate of change of dye concentration with respect to time. Its units are \(\frac{\text{g}}{\text{L}\cdot\text{h}}\). 2. Differentiate using the quotient rule: \(c'(t)=\frac{10(t+4)^2-20t(t+4)}{(t+4)^4}=\frac{40-10t}{(t+4)^3}\). 3. Evaluate the derivative: \(c'(0)=\frac{5}{8}=0.625\), \(c'(4)=0\), and \(c'(8)=-\frac{5}{216}\approx -0.0231\), all in \(\frac{\text{g}}{\text{L}\cdot\text{h}}\). Thus the concentration is increasing initially, is momentarily unchanged at \(t=4\), and is decreasing at \(t=8\). 4. The average rate of change over \([0,8]\) is \(\frac{c(8)-c(0)}{8}=\frac{5}{72}\approx 0.0694\,\frac{\text{g}}{\text{L}\cdot\text{h}}\). 5. The average rate is positive because the concentration at \(t=8\) is greater than at \(t=0\), while \(c'(4)=0\) describes only the instantaneous change at \(t=4\).

a) \(c'(t)\) is the instantaneous rate of change of dye concentration with respect to time, measured in \(\frac{\text{g}}{\text{L}\cdot\text{h}}\). b) \(c'(0)=0.625\,\frac{\text{g}}{\text{L}\cdot\text{h}}\), \(c'(4)=0\,\frac{\text{g}}{\text{L}\cdot\text{h}}\), and \(c'(8)=-\frac{5}{216}\,\frac{\text{g}}{\text{L}\cdot\text{h}}\approx -0.0231\,\frac{\text{g}}{\text{L}\cdot\text{h}}\). These mean the concentration is increasing initially, momentarily unchanged at \(t=4\), and decreasing at \(t=8\). c) The average rate is \(\frac{5}{72}\,\frac{\text{g}}{\text{L}\cdot\text{h}}\approx 0.0694\,\frac{\text{g}}{\text{L}\cdot\text{h}}\). It is positive even though \(c'(4)=0\) because the two rates describe different kinds of change.

- Rewrite the function so the coefficient and vertical shift are easy to identify. - What happens to the exponential term as time increases without bound? - Use the chain rule to differentiate the exponential expression. - In a time-dependent process, what does the first derivative represent?

1. Rewrite the function as \(U(t)=-15e^{-0.4t}+15\). Starting with \(g(t)=e^{-0.4t}\), reflect the graph across the \(t\)-axis, stretch it vertically by a factor of \(15\), and shift it up \(15\) units. 2. Since \(e^{-0.4t}\to 0\) as \(t\to\infty\), \(U(t)\to 15\). The horizontal asymptote is \(U=15\). It represents the limiting voltage of the capacitor. 3. Differentiate using the chain rule: \(U'(t)=15\left(0.4e^{-0.4t}\right)=6e^{-0.4t}\). 4. At \(t=2\), \(U'(2)=6e^{-0.8}\approx 2.70\,\text{V/s}\). At that instant, the voltage is increasing at about \(2.70\) volts per second.

a) Reflect across the \(t\)-axis, stretch vertically by a factor of \(15\), and shift up \(15\) units. b) The horizontal asymptote is \(U=15\). The capacitor voltage approaches \(15\,\text{V}\). c) \(U'(t)=6e^{-0.4t}\), and \(U'(2)\approx 2.70\,\text{V/s}\).