- Treat the stated rate of change of velocity as \(v'(t)\). - Keep the constant contribution \(9\) separate from the velocity-dependent term \(-0.4v(t)\). - The task asks only for the model, so no initial condition or solution is needed.

1. Translate each named rate into the corresponding derivative. 2. Preserve the stated signs and proportional relationships to obtain \(v'(t)=9-0.4v(t)\).

\(v'(t)=9-0.4v(t)\)

- Acceleration is the second derivative \(x''(t)\). - Translate “the negative of \(5\) times displacement” with the coefficient and sign unchanged. - Check that the equation relates acceleration directly to displacement.

1. Translate each named rate into the corresponding derivative. 2. Preserve the stated signs and proportional relationships to obtain \(x''(t)=-5x(t)\).

\(x''(t)=-5x(t)\)

- Write acceleration, velocity, and position as \(s''(t)\), \(s'(t)\), and \(s(t)\). - Translate “\(3\) times velocity minus \(2\) times position” term by term. - Keep the product coefficients attached to the corresponding derivatives.

1. Translate each named rate into the corresponding derivative. 2. Preserve the stated signs and proportional relationships to obtain \(s''(t)=3s'(t)-2s(t)\).

\(s''(t)=3s'(t)-2s(t)\)

- Angular acceleration is \(\theta''(t)\). - Keep \(\sin(\theta(t))\) as a function of the angular displacement, not of time alone. - The word “negative” determines the sign of the restoring term.

1. Translate each named rate into the corresponding derivative. 2. Preserve the stated signs and proportional relationships to obtain \(\theta''(t)=-4\sin(\theta(t))\).

\(\theta''(t)=-4\sin(\theta(t))\)

- Use \(v'(t)\) for acceleration. - The constant acceleration contributes a positive term, while drag contributes a term proportional to \(v(t)\). - Since drag opposes the increase, write its contribution with a negative sign and \(k>0\).

1. Translate each named rate into the corresponding derivative. 2. Preserve the stated signs and proportional relationships to obtain \(v'(t)=12-kv(t)\).

\(v'(t)=12-kv(t)\)

- Determine the sign from the direction of change. - Preserve the stated power of the speed. - Choose the equation containing both features.

1. “Decreases” requires a negative rate. 2. “Proportional to the square” requires the factor \(kv^2\). 3. Therefore, choice A is \(v'=-kv^2\).

A. \(v'=-kv^2\)

- Match acceleration and velocity to the correct derivatives. - Distinguish a product from a sum. - Check the derivative order on both quantities.

1. Acceleration is \(s''(t)\), and velocity is \(s'(t)\). 2. “Product” indicates multiplication by time \(t\). 3. Therefore, \(s''(t)=t s'(t)\).

\(s''(t)=t s'(t)\)

- Does the signal become stronger or weaker over nonnegative distance? - Which option uses the product of both quantities named in the rate relationship?

1. The signal strength decreases, so its derivative is nonpositive for \(x\ge0\). 2. The proportional factors are distance \(x\) and current strength \(S(x)\), multiplied together. 3. Therefore, choice D is \(S'=-kxS\).

D. \(S'=-kxS\)

- Read the derivative as a rate. - Use the sign to determine the direction of change. - Include the time-dependent magnitude in the description.

1. Read \(r'(t)\) as the rate of change of the radius. 2. The negative sign indicates a decrease. 3. The radius decreases at a rate of \(\frac{2}{1+t}\) centimeters per minute.

The radius decreases at a rate of \(\frac{2}{1+t}\) centimeters per minute.

- Read the derivative as a rate with respect to distance. - Use the sign to determine the direction of change. - Describe every factor in the product.

1. Read \(F'(x)\) as the rate of change of signal intensity with distance. 2. The negative sign indicates a decrease. 3. The magnitude is \(0.03\) times the distance traveled and the current intensity.

Signal intensity decreases at a rate equal to \(0.03\) times the distance traveled and the current intensity.

- Label each rate as an input or an output. - Inputs contribute positively and outputs negatively to the net rate. - Match the initial condition to the stated starting charge.

1. Solar input adds charge, so \(4+2\sin(\pi t/6)\) enters with a positive sign. 2. Phone use removes charge, so \(0.15B\) enters with a negative sign. 3. The starting charge is \(B(0)=60\). Therefore, choice B is correct.

B. \(B'=4+2\sin(\pi t/6)-0.15B\), \(B(0)=60\). Solar input raises the charge, while phone use lowers it.

- Compare the amount at the start and end of the hour. - Divide the change in amount by the elapsed time. - An interval average and a rate at a single moment are different quantities.

1. The average rate is \(\frac{135-120}{1}=15\) gallons per hour, so it is positive. 2. An average rate over an interval does not determine the instantaneous rate at its start.

Positive, \(15\,\text{gal/h}\). No; the instantaneous rate at noon is not determined.

- Substitute the given temperature into the rate expression. - A positive temperature derivative means the temperature rises.

1. At \(T=80\), the rate is \(6-0.3(80-70)=3\) degrees Fahrenheit per hour. 2. Since the derivative is positive, the temperature is increasing.

Warming, because \(T'=3\,\text{°F/h}>0\).

- Use \(x''(t)\), \(x'(t)\), and \(x(t)\) for acceleration, velocity, and displacement. - Combine the restoring, damping, and forcing contributions with their stated signs. - Translate “\(2\,\text{ft}\) above equilibrium and momentarily at rest” into the two initial conditions required for a second-order equation.

1. Translate acceleration as \(x''(t)\), velocity as \(x'(t)\), and displacement as \(x(t)\). 2. Combine the three stated acceleration contributions: \(x''(t)=-6x(t)-1.5x'(t)+4\cos(2t)\). 3. Translate the starting data as \(x(0)=2\) and \(x'(0)=0\).

\(x''(t)=-6x(t)-1.5x'(t)+4\cos(2t),\quad x(0)=2,\quad x'(0)=0\)

- Write the adjustment rate as \(0.3\) times the target-minus-current gap. - At \(t=0\), evaluate the target using \(\sin0\) before substituting \(H(0)=55\). - Use only the sign of the resulting derivative to decide whether humidity is initially rising or falling.

1. The target-minus-current gap is \(60+5\sin t-H(t)\). 2. The model is \(H'(t)=0.3\bigl(60+5\sin t-H(t)\bigr)\). 3. At \(t=0\) and \(H=55\), \(H'(0)=0.3\cdot(60-55)=1.5>0\).

\(H'(t)=0.3\bigl(60+5\sin t-H(t)\bigr)\), and \(H'(0)=1.5>0\)

- With downward positive, gravity contributes a positive constant to \(v'(t)\). - Quadratic drag opposes the downward velocity, so its contribution has the form \(-kv(t)^2\) with \(k>0\). - Translate “released from rest” as the initial condition on velocity.

1. With downward positive, gravity contributes \(+32\) to acceleration. 2. Quadratic drag opposes the downward velocity, so it contributes \(-kv(t)^2\). 3. The initial-value problem is \(v'(t)=32-kv(t)^2,\quad v(0)=0\).

\(v'(t)=32-kv(t)^2,\quad v(0)=0,\quad k>0\)

- The derivative \(N'(t)\) has units of users per day. - Divide those units by the units of \(N(t)^2\) to obtain the units of \(k\). - Multiply the proposed units of \(k\) by users squared to verify dimensional consistency.

1. The left side \(N'(t)\) has units of users per day. 2. The factor \(N(t)^2\) has units of users squared. 3. Therefore, \(k\) must have units \((\text{user}\cdot\text{day})^{-1}\) so that \(kN^2\) has units of users per day.

\(k\) has units \((\text{user}\cdot\text{day})^{-1}\), not users per day.

- What sign must the derivative of speed have while the sled slows? - Which quantities affect the rate of slowing down? - How does using \(k>0\) help keep the sign of the rate clear?

1. The speed decreases, so its derivative is negative while the sled moves. 2. Proportionality to the product of the time factor and squared speed gives \(v'(t)=-k(1+t)v(t)^2\).

\(v'(t)=-k(1+t)v(t)^2\), for \(v(t)\ge0\)

- Translate acceleration, velocity, and displacement as \(x''(t)\), \(x'(t)\), and \(x(t)\). - Keep the external forcing term \(3\cos t\) separate from damping and restoring terms. - Use the stated minus signs on both the velocity and displacement contributions.

1. Translate acceleration as \(x''(t)\), velocity as \(x'(t)\), and displacement as \(x(t)\). 2. Preserve the forcing, damping, and restoring signs to obtain \(x''(t)=3\cos t-2x'(t)-5x(t)\).

\(x''(t)=3\cos t-2x'(t)-5x(t)\)

- Identify the quantity whose rate is described. - Include every factor named in the proportional relationship. - Check whether the rate should be positive or negative.

1. Represent the rate of change of brightness by \(B'(x)\). 2. Include both proportional factors and the negative sign for a decrease to obtain \(B'(x)=-kd(x)B(x)\).

\(B'(x)=-kd(x)B(x)\)

- Use \(r'(t)\) for the rate of change of the radius. - Keep the proportional factor exactly as \(1+r(t)^2\); it is a sum inside one factor. - Because the radius decreases and \(k>0\), the modeled rate must be negative.

1. Represent the rate of change of the radius by \(r'(t)\). 2. Use a negative sign for the decrease and the stated proportional factor to obtain \(r'(t)=-k\bigl(1+r(t)^2\bigr)\).

\(r'(t)=-k\bigl(1+r(t)^2\bigr)\)

- Identify the quantity whose rate is described. - Express the number of tickets still available in terms of \(N(t)\). - Check whether the rate should be positive or negative.

1. Represent the rate of change of cumulative ticket sales by \(N'(t)\). 2. Since the sales increase in proportion to the remaining tickets, obtain \(N'(t)=k\bigl(C-N(t)\bigr)\).

\(N'(t)=k\bigl(C-N(t)\bigr)\)

- Use \(h'(t)\) for the rate of change of water depth. - The magnitude of the rate is proportional to \(\sqrt{h(t)}\), not to \(h(t)\) or \(h(t)^2\). - A dropping depth requires a negative sign when \(k>0\).

1. Represent the rate of change of the water depth by \(h'(t)\). 2. Use a negative sign for the drop and the stated proportional factor to obtain \(h'(t)=-k\sqrt{h(t)}\).

\(h'(t)=-k\sqrt{h(t)}\)

- Which process adds to the amount of paper, and which removes it? - Express a percentage of the current amount as a rate per day. - Combine the signed contributions for the net rate.

1. Paper enters at a rate of \(8\,\frac{\text{lb}}{\text{day}}\), increasing \(P\). 2. Removal accounts for \(0.15P\) pounds per day, decreasing \(P\). 3. The net rate is \(P'=8-0.15P\), which is choice C.

C. \(P'=8-0.15P\)

- Express the number of unreserved seats in terms of \(Q\). - Apply the stated square root to that quantity. - Check that the rate is positive while seats remain available.

1. The number of unreserved seats is \(900-Q\). 2. Taking its square root gives the rate factor \(\sqrt{900-Q}\). 3. Therefore, choice D is \(Q'=k\sqrt{900-Q}\).

D. \(Q'=k\sqrt{900-Q}\)

- The fixed arrival of \(25\) members contributes a positive constant term. - The loss term must contain the complete quantity \(P+10\) multiplied by \(k\). - Choose the equation that subtracts the loss term rather than multiplying the arrival term by \(P\).

1. The fixed arrival contributes \(+25\). 2. The proportional loss contributes \(-k(P+10)\). 3. Therefore, choice B is \(P'=25-k(P+10)\).

B. \(P'=25-k(P+10)\)

- Combine inflow and outflow to form the net rate. - Follow the stated division by the average depth. - Verify that the resulting units are area per unit time.

1. Form the net volume rate as \(I(t)-O(t)\). 2. Divide that rate by the constant depth \(d\). 3. Therefore, choice A is \(A'=\frac{I(t)-O(t)}{d}\).

A. \(A'=\frac{I(t)-O(t)}{d}\)

- Interpret the constant term \(14\) as a fixed contribution to the wristband rate. - Interpret \(-0.07C(t)\) as a loss proportional to the current number of valid wristbands. - State both contributions with per-minute language and preserve their signs.

1. Read the left side as the rate of change of the number of valid concert wristbands. 2. Interpret each term on the right side separately. 3. Wristbands are issued at \(14\) per minute and invalidated at \(7\%\) of the current number per minute.

Wristbands are issued at \(14\) per minute and invalidated at \(7\%\) of the current number per minute.

- Which term depends explicitly on time, and which depends on the stored energy? - Interpret the positive and negative signs separately. - Express both contributions as energy per unit time.

1. The derivative \(H'(t)\) is the rate of change of stored thermal energy. 2. The term \(0.5t\) adds energy at a rate that increases with time. 3. The term \(-\frac{H(t)}{20}\) removes energy at a rate proportional to the energy stored, both measured in joules per second.

Thermal energy is added at a rate of \(0.5t\) joules per second and lost at a rate of \(\frac{H(t)}{20}\) joules per second.

- Read \(v'(t)\) as acceleration. - The constant \(6\) contributes acceleration in the positive direction. - The term \(-0.2v(t)^2\) is a negative contribution proportional to the square of the velocity.

1. Read the left side as the rate of change of the modeled velocity. 2. Interpret each term on the right side separately. 3. Acceleration has a constant \(6\,\frac{\text{ft}}{\text{s}^2}\) contribution and a negative quadratic-velocity contribution of \(0.2v(t)^2\).

Acceleration has a constant \(6\,\frac{\text{ft}}{\text{s}^2}\) contribution and a negative quadratic-velocity contribution of \(0.2v(t)^2\).

- Match displacement, velocity, and acceleration with derivatives of \(y(t)\). - Which force tends to bring a displaced object toward equilibrium? - Which term opposes the current direction of motion?

1. Read \(y''(t)\) as acceleration, \(y'(t)\) as velocity, and \(y(t)\) as displacement. 2. The term \(-3y(t)\) is a restoring contribution opposite displacement. 3. The term \(-0.4y'(t)\) is a damping contribution opposite velocity.

Acceleration has a restoring term equal to \(-3\) times displacement and a damping term equal to \(-0.4\) times velocity.

- Interpret \(1200-N(t)\) before describing the positive term. - Describe each right-hand term separately. - Use the sign of each term to decide whether it increases or decreases the quantity.

1. The number of potential subscribers who have not subscribed is \(1200-N(t)\). 2. The positive term represents new subscriptions at \(4\%\) of that remaining pool per day. 3. The negative term represents \(5\) cancellations per day.

New subscriptions occur at \(4\%\) of the remaining potential-subscriber pool per day, while \(5\) subscriptions are canceled per day.

- Separate the contribution that increases the balance from the contribution that decreases it. - Translate the continuous percentage into a term involving the current balance. - Evaluate the model at the initial time after writing the initial-value problem.

1. Interest contributes \(0.05B\) dollars per year, while scholarships subtract \(2000+100t\) dollars per year. 2. Therefore, \(B'=0.05B-(2000+100t)\), with \(B(0)=50{,}000\). 3. The initial rate is \(B'(0)=0.05\cdot50{,}000-2000=500\) dollars per year.

\(B'(t)=0.05B(t)-(2000+100t)\), \(B(0)=50{,}000\), and \(B'(0)=500\) dollars per year.

- Write the net rate as arrivals minus completed jobs. - Translate the percentage completion rate into a term involving the current queue. - Evaluate the forcing term and queue term at the initial time.

1. The queue rate equals arrivals minus completions, so \(Q'=80+20\sin\left(\frac{\pi t}{12}\right)-0.1Q\). 2. The initial condition is \(Q(0)=500\). 3. At \(t=0\), \(Q'(0)=80+20\sin0-0.1\cdot500=30\) jobs per minute.

\(Q'(t)=80+20\sin\left(\frac{\pi t}{12}\right)-0.1Q(t)\), \(Q(0)=500\), and \(Q'(0)=30\) jobs per minute.

- Form excess demand by subtracting supply from demand. - Apply the stated adjustment factor to that difference. - Set the rate equal to zero for equilibrium and evaluate it at the listed price.

1. Excess demand is \(D-S=(500-4p)-(80+2p)=420-6p\). 2. Therefore, \(p'=0.02(420-6p)=8.4-0.12p\). 3. Equilibrium occurs when \(8.4-0.12p=0\), giving \(p=70\). 4. At \(p=50\), \(p'=8.4-0.12\cdot50=2.4\), so the price is increasing at \(\$2.40\) per day.

\(p'=8.4-0.12p\); the equilibrium price is \(\$70\); at \(p=50\), the price is increasing at \(\$2.40\) per day.

- Use “rate in minus rate out” for the salt amount. - Find the outgoing concentration from the current salt amount and the constant tank volume. - Evaluate the completed model at the initial amount to find the initial rate.

1. Salt enters at \(0.5\cdot3=1.5\,\text{lb/min}\). 2. Because the tank volume remains \(100\,\text{gal}\), the salt concentration in the tank is \(\frac{A}{100}\,\text{lb/gal}\). Salt leaves at \(\frac{3A}{100}=0.03A\,\text{lb/min}\). 3. Therefore, the initial-value problem is \(A'=1.5-0.03A\), \(A(0)=20\). 4. The initial rate is \(A'(0)=1.5-0.03\cdot20=0.9\,\text{lb/min}\).

\(A'=1.5-0.03A\), \(A(0)=20\), and \(A'(0)=0.9\,\text{lb/min}\).

- Separate the pump's addition from the body's removal. - What value of the amount belongs in the initial condition? - Evaluate the two rate terms using the starting time and amount.

1. Input minus removal gives \(M'=12+3t-0.04M^{3/2}\). 2. The initial condition is \(M(0)=25\). 3. The initial net rate is \(M'(0)=12-0.04\cdot25^{3/2}=12-5=7\) milligrams per hour.

\(M'=12+3t-0.04M^{3/2}\), \(M(0)=25\), and \(M'(0)=7\) milligrams per hour.

- Combine one input term with two distinct removal terms. - Keep the time-dependent arrival rate unchanged in the model. - Evaluate the model at the initial time and queue size.

1. The net rate is arrival minus both loss terms: \(C'=500e^{-0.1t}-0.03C-0.0002C^2\). 2. The initial condition is \(C(0)=1000\). 3. The initial rate is \(C'(0)=500-0.03\cdot1000-0.0002\cdot1000^2=500-30-200=270\) megabytes per minute.

\(C'=500e^{-0.1t}-0.03C-0.0002C^2\), \(C(0)=1000\), and \(C'(0)=270\) megabytes per minute.

- Label each process as bringing batteries in or taking them out. - Use the starting stock as the initial condition. - Calculate the net rate at the start after correcting the signs.

1. Shipments add batteries, so \(120e^{-0.1t}\) is positive. Orders and damage remove batteries, so both terms are negative. 2. The correct equation is \(I'=120e^{-0.1t}-0.08I-5\). 3. The initial condition is \(I(0)=500\). 4. The initial rate is \(I'(0)=120-0.08\cdot500-5=75\) batteries per day.

Lee reversed all three signs. The correct model is \(I'=120e^{-0.1t}-0.08I-5\), \(I(0)=500\), with \(I'(0)=75\) batteries per day.

- Decide which process adds to the queue and which removes from it. - Form the net rate as input minus output. - Evaluate the trigonometric input at \(t=0\) before subtracting admissions.

1. Arrivals add people at rate \(40+10\cos t\), so that term is positive. 2. Admissions remove people at rate \(0.20N\), so that term is negative. 3. The IVP is \(N'=40+10\cos t-0.20N\), \(N(0)=50\). 4. The initial net rate is \(N'(0)=40+10-0.20\cdot50=40\) people per minute.

The completed rows are: arrivals, \(40+10\cos t\), positive; admissions, \(0.20N\), negative. The IVP is \(N'=40+10\cos t-0.20N\), \(N(0)=50\), and \(N'(0)=40\) people per minute.

- Express the uninformed group in terms of \(R\). - Translate “proportional to the product” directly into a rate equation. - Use dimensional analysis to determine the constant's units.

1. The number who have not heard the announcement is \(1200-R\). 2. The model is \(R'=0.00002R(1200-R)\), with \(R(0)=100\). 3. The initial rate is \(R'(0)=0.00002\cdot100\cdot1100=2.2\) students per hour. 4. Since \(R(1200-R)\) has units of students squared and \(R'\) has units of students per hour, the constant has units of \(1\) per student-hour.

\(R'=0.00002R(1200-R)\), \(R(0)=100\), and \(R'(0)=2.2\) students per hour. The proportionality constant has units of \(1\) per student-hour.

- Treat power input as positive and system use as negative. - Use the stated energy-rate relationship before evaluating the initial state. - Evaluate the cosine and stored-energy term at the initial state.

1. Stored energy changes by input power minus output power: \(B'=8+2\cos\left(\frac{\pi t}{12}\right)-5-0.01B^2\). 2. The initial condition is \(B(0)=20\). 3. The initial rate is \(B'(0)=8+2-5-0.01\cdot400=1\), so the initial net power is \(1\,\text{W}\).

\(B'=3+2\cos\left(\frac{\pi t}{12}\right)-0.01B^2\), \(B(0)=20\), and the initial net power is \(1\,\text{W}\).

- Treat invitations as an input and departures as an output. - Keep the interaction term in its given factored form. - Evaluate the square root and interaction term at the initial value.

1. The net rate is invitation-based growth minus departures: \(A'=0.02A(500-A)-3\sqrt{A}\). 2. The initial condition is \(A(0)=100\). 3. The initial rate is \(A'(0)=0.02\cdot100\cdot400-3\cdot10=800-30=770\) users per day.

\(A'=0.02A(500-A)-3\sqrt{A}\), \(A(0)=100\), and \(A'(0)=770\) users per day.

- Which process increases the amount of chemical? - Convert the hourly percentage breakdown into an amount-dependent removal rate. - Evaluate the rate at the initial time and starting amount.

1. Production adds chemical, while breakdown removes it. Therefore, \(A'=12e^{-0.2t}-0.10A\). 2. The initial condition is \(A(0)=80\). 3. The initial rate is \(A'(0)=12-0.10\cdot80=4\) grams per hour.

\(A'=12e^{-0.2t}-0.10A\), \(A(0)=80\), and \(A'(0)=4\) grams per hour.

- Which term represents the external source and which represents treatment? - Why is the ratio \(C/C_*\) unitless? - Check both rates at the starting time before finding the net rate.

1. The net concentration rate is the source minus treatment: \(C'=0.5+0.2\sin t-\frac{0.1C}{1+C/C_*}\), where \(C_*=1\,\text{mg/L}\). 2. The initial condition is \(C(0)=1\,\text{mg/L}\). 3. The initial rate is \(C'(0)=0.5-\frac{0.1}{2}=0.45\) milligram per liter per hour.

\(C'=0.5+0.2\sin t-\frac{0.1C}{1+C/C_*}\), \(C_*=1\,\text{mg/L}\), \(C(0)=1\,\text{mg/L}\), and \(C'(0)=0.45\,\frac{\text{mg}}{\text{L}\cdot\text{h}}\).

- Decide which process adds to the unresolved queue. - A resolution rate removes reports and therefore has a negative sign. - Evaluate both rates at \(t=0\) and \(M=100\).

1. New reports add to the queue, so \(\frac{60}{t+1}\) is positive. Resolved reports leave the queue, so \(0.30M\) is negative. 2. The correct IVP is \(M'=\frac{60}{t+1}-0.30M\), with \(M(0)=100\). 3. The initial net rate is \(M'(0)=60-0.30\cdot100=30\) reports per hour.

The resolution term must be negative. The correct IVP is \(M'=\frac{60}{t+1}-0.30M\), \(M(0)=100\), and \(M'(0)=30\) reports per hour.

- Treat added sugar as an input and consumption as an output. - Keep the saturating consumption rate as one fraction. - Evaluate the consumption term at \(S=40\).

1. Sugar input is positive and yeast consumption is negative, so \(S'=5-\frac{6S}{20+S}\). 2. The initial condition is \(S(0)=40\). 3. The initial consumption rate is \(\frac{6\cdot40}{20+40}=4\) grams per hour, so \(S'(0)=5-4=1\) gram per hour.

\(S'=5-\frac{6S}{20+S}\), \(S(0)=40\), and \(S'(0)=1\) gram per hour.

- Substitute each physical parameter into the stated circuit balance. - Keep the voltage's time dependence unchanged. - Evaluate the equation at the initial time and current.

1. Substituting the circuit data gives \(2I'+5I=10e^{-t}\). 2. The initial condition is \(I(0)=1\). 3. At \(t=0\), \(2I'(0)+5=10\), so \(I'(0)=2.5\) amperes per second.

\(2I'+5I=10e^{-t}\), \(I(0)=1\), and \(I'(0)=2.5\) amperes per second.

- Add buoyancy, weight, and drag with their stated directions. - Use the supplied force law to convert net force into acceleration. - Evaluate the signed quadratic drag carefully at a negative velocity.

1. Newton's second law gives \(100v'=1200-980-30v|v|\). 2. Therefore, \(v'=2.2-0.3v|v|\), with \(v(0)=-1\). 3. Since \(v(0)|v(0)|=-1\), \(v'(0)=2.2+0.3=2.5\) meters per second squared. The positive sign means the acceleration is upward.

\(v'=2.2-0.3v|v|\), with \(v(0)=-1\). The initial acceleration is \(2.5\) meters per second squared upward.

- The height of the line gives the derivative for each amount on the horizontal axis. - Identify the line's vertical intercept and slope. - At an equilibrium amount, the derivative is zero.

1. The straight line has vertical intercept \(4\) and slope \(-1\), so the rate is \(m'(t)=4-m(t)\). 2. Setting the rate equal to zero gives \(m=4\) kilograms.

\(m'(t)=4-m(t)\); equilibrium at \(4\,\text{kg}\).

- What quantity does the stated percentage apply to? - The left side measures liters per minute; check the units of each term on the right.

1. The loss depends on the amount in the tank, not on the elapsed time. 2. The correct model is \(V'(t)=4-0.1V(t)\), with \(V\) in liters and \(t\) in minutes.

The student used time instead of the current volume. \(V'(t)=4-0.1V(t)\).

- A zero derivative means a horizontal tangent. - Locate the highest part of the curve. - Compare the signs of slopes before and after that point.

1. The curve has a horizontal tangent at its highest point, \(t=3\). 2. It increases before \(t=3\) and decreases afterward, so the change from filling to emptying occurs at \(t=3\).

Both occur at point \(B\), approximately \(t=3\,\text{h}\).

- Examine whether the target is constant or changes with time. - What happens to the target after one or two hours? - A fixed equilibrium would require the rate to remain zero at the same temperature.

1. At \(t=2\) hours, the target is \(21\) degrees Celsius. 2. If \(T(2)=20\), then \(T'(2)=0.4(21-20)=0.4\) degrees Celsius per hour, not zero. 3. The target keeps rising, so \(20\) cannot be a fixed equilibrium temperature of this model.

\(T'(2)=0.4\,\text{°C/h}\). The temperature would still be rising at \(20\) degrees Celsius after two hours; the target is not fixed at \(20\).

- Write the net rate as inflow minus outflow. - The initial amount is zero because the tank begins with pure water. - Find the equilibrium amount by setting the derivative equal to zero. - Examine the exponential term as time becomes large.

1. The rate of change equals the constant inflow minus the proportional outflow: \(m'(t)=15-0.025m(t)\), with \(m(0)=0\). 2. At equilibrium, \(15-0.025S=0\), so \(S=600\). The solution satisfying \(m(0)=0\) is \(m(t)=600(1-e^{-0.025t})\). 3. After \(60\) minutes, \(m(60)=600(1-e^{-1.5})\approx 466.12\,\text{g}\). 4. As \(t\to\infty\), \(e^{-0.025t}\to 0\), so \(m(t)\to 600\,\text{g}\).

a) \(m'(t)=15-0.025m(t)\), with \(m(0)=0\) b) \(m(t)=600(1-e^{-0.025t})\) c) \(m(60)\approx 466.12\,\text{g}\) d) \(600\,\text{g}\)

- What happens to units when a quantity is differentiated with respect to time? - What would it mean if twice as much bacterial mass produced twice as much new mass? - What happens to the slope of a curve at a logistic inflection point? - Can a culture in a closed dish grow exponentially forever?

1. Since mass is measured in grams and time in hours, \(m'(t)\) has units \(\frac{\text{g}}{\text{h}}\) and represents the instantaneous growth rate. The second derivative \(m''(t)\) has units \(\frac{\text{g}}{\text{h}^2}\) and represents how the growth rate changes over time. 2. The equation \(m'(t)=km(t)\) states that the growth rate is proportional to the current mass. Biologically, it assumes that each bacterium contributes to reproduction at a constant average rate, producing unrestricted exponential growth. 3. A sign change in \(m''(t)\) identifies an inflection point. At this time, \(m'(t)\) reaches its maximum, so the culture is growing as rapidly as possible before its growth rate begins to decrease. 4. Exponential growth assumes unlimited resources. In a petri dish, limited space and nutrients and the accumulation of waste eventually slow and stop growth.

a) \(m'(t)\) has units \(\frac{\text{g}}{\text{h}}\) and is the growth rate. \(m''(t)\) has units \(\frac{\text{g}}{\text{h}^2}\) and is the rate of change of the growth rate. b) The growth rate is proportional to the current bacterial mass. c) It is the time of maximum growth rate, where growth changes from accelerating to slowing. d) Real cultures have limited space and nutrients, so unrestricted exponential growth cannot continue indefinitely.

- A decreasing concentration has a negative derivative. How does the stated proportionality affect the expression for that derivative? - Use the observed concentration and its rate to find the coefficient, including its units. - Use the starting concentration as the initial condition. There is no need to solve the equation.

1. Write the rate law as \(C'(t)=-kC(t)^2\), where \(k>0\). 2. Use \(C=4\) and \(C'=-0.8\): \(-0.8=-16k\), so \(k=0.05\,\frac{\text{L}}{\text{g}\cdot\text{min}}\). 3. The initial-value problem is \(C'(t)=-0.05C(t)^2,\quad C(0)=4\).

\(k=0.05\,\frac{\text{L}}{\text{g}\cdot\text{min}}\), and \(C'(t)=-0.05C(t)^2,\quad C(0)=4\)

- The radius changes with time, so which differentiation rule applies to the volume formula? - Compare the volume's rate of change with the rate described using surface area. - Which values of the radius allow you to divide by the surface-area expression?

1. Differentiate \(V=\frac{4}{3}\pi r^3\) with respect to \(t\): \(V'=4\pi r^2r'\). 2. The stated model gives \(V'=k(4\pi r^2)\). 3. For \(r>0\), equating the rates and canceling \(4\pi r^2\) gives \(r'(t)=k\).

\(r'(t)=k\) for \(r(t)>0\)

- Joint proportionality gives a rate containing both \(R(t)\) and \(1/(1+t)\). - Use the negative sign for decay, then substitute the initial value and initial derivative to determine \(k\). - Insert the resulting constant and \(R(0)=90\) into the IVP without solving it.

1. Translate the rate statement as \(R'(t)=-\frac{k}{1+t}R(t)\). 2. At \(t=0\), \(R'(0)=-kR(0)=-90k\). 3. Using \(R'(0)=-18\) gives \(k=0.2\). 4. The initial-value problem is \(R'(t)=-\frac{0.2}{1+t}R(t),\quad R(0)=90\).

\(k=0.2\), and \(R'(t)=-\frac{0.2}{1+t}R(t),\quad R(0)=90\)

- Which geometric quantity is changing, and which measures the boundary of the circle? - Both area and circumference depend on radius. How could you remove radius from the final equation? - Should the area derivative be positive or negative?

1. For a circle, \(A=\pi r^2\), so \(r=\sqrt{\frac{A}{\pi}}\). 2. The circumference is \(2\pi r=2\pi\sqrt{\frac{A}{\pi}}\). 3. The rate model is \(A'(t)=k\cdot2\pi\sqrt{\frac{A(t)}{\pi}}=2k\sqrt{\pi A(t)}\).

\(A'(t)=2k\sqrt{\pi A(t)}\)

- Identify the quantity whose rate is described. - Rewrite the area in terms of the side length. - Check whether the rate should be positive or negative.

1. The area is \(A(t)=s(t)^2\), so \(A'(t)=2s(t)s'(t)\). 2. Since the area decreases at a constant rate of \(6\), set \(A'(t)=-6\) to obtain \(2s(t)s'(t)=-6\).

\(2s(t)s'(t)=-6\)

- How many lamps are expected to be unlit if \(L(t)\) are lit? - Which change increases the lit count and which decreases it? - Combine the two signed rates before simplifying.

1. The expected number of unlit lamps is \(500-L(t)\). 2. Turning on adds to the lit count and turning off subtracts from it, so \(L'(t)=0.3\bigl(500-L(t)\bigr)-0.1L(t)\). 3. Simplifying gives \(L'(t)=150-0.4L(t)\).

\(L'(t)=150-0.4L(t)\)

- Decide separately whether each process increases or decreases the amount present. - The removal rate depends on how much medicine is currently in the body. - Check that every term in the rate equation has units of milligrams per hour.

1. The infusion adds medicine, so its contribution is \(+12\). The body removes medicine, so its contribution is \(-0.15A\). 2. Therefore, the differential equation is \(A'=12-0.15A\). 3. With the initial amount included, the initial-value problem is \(A'=12-0.15A\), \(A(0)=40\). 4. The constant \(12\) has units \(\text{mg/h}\), and \(0.15\) has units \(\text{h}^{-1}\), so \(0.15A\) has units \(\text{mg/h}\).

a) \(A'=12-0.15A\). Infusion increases the amount, and removal decreases it. b) \(A'=12-0.15A\), \(A(0)=40\) c) \(12\) has units \(\text{mg/h}\), and \(0.15\) has units \(\text{h}^{-1}\).

- Express the cone's volume using only its height. - Differentiate the volume with respect to time before using the input and compaction rates. - Evaluate the resulting height rate at the stated height to determine its sign.

1. The cone's volume is \(V=\frac{1}{3}\pi r^2h=\frac{4}{3}\pi h^3\), so \(V'=4\pi h^2h'\). 2. The net volume rate is \(V'=4\pi-\frac{V}{25}=4\pi-\frac{4\pi h^3}{75}\). 3. Equating the two expressions and dividing by \(4\pi h^2\) gives \(h'=\frac{1}{h^2}-\frac{h}{75}\). 4. At \(h=5\), \(h'=\frac{1}{25}-\frac{1}{15}=-\frac{2}{75}\,\text{m/min}\), so the height is decreasing.

\(h'(t)=\frac{1}{h(t)^2}-\frac{h(t)}{75}\). At \(h=5\,\text{m}\), \(h'=-\frac{2}{75}\,\text{m/min}\), so the height is decreasing.

- Convert cable speed into a volume rate using its cross-sectional area. - Differentiate the given reel-volume expression with respect to time. - Equate the incoming volume rate to the reel's volume rate.

1. The incoming cable volume rate is \(0.001\cdot2=0.002\,\text{m}^3/\text{s}\). 2. Differentiating the reel volume gives \(V'=0.8\pi r r'\). 3. Equating rates gives \(0.8\pi r r'=0.002\), so \(r'=\frac{1}{400\pi r}\). 4. At \(r=0.2\), \(r'=\frac{1}{80\pi}\,\text{m/s}\).

\(r'(t)=\frac{1}{400\pi r(t)}\), and when \(r=0.2\,\text{m}\), \(r'=\frac{1}{80\pi}\,\text{m/s}\).

- Differentiate the stored-energy expression with respect to time. - Use the stated power balance to relate the energy rate to supplied and dissipated power. - Use the positive-voltage condition when simplifying.

1. Since \(C=2\), the stored energy is \(U=V^2\), so \(U'=2VV'\). 2. Net power equals input minus loss: \(U'=3V-\frac{V^2}{4}\). 3. For \(V>0\), \(2VV'=3V-\frac{V^2}{4}\), so \(V'=\frac{3}{2}-\frac{V}{8}\). 4. At \(V=4\), \(V'=\frac{3}{2}-\frac{1}{2}=1\,\text{V/s}\).

\(V'(t)=\frac{3}{2}-\frac{V(t)}{8}\), and when \(V=4\,\text{V}\), \(V'=1\,\text{V/s}\).

- Express the conserved salt mass as concentration times volume. - Differentiate that product while accounting for the changing volume. - Require the remaining water volume to stay positive. - Compute the initial concentration before evaluating its rate.

1. The volume is \(V(t)=100-2t\), while the salt mass is constant. Since salt mass equals \(CV\), \((CV)'=0\). 2. Thus \(C'V+CV'=0\), so \(C'=\frac{2C}{100-2t}\). 3. Initially, \(C(0)=\frac{5}{100}=0.05\,\text{kg/L}\). 4. The model is physically meaningful while the volume is positive, so \(0\le t<50\) minutes. 5. Therefore, \(C'(0)=\frac{2\cdot0.05}{100}=0.001\,\frac{\text{kg}}{\text{L}\cdot\text{min}}\).

\(C'(t)=\frac{2C(t)}{100-2t}\), \(C(0)=0.05\,\text{kg/L}\), for \(0\le t<50\), and \(C'(0)=0.001\,\frac{\text{kg}}{\text{L}\cdot\text{min}}\).

- Which process raises the dye amount, and which lowers it? - Use the initial amount and rate to find the unknown coefficient. - What units must the coefficient have when multiplied by a squared mass to give a mass per second?

1. The net rate is \(R'=40e^{-0.2t}-kR^2\). 2. At \(t=0\), \(15=40-k(10)^2\). 3. Thus \(100k=25\), so \(k=0.25\,\frac{1}{\text{mg}\cdot\text{s}}\). 4. The initial-value problem is \(R'=40e^{-0.2t}-0.25R^2\), \(R(0)=10\).

\(k=0.25\,\frac{1}{\text{mg}\cdot\text{s}}\), and \(R'(t)=40e^{-0.2t}-0.25R(t)^2\), with \(R(0)=10\,\text{mg}\).

- Compare the chamber temperature with the changing surrounding temperature. - The heater adds to the rate while cooling toward the air temperature opposes the difference. - Use the initial temperature and rate to find the unknown coefficient, including its units.

1. The model is \(T'=-k[T-(20+2t)]+4\). 2. At \(t=0\), \(-8=-k(80-20)+4\). 3. Thus \(-12=-60k\), so \(k=0.20\,\text{min}^{-1}\). 4. The initial-value problem is \(T'=-0.20[T-(20+2t)]+4\), \(T(0)=80\).

\(k=0.20\,\text{min}^{-1}\), and \(T'=-0.20[T-(20+2t)]+4\), \(T(0)=80\).

- Differentiate the given volume-depth relation with respect to time. - Set the resulting volume rate equal to inflow minus outflow. - Substitute the initial depth after isolating \(h'\).

1. Differentiating the volume relation gives \(\frac{dV}{dt}=(100+20h)h'\). 2. The net volume rate is \(50-5h\), so \((100+20h)h'=50-5h\). 3. Therefore, \(h'=\frac{50-5h}{100+20h}\), with \(h(0)=2\). 4. The initial rate is \(h'(0)=\frac{40}{140}=\frac27\) meters per minute.

\(h'=\frac{50-5h}{100+20h}\), \(h(0)=2\), and \(h'(0)=\frac27\) meters per minute.

- Differentiate the sphere volume with respect to time. - Set volume change equal to inflow minus leakage. - Use the initial radius after expressing the volume rate in terms of the radius change.

1. The bubble volume is \(V=\frac{4}{3}\pi r^3\), so \(V'=4\pi r^2r'\). 2. The net volume rate is \(12\pi r-3\pi r^2\). 3. Thus \(4\pi r^2r'=12\pi r-3\pi r^2\), so \(r'=\frac{3}{r}-\frac{3}{4}\). 4. With \(r(0)=2\), \(r'(0)=\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\) centimeter per second.

\(r'=\frac{3}{r}-\frac{3}{4}\), \(r(0)=2\), and \(r'(0)=\frac{3}{4}\) centimeter per second.

- Express mass using the cube's volume and given density. - Differentiate the mass expression as the side length changes. - Equate mass change to deposited material minus material worn away, then use the initial value.

1. The mass is \(M=2s^3\), so \(M'=6s^2s'\). 2. The net mass rate is \(12s^2-4s\). 3. Thus \(6s^2s'=12s^2-4s\), so \(s'=2-\frac{2}{3s}\). 4. With \(s(0)=1\), \(s'(0)=2-\frac{2}{3}=\frac{4}{3}\) centimeters per minute.

\(s'=2-\frac{2}{3s}\), \(s(0)=1\), and \(s'(0)=\frac{4}{3}\) centimeters per minute.

- Differentiate circular area with respect to time. - Set that derivative equal to area added minus area removed. - Divide by the factor multiplying the radial rate.

1. The area is \(A=\pi r^2\), so \(A'=2\pi rr'\). 2. The net area rate is \(3\pi r-0.4\pi r^2\). 3. Thus \(2\pi rr'=3\pi r-0.4\pi r^2\), so \(r'=1.5-0.2r\). 4. At \(r=4\), \(r'(0)=1.5-0.8=0.7\) centimeter per minute.

\(r'=1.5-0.2r\), \(r(0)=4\), and \(r'(0)=0.7\) centimeter per minute.

- Differentiate the product of mass and velocity. - Include the momentum needed to accelerate newly collected sand. - Use the initial speed after isolating the acceleration.

1. Differentiate momentum: \(\frac{d}{dt}(Mv)=M v'+M'v=(50+2t)v'+2v\). 2. Set this equal to the net external force: \((50+2t)v'+2v=100-4v\). 3. Thus \(v'=\frac{100-6v}{50+2t}\), with \(v(0)=0\). 4. The initial acceleration is \(v'(0)=\frac{100}{50}=2\) meters per second squared.

\(v'=\frac{100-6v}{50+2t}\), \(v(0)=0\), and \(v'(0)=2\) meters per second squared.

- Relate volume change to cross-sectional area times depth change. - Set that rate equal to inflow minus outflow. - Substitute the initial depth after isolating \(h'\).

1. Volume change satisfies \(V'=(20+5h)h'\). 2. The net volume rate is \(30-4\sqrt{h}\). 3. Thus \(h'=\frac{30-4\sqrt{h}}{20+5h}\), with \(h(0)=4\). 4. The initial rate is \(h'(0)=\frac{30-8}{40}=0.55\) meter per minute.

\(h'=\frac{30-4\sqrt{h}}{20+5h}\), \(h(0)=4\), and \(h'(0)=0.55\) meter per minute.

- Write the net change as inflow minus removal. - Find the equilibrium amount from the differential equation. - The long-term value is the equilibrium amount. - For part d), express the equilibrium in terms of the unknown inflow.

1. The net rate is inflow minus proportional removal: \(P'(t)=20-0.08P(t)\), with \(P(0)=50\). 2. The equilibrium amount is \(S=\frac{20}{0.08}=250\). Therefore, \(P(t)=250+(50-250)e^{-0.08t}=250-200e^{-0.08t}\). 3. \(P(10)=250-200e^{-0.8}\approx 160.13\,\text{kg}\). 4. The long-term amount is \(250\,\text{kg}\), which is below \(300\,\text{kg}\), so the requirement is satisfied. 5. If the inflow is \(Z\,\text{kg/day}\), the equilibrium amount is \(\frac{Z}{0.08}\). Requiring \(\frac{Z}{0.08}\le 200\) gives \(Z\le 16\).

a) \(P'(t)=20-0.08P(t)\), with \(P(0)=50\) b) \(P(10)\approx 160.13\,\text{kg}\) c) Yes; the long-term amount is \(250\,\text{kg}\). d) At most \(16\,\text{kg/day}\)