Logarithmic models in context

How do you find the instantaneous rate for this model?

Move the time marker. Compare the height of P with the slope of its dashed tangent. The model applies only on the displayed time interval

What happens as time increases from 0 to 4 hours?

Logarithmic rate models

A logarithmic model gives a quantity at a time. Differentiate with respect to time to find its instantaneous rate, then substitute the time and attach the correct units

1. 1Read the baseline At t=0 the logarithm input is 1, so the model value is its baseline constant \(\displaystyle C(0)=20+6\log_e\left(1\right)=20\)
2. 2Differentiate with the chain rule Differentiate the logarithm and multiply by the derivative of its inner linear expression \(\displaystyle C'(t)=\frac{6\times\frac{1}{2}}{1+\frac{1}{2}t}=\frac{3}{1+\frac{1}{2}t}\)
3. 3Attach rate units The derivative has units of quantity per unit of time because it measures change in the model for each unit of time \(\displaystyle C'(2)=\frac{3}{2}=1.5\text{ mg/L per hour}\)
4. 4Solve a target value To find a time for a target, isolate the logarithm and then exponentiate both sides \(\displaystyle C(t)=26\quad\Rightarrow\quad \log_e\left(1+\frac{1}{2}t\right)=1\quad\Rightarrow\quad 1+\frac{1}{2}t=e\)

Concentration model

\(\displaystyle C(t)=20+6\ln(1+t/2)\)

Differentiate the logarithmic term and find the numerator coefficient

Concentration model

\(\displaystyle C(t)=20+6\ln(1+t/2)\)

Find the instantaneous rate at t=2.

Calling the model value a rate. One of these lines is right. Which one?

Read the context interval before solving

A tank concentration C is measured in mg/L and time t in hours. The mathematical logarithm is defined for t>−2, but this model is specified only for 0≤t≤4. Do not extend the model automatically to every mathematical input

Model data

\(\displaystyle C(t)=32-6\ln(1+t/2),\quad t=0\)

Find the concentration at the starting time

Model data

\(\displaystyle C(t)=32-6\ln(1+t/2)\)

Find k in the derivative C′(t)=k/(1+t/2), including its sign

Model data

\(\displaystyle C'(t)=\frac{-3}{1+t/2}\)

Find the signed instantaneous rate at t=2 hours

How should the rate C′(2)=−3/2 be interpreted?

Keep rate sign separate from quantity sign

The denominator is positive throughout the stated interval. The rate is negative, so concentration decreases, but the rate magnitude becomes smaller as the denominator grows. The quantity stays positive on the whole model interval

Integrate the rate for signed change

The baseline cancels when subtracting endpoint values. Since the rate is negative, its integral is a negative change in concentration, not a negative geometric area

Model data

\(\displaystyle C(t)=32-6\ln(1+t/2),\quad [0,4]\)

Write ΔC=k ln(q), keeping k equal to the signed logarithm coefficient in the model. Find k.

Model data

\(\displaystyle C(t)=32-6\ln(1+t/2),\quad [0,4],\quad \Delta C=-6\ln(q)\)

Find q>1 using the upper inner value divided by the lower inner value

Report change and decrease differently

The signed change is −6 ln3 mg/L. The concentration decreases by 6 ln3 mg/L, a positive magnitude. Neither expression is an instantaneous rate; no per-hour unit belongs on this accumulated change

The entry form fixes k=−6 and uses the endpoint ratio as q. Which pair follows that convention?

Solve a target, then check its time

To find when the concentration is 26 mg/L, first isolate the logarithm. A correct algebraic solution still needs to lie in the stated model interval

Model data

\(\displaystyle 32-6\ln(1+t/2)=26\)

Find u after isolating ln(1+t/2)=u for this target

Model data

\(\displaystyle 1+t/2=e\)

After exponentiating, write the time as k(e−1). Find k.

Which final step is needed after finding t=2(e−1)?

A reachable target has a model-valid time

The exact time is inside the context interval. Substituting it gives an inner value of e, whose logarithm is 1, so the concentration is 26 mg/L as required

The model is specified only for 0≤t≤4. What can it establish about a target concentration of 20 mg/L?

Average rate is change divided by elapsed time

An average rate uses the entire interval, unlike C′(2), which describes one instant. Both have concentration-per-hour units, but their values need not be equal

Model data

\(\displaystyle \frac{C(4)-C(0)}4=\frac{-6\ln3}{4}\)

Find k in the average-rate form k ln3 over [0,4]

Construct and interpret a complete model solution, ungraded

A recovered water volume R is measured in litres and t in hours. On paper, state the baseline, differentiate, find and interpret R′(4), calculate the signed change from 0 to 8 hours, and solve R(t)=18 while checking the model interval

Model data

\(\displaystyle R(t)=10+8\ln(1+t/4)\)

After completing the full solution, enter the signed rate R′(4)

Model data

\(\displaystyle R(t)=10+8\ln(1+t/4),\quad [0,8]\)

Enter q in the change form ΔR=8 ln(q) on [0,8], retaining coefficient 8.

Check the whole interpretation

The recovered volume increases throughout the stated interval, but its positive recovery rate decreases. Check the units and target-time validity as well as the numerical entries

1. 1Baseline and derivative The baseline is the amount already represented at t=0. The derivative measures change in that amount per hour \(\displaystyle R(0)=10\text{ L},\quad R'(t)=\frac2{1+t/4},\quad R'(4)=1\text{ L per hour}\)
2. 2Signed accumulated change The positive change has litre units, not litre-per-hour units \(\displaystyle \Delta R=R(8)-R(0)=8\ln3\text{ L}\)
3. 3Reach the target within the model interval The target gives ln(1+t/4)=1. Since 1<e<3, the resulting time lies between 0 and 8 hours \(\displaystyle R(t)=18\Rightarrow t=4(e-1),\quad 0<4(e-1)<8\)

Use the requested quantity and its valid interval

Differentiate for a signed instantaneous rate, integrate the rate for signed change, and divide change by elapsed time for average rate. Interpret units and signs separately. Solve targets algebraically, then check the stated model interval; a positive real time can still be outside it