Appendix H — Problem Set 6 | Differentiation

Homework — CHANGE: the rate of anything, right now

Purpose

You already compute average rates: a change divided by an interval. A calibration slope is one; so is “the concentration dropped 0.60 M in 10 minutes.” But a reaction does not have one speed per ten minutes — it has a speed now, and a different speed a moment later. The derivative is the tool that turns “how much changed over an interval” into “how fast it is changing at this instant.” Every problem below is a chemical or physical function of a single variable; everything else in it is a constant. For each derivative you take, write one sentence saying what the result means physically, and give its units.

The five rules you need (nothing else is required):

\frac{d}{dx}x^n = nx^{n-1} \qquad \frac{d}{dx}e^{ax} = ae^{ax} \qquad \frac{d}{dx}\ln x = \frac{1}{x}

\frac{d}{dx}\big[f(x)+g(x)\big] = f'(x)+g'(x) \qquad \frac{d}{dx}\big[c\,f(x)\big] = c\,f'(x)

A. Before computing: what a derivative is

1. Secant first, tangent second

A reactant follows [\mathrm{A}](t) = 0.80\,e^{-0.139t} (concentration in M, time in minutes).

Compute the average rate of change between t = 0 and t = 5.0 min: \big([\mathrm{A}](5) - [\mathrm{A}](0)\big)/5.0. Give units.
Now shrink the interval: compute the average rate between t = 0 and t = 1.0, then between t = 0 and t = 0.1 min.
The three numbers you computed are approaching something. Predict what the instantaneous rate at t = 0 is, before using any differentiation rule. (You will check this prediction in Problem 7.)

2. Units and signs without any calculation

For each quantity, state the units and whether you expect it to be positive, negative, or either — no computation allowed:

\dfrac{d[\mathrm{A}]}{dt} for a reactant being consumed.
\dfrac{dA}{dc} for absorbance A (dimensionless) versus concentration c in mM.
\dfrac{dV}{dr} for the volume of a growing droplet, V = \tfrac{4}{3}\pi r^3.
\dfrac{dP}{dV} for a fixed amount of ideal gas at constant temperature.

3. Read the graph

Sketch (or imagine) the curve [\mathrm{A}](t) = 0.80\,e^{-0.139t} from t = 0 to t = 20 min.

Where is the derivative most negative — early or late? Say why in one sentence about the chemistry, not the mathematics.
Does the derivative ever reach exactly zero at finite time? What does your answer say about “the reaction finishing”?

B. The rules, on functions you have already met

Differentiate; then give units and one interpretation sentence.

4. A calibration line is a constant derivative

A(c) = 0.140\,c + 0.020 (absorbance versus concentration in mM). Find \dfrac{dA}{dc}. You have been using this derivative all semester — what did you call it?

5. Product formed, production slowing

Moles of product in a batch reactor: n(t) = 3.0\,t - 0.25\,t^{2} (n in mol, t in hours, valid for 0 \le t \le 6).

Find \dfrac{dn}{dt} and evaluate it at t = 1.0 and t = 5.0 h.
At what time does production stop (dn/dt = 0)? What is happening chemically after that time, according to this model?

6. Power laws from physics

Ideal gas, fixed n and T: P(V) = \dfrac{nRT}{V}. Write P = nRT\,V^{-1}, find \dfrac{dP}{dV}, and state its sign. Why must it have that sign physically?
Coulomb potential energy between two charges: V(r) = \dfrac{kq_1q_2}{r}. Find -\dfrac{dV}{dr} (this is the force). For two like charges, is the force positive (repulsive) or negative at every r?

7. The exponential — the chemistry workhorse

[\mathrm{A}](t) = 0.80\,e^{-0.139t} (M, minutes).

Find \dfrac{d[\mathrm{A}]}{dt}. Evaluate it at t = 0 — and compare with your prediction from Problem 1c.
Show that your derivative can be rewritten as \dfrac{d[\mathrm{A}]}{dt} = -0.139\,[\mathrm{A}](t). In words: the rate is proportional to how much is left. Why does that make chemical sense for molecules reacting independently?

8. The logarithm — why the straight-line plot works

Let y(t) = \ln[\mathrm{A}](t) = \ln(0.80) - 0.139\,t.

Find \dfrac{dy}{dt}.
One sentence: you plot \ln[\mathrm{A}] against t and lay a ruler on it. What quantity is the slope of that ruler? (This is the k-extraction you have done before — now named as a derivative.)

C. Medium: combining and checking

9. Position, velocity, acceleration

A molecule in a flow tube: x(t) = 2.0 + 15.0\,t - 4.9\,t^{2} (x in m, t in s).

Find the velocity v(t) = dx/dt and acceleration a(t) = dv/dt, with units.
At what time is the molecule momentarily at rest? How far has it traveled by then?

10. The tempting error (explain, don’t just fix)

A classmate differentiates [\mathrm{A}](t) = 0.80\,e^{-0.139t} and writes: “the exponential is its own derivative, so \dfrac{d[\mathrm{A}]}{dt} = 0.80\,e^{-0.139t}.”

Two independent reasons this must be wrong before any rule is applied: one from the sign (what is the concentration doing?) and one from the units (what units must d[\mathrm{A}]/dt carry, and where can “per minute” possibly come from?).
State the correct rule for e^{at} and the correct derivative.

11. The derivative at zero is where the shortcut comes from

You showed on the exam that e^{x} \approx 1 + x near x = 0.

Compute the derivative of e^{x} at x = 0.
One sentence: the approximation 1 + x is a straight line — what is that line’s slope, and why must it equal your answer from (a)? (A linear approximation is the tangent line.)

12. Choose the model from its derivative (inverse direction)

An instrument reports that a signal’s rate of change is constant at -0.025 units per second, at every moment it measures.

Which function family must the signal belong to: y = y_0 e^{-kt}, y = y_0 - kt, or y = y_0/t? Rule out the other two by differentiating them.
What single later observation would expose the chosen model’s breakdown? (Hint: what does this model predict the signal eventually does that a real physical signal usually cannot?)

Expectations

State the rule you are using before using it.
Carry units through every derivative — a derivative’s units are always (units of output)/(units of input).
End each problem with one sentence of physical meaning.
Submit on the course submission page as Set 6, or hand in on paper.