Six Pillar Quick Check
Quick check — pick the correct option in each question. These test the six pillars of the integration & DE toolkit.
Integrating $\dfrac{1}{ax+b}$ gives $\dfrac{1}{a}\ln|ax+b| + C$ — the factor $\dfrac{1}{a}$ compensates for the inner derivative $a$.
$\cos^2 x \equiv \dfrac{1 + \cos 2x}{2}$, so $\displaystyle\int \cos^2 x\,\text{d}x = \dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$.
$\displaystyle\int u\,\dfrac{\text{d}v}{\text{d}x}\,\text{d}x = uv - \int v\,\dfrac{\text{d}u}{\text{d}x}\,\text{d}x$. The LIATE rule helps choose which factor to call $u$.
The trapezium rule uses $\dfrac{h}{2}\Big[f(x_0) + 2\big(f(x_1) + \cdots + f(x_{n-1})\big) + f(x_n)\Big]$ — the two end ordinates once each, the middle ordinates twice.
Collect the $y$ terms with $\text{d}y$ on one side and the $x$ terms with $\text{d}x$ on the other: $\dfrac{\text{d}y}{g(y)} = f(x)\,\text{d}x$, then integrate both sides.
As an object cools, $T$ decreases toward $T_a$, so $\dfrac{\text{d}T}{\text{d}t}$ is negative when $T > T_a$. Hence the model is $\dfrac{\text{d}T}{\text{d}t} = -k(T - T_a)$ with $k > 0$.
A student solves the cooling problem “a cup of tea at $80^{\circ}\text{C}$ cools in a room at $20^{\circ}\text{C}$, with rate of cooling proportional to $(T - 20)$” and writes:
DE: $\dfrac{\text{d}T}{\text{d}t} = k(T - 20)$
Separate: $\dfrac{\text{d}T}{T - 20} = k\,\text{d}t$
Integrate: $\ln|T - 20| = kt$
So: $T = 20 + e^{kt}$
What went wrong? (Three errors — find them all.)
Hint: the temperature should decrease, the integration is missing a constant, and the final form is wrong.
👁 Reveal the three errors
Error 1 (Sign): The DE should be $\dfrac{\text{d}T}{\text{d}t} = -k(T - 20)$ with $k > 0$. The tea is cooling, so $T$ decreases — the negative sign is essential. Without it, the model predicts the tea heats up toward infinity.
Error 2 (Missing $+C$): The integration $\ln|T - 20| = kt$ drops the constant of integration. It should be $\ln|T - 20| = -kt + C$. This constant is what allows the initial condition to be applied later.
Error 3 (Algebra): Even ignoring the sign, $T = 20 + e^{kt}$ is wrong. From $\ln|T - 20| = -kt + C$ we get $T - 20 = e^{-kt + C} = Ae^{-kt}$ where $A = e^C > 0$, so $T = 20 + Ae^{-kt}$. The student forgot to introduce $A$ and kept the wrong sign in the exponent.
Correct solution: $T = 20 + 60e^{-kt}$ (with $A = 60$ from $T(0) = 80$). The tea approaches $20^{\circ}\text{C}$ as $t \to \infty$ — matching physical intuition.
Six pillars connect everything we have studied over the past four weeks. Each pillar feeds into the next — especially into differential equations, which consume integration skills.
Choose the technique by inspecting the integrand's shape:
• $\displaystyle\int (ax+b)^n\,\text{d}x$ → reverse differentiation (linear substitution).
• $\displaystyle\int \cos^2 x\,\text{d}x$ → use identity $\cos^2 x = \dfrac{1 + \cos 2x}{2}$ first.
• $\displaystyle\int x e^x\,\text{d}x$ or $\displaystyle\int x \ln x\,\text{d}x$ → by parts (LIATE: pick $u$ = Log > Inverse Tri > Alg > Tri > Exp).
• $\displaystyle\int f(g(x))\,g'(x)\,\text{d}x$ → substitution $u = g(x)$.
Always add $+C$ on indefinite integrals — losing it costs marks in the DE particular-solution step.
• Area under curve: $A = \displaystyle\int_a^b y\,\text{d}x$ (split into parts if the curve crosses the axis).
• Area between two curves: $A = \displaystyle\int_a^b \big(\,y_{\text{upper}} - y_{\text{lower}}\,\big)\,\text{d}x$.
• Volume of revolution (about $x$-axis): $V = \pi\displaystyle\int_a^b y^2\,\text{d}x$.
• Sign change: if $f(a)f(b) < 0$ and $f$ is continuous, a root lies in $(a, b)$.
• Iteration: rewrite $f(x) = 0$ as $x = g(x)$ and iterate $x_{n+1} = g(x_n)$.
• Trapezium rule with $n$ strips of width $h = \dfrac{b-a}{n}$:
The trapezium rule overestimates when the curve is concave-up and underestimates when concave-down.
• Formulate: translate “rate of change is proportional to …” into $\dfrac{\text{d}y}{\text{d}x} = k \cdot f(y)$ (or $f(x)$).
• Separate: rearrange $\dfrac{\text{d}y}{\text{d}x} = f(x)\,g(y)$ as $\dfrac{\text{d}y}{g(y)} = f(x)\,\text{d}x$.
• Integrate both sides to get the general solution (containing $+C$).
• Apply the initial condition $y(x_0) = y_0$ to find $C$ and obtain the particular solution.
Always ask: what does the solution tell us about the real situation?
• Exponential growth/decay: $y = Ae^{kt}$. If $k > 0$, $y \to \infty$ (growth); if $k < 0$, $y \to 0$ (decay). Half-life $= \dfrac{\ln 2}{|k|}$.
• Logistic model: $N = \dfrac{M}{1 + Ae^{-kt}}$. As $t \to \infty$, $N \to M$ (carrying capacity) — growth slows as the population approaches the limit.
• Cooling: $T = T_a + (T_0 - T_a)e^{-kt}$. As $t \to \infty$, $T \to T_a$ (ambient temperature).
Solving a separable DE $\dfrac{\text{d}y}{\text{d}x} = f(x)g(y)$ reduces to two integrations: $\displaystyle\int \dfrac{\text{d}y}{g(y)} = \int f(x)\,\text{d}x$. The DE is only as solvable as the integrations it requires.
• If $g(y) = y$ → $\ln|y|$ (logarithmic integration).
• If $g(y) = y^2 + 1$ → $\arctan y$ (inverse trig).
• If the RHS is $x\sin x$ → integration by parts is needed.
• If $g(y)$ factorises → partial fractions unlock the integral (logistic DEs).
Split into four experts. Each expert holds one piece of the §3.8 toolkit and must complete their own card to certify it — the other three cannot see what is missing. Only when all four pieces are certified does the combined problem unlock on this page: that problem needs every piece at once, so the group has to pool its information.
Reporting rule: when your piece is certified, explain it to the other three in one sentence before the next piece is attempted.
On your own. Each level shows three differential equations. Your job is to name the rule they share, not to solve them. Level 3 only opens when level 2 is cleared, and only a perfect run on the first attempt at each level earns all three stars.
A tank containing water is in the form of a cone with vertex $C$. The axis is vertical and the semi-vertical angle is $60^{\circ}$, as shown in the diagram. At time $t = 0$, the tank is full and the depth of water is $H$. At this instant, a tap at $C$ is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to $\sqrt{h}$, where $h$ is the depth of water at time $t$. The tank becomes empty when $t = 60$.
(i) Show that $h$ and $t$ satisfy a differential equation of the form $\dfrac{\text{d}h}{\text{d}t} = -Ah^{-\frac{3}{2}}$, where $A$ is a positive constant.
(ii) Solve the differential equation given in part (i) and obtain an expression for $t$ in terms of $h$ and $H$.
(iii) Find the time at which the depth reaches $\dfrac{1}{2}H$.
[The volume $V$ of a cone of vertical height $h$ and base radius $r$ is given by $V = \dfrac{1}{3}\pi r^{2}h$.]
👁 Reveal Solution
Semi-vertical angle $60^{\circ}$: $r = h\tan 60^{\circ} = h\sqrt{3}$.
$V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi(3h^2)h = \pi h^3$.
So $\dfrac{\text{d}V}{\text{d}h} = 3\pi h^2$.
Volume decreases at a rate proportional to $\sqrt{h}$: $\dfrac{\text{d}V}{\text{d}t} = -k\sqrt{h}$ for some $k > 0$.
By the chain rule: $\dfrac{\text{d}V}{\text{d}t} = \dfrac{\text{d}V}{\text{d}h}\cdot\dfrac{\text{d}h}{\text{d}t} = 3\pi h^2\cdot\dfrac{\text{d}h}{\text{d}t}$.
So $3\pi h^2 \dfrac{\text{d}h}{\text{d}t} = -k\sqrt{h} = -k h^{1/2}$.
$\dfrac{\text{d}h}{\text{d}t} = -\dfrac{k}{3\pi} h^{1/2 - 2} = -\dfrac{k}{3\pi} h^{-3/2}$.
Setting $A = \dfrac{k}{3\pi} > 0$: $\;\boxed{\dfrac{\text{d}h}{\text{d}t} = -Ah^{-3/2}}$
Separate: $h^{3/2}\,\text{d}h = -A\,\text{d}t$.
Integrate: $\displaystyle\int h^{3/2}\,\text{d}h = -A\displaystyle\int \text{d}t \;\Rightarrow\; \dfrac{2}{5}h^{5/2} = -At + C$.
Apply $h = H$ at $t = 0$: $\dfrac{2}{5}H^{5/2} = C$.
So $\dfrac{2}{5}h^{5/2} = -At + \dfrac{2}{5}H^{5/2}$, giving $At = \dfrac{2}{5}\big(H^{5/2} - h^{5/2}\big)$.
Apply $h = 0$ at $t = 60$: $60A = \dfrac{2}{5}H^{5/2}$, so $A = \dfrac{H^{5/2}}{150}$.
Substitute back: $\dfrac{H^{5/2}}{150}\,t = \dfrac{2}{5}\big(H^{5/2} - h^{5/2}\big)$.
$\boxed{t = 60\!\left(1 - \left(\dfrac{h}{H}\right)^{5/2}\right)}$
Substitute $h = \dfrac{H}{2}$: $\left(\dfrac{h}{H}\right)^{5/2} = \left(\dfrac{1}{2}\right)^{5/2} = \dfrac{1}{2^{5/2}} = \dfrac{1}{4\sqrt{2}}$.
$t = 60\!\left(1 - \dfrac{1}{4\sqrt{2}}\right) = 60 - \dfrac{60}{4\sqrt{2}} = 60 - \dfrac{15}{\sqrt{2}} = 60 - \dfrac{15\sqrt{2}}{2}$.
$\boxed{t = 60 - \dfrac{15\sqrt{2}}{2} \approx 49.4 \text{ seconds}}$
Interpretation: The depth reaches half its initial value after about 49.4 s, not 30 s. The tank drains slowly at first (large $h$ means large cross-section, but the rate depends only on $\sqrt{h}$, so the volume drops faster than the depth). Then drainage accelerates as $h$ shrinks. This is the opposite of linear intuition.
(i) Show that $\displaystyle\int_{6}^{16} \dfrac{6}{2x - 7}\,\text{d}x = \ln 125$.
(ii) Use the trapezium rule with four intervals to find an approximation to $\displaystyle\int_{1}^{17} \log_{10} x\,\text{d}x$, giving your answer correct to 3 significant figures.
👁 Reveal Solution
$\displaystyle\int \dfrac{6}{2x - 7}\,\text{d}x = 6 \cdot \dfrac{1}{2}\ln|2x - 7| + C = 3\ln|2x - 7| + C$.
Evaluate from 6 to 16:
$\big[3\ln|2x - 7|\big]_{6}^{16} = 3\ln|2(16) - 7| - 3\ln|2(6) - 7| = 3\ln 25 - 3\ln 5$.
$= 3(\ln 25 - \ln 5) = 3\ln\!\left(\dfrac{25}{5}\right) = 3\ln 5 = \ln 5^3 = \boxed{\ln 125}$.
Strip width: $h = \dfrac{17 - 1}{4} = 4$.
Ordinates: $x_0 = 1,\, x_1 = 5,\, x_2 = 9,\, x_3 = 13,\, x_4 = 17$.
Values of $y = \log_{10} x$:
• $y_0 = \log_{10} 1 = 0$
• $y_1 = \log_{10} 5 \approx 0.6990$
• $y_2 = \log_{10} 9 \approx 0.9542$
• $y_3 = \log_{10} 13 \approx 1.1139$
• $y_4 = \log_{10} 17 \approx 1.2304$
Trapezium rule:
$\displaystyle\int_1^{17} \log_{10} x\,\text{d}x \approx \dfrac{h}{2}\big[y_0 + y_4 + 2(y_1 + y_2 + y_3)\big]$
$= \dfrac{4}{2}\big[0 + 1.2304 + 2(0.6990 + 0.9542 + 1.1139)\big]$
$= 2\big[1.2304 + 2(2.7671)\big] = 2\big[1.2304 + 5.5343\big] = 2 \times 6.7647$
$\boxed{\approx 13.5 \text{ (3 s.f.)}}$
Check: the curve $y = \log_{10} x$ is concave-down on $[1, 17]$ (since $\dfrac{\text{d}^2 y}{\text{d}x^2} = -\dfrac{1}{x^2 \ln 10} < 0$), so the trapezium rule gives an underestimate of the true value.
Test your understanding of all four lesson objectives. Each question targets one objective — try each before moving on!
Which technique should you use to integrate $\displaystyle\int x e^{x^2}\,dx$?
Let $u = x^2$, then $du = 2x\,dx$, so $\displaystyle\int x e^{x^2}\,dx = \frac{1}{2}\int e^u\,du = \frac{1}{2}e^{x^2} + C$. The presence of $x$ (the derivative of $x^2$) alongside $e^{x^2}$ signals substitution.
The trapezium rule with 4 strips approximates $\displaystyle\int_1^9 f(x)\,dx$. What is the strip width $h$?
$h = \dfrac{b - a}{n} = \dfrac{9 - 1}{4} = \dfrac{8}{4} = 2$. The ordinates are at $x = 1, 3, 5, 7, 9$.
A body cools according to Newton's law: $\dfrac{dT}{dt} = -k(T - T_a)$. After separating variables and integrating, what form does the solution take?
Separate: $\dfrac{dT}{T - T_a} = -k\,dt$. Integrate: $\ln|T - T_a| = -kt + C$. So $T - T_a = Ae^{-kt}$, giving $T = T_a + Ae^{-kt}$. The negative sign in $-k$ produces exponential decay toward $T_a$.
For the cooling model $T(t) = T_a + (T_0 - T_a)e^{-kt}$ with $k > 0$, what is the long-term temperature as $t \to \infty$?
As $t \to \infty$, $e^{-kt} \to 0$ (since $k > 0$), so $T \to T_a + 0 = T_a$. The body approaches the ambient temperature, which makes physical sense for a cooling process.
Integration review — trig identity and definite integral.
(i) Express $\cos^2 x$ in terms of $\cos 2x$.
(ii) Hence show that $\displaystyle\int_{0}^{\frac{1}{3}\pi} \cos^2 x\,\text{d}x = \dfrac{1}{6}\pi + \dfrac{1}{8}\sqrt{3}$.
(iii) By using an appropriate trigonometrical identity, deduce the exact value of $\displaystyle\int_{0}^{\frac{1}{3}\pi} \sin^2 x\,\text{d}x$.
💡 Hint
Use the double-angle identity: $\cos 2x = 2\cos^2 x - 1$, so $\cos^2 x = \dfrac{1 + \cos 2x}{2}$.
For part (iii), use $\sin^2 x + \cos^2 x = 1$, so $\sin^2 x = 1 - \cos^2 x$. Integrate this on the same interval.
👁 Reveal Solution
From $\cos 2x = 2\cos^2 x - 1$:
$\boxed{\cos^2 x = \dfrac{1 + \cos 2x}{2}}$
$\displaystyle\int_0^{\pi/3} \cos^2 x\,\text{d}x = \int_0^{\pi/3} \dfrac{1 + \cos 2x}{2}\,\text{d}x = \dfrac{1}{2}\!\left[x + \dfrac{\sin 2x}{2}\right]_0^{\pi/3}$
$= \dfrac{1}{2}\!\left(\dfrac{\pi}{3} + \dfrac{\sin(2\pi/3)}{2} - 0 - 0\right) = \dfrac{1}{2}\!\left(\dfrac{\pi}{3} + \dfrac{\sqrt{3}/2}{2}\right) = \dfrac{1}{2}\!\left(\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{4}\right)$
$= \boxed{\dfrac{\pi}{6} + \dfrac{\sqrt{3}}{8}}$ ✓
Using $\sin^2 x = 1 - \cos^2 x$:
$\displaystyle\int_0^{\pi/3} \sin^2 x\,\text{d}x = \int_0^{\pi/3} (1 - \cos^2 x)\,\text{d}x = \int_0^{\pi/3} 1\,\text{d}x - \int_0^{\pi/3} \cos^2 x\,\text{d}x$
$= \dfrac{\pi}{3} - \left(\dfrac{\pi}{6} + \dfrac{\sqrt{3}}{8}\right) = \boxed{\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{8}}$
DE review — separation of variables with a trigonometric twist.
The variables $x$ and $t$ are related by the differential equation $e^{2t}\dfrac{\mathrm{d}x}{\mathrm{d}t} = \cos^2 x$, where $t \geq 0$. When $t = 0$, $x = 0$.
(i) Solve the differential equation, obtaining an expression for $x$ in terms of $t$.
(ii) State what happens to the value of $x$ when $t$ becomes very large.
(iii) Explain why $x$ increases as $t$ increases.
💡 Hint
For part (i): separate as $\sec^2 x\,\text{d}x = e^{-2t}\,\text{d}t$. The LHS integrates to $\tan x$; the RHS to $-\dfrac{1}{2}e^{-2t}$.
Apply $x = 0$ at $t = 0$ to find the constant. Solve for $x$ using arctan.
For part (ii): as $t \to \infty$, $e^{-2t} \to 0$, so $\tan x \to$ the constant. For part (iii): inspect the sign of $\dfrac{\text{d}x}{\text{d}t}$.
👁 Reveal Solution
Rearrange: $\dfrac{\text{d}x}{\text{d}t} = e^{-2t}\cos^2 x$, so $\sec^2 x\,\text{d}x = e^{-2t}\,\text{d}t$.
Integrate: $\tan x = -\dfrac{1}{2}e^{-2t} + C$.
Apply $x = 0$ at $t = 0$: $0 = -\dfrac{1}{2} + C$, so $C = \dfrac{1}{2}$.
$\tan x = \dfrac{1}{2} - \dfrac{1}{2}e^{-2t} = \dfrac{1 - e^{-2t}}{2}$.
$\boxed{x = \arctan\!\left(\dfrac{1 - e^{-2t}}{2}\right)}$
As $t \to \infty$, $e^{-2t} \to 0$, so $\tan x \to \dfrac{1}{2}$.
$\boxed{x \to \arctan\!\left(\dfrac{1}{2}\right) \approx 0.4636 \text{ rad}}$
From the DE: $\dfrac{\text{d}x}{\text{d}t} = e^{-2t}\cos^2 x$.
Since $e^{-2t} > 0$ for all $t$ and $\cos^2 x \geq 0$ (with equality only at isolated points), we have $\dfrac{\text{d}x}{\text{d}t} \geq 0$ for all $t \geq 0$, with strict inequality except where $\cos x = 0$.
Therefore $x$ is a non-decreasing function of $t$, and (since $\cos x \neq 0$ near $x = 0$) $x$ is strictly increasing.
Numerical methods review — trapezium rule with multiple strip widths.
The trapezium rule is used to estimate the value of $\displaystyle\int_{0}^{1} \dfrac{4}{1+x^{2}}\,\text{d}x$.
(i) Draw the graph of $y = \dfrac{4}{1+x^{2}}$ for $0 \leqslant x \leqslant 1$.
(ii) Use strip widths of $1$, $0.5$, $0.25$ and $0.125$ to find approximations to the value of the integral.
(iii) State the value of the integral to as many decimal places as you can justify.
💡 Hint 1
For each strip width $h$, the $x$-values are $0, h, 2h, \ldots, 1$, and the ordinates are $y_i = \dfrac{4}{1 + x_i^2}$.
The trapezium formula: $\text{T} = \dfrac{h}{2}[y_0 + y_n + 2(y_1 + \cdots + y_{n-1})]$.
For $h = 0.5$: $x = 0, 0.5, 1$ with $y = 4, 3.2, 2$. Compute $\text{T} = \dfrac{0.5}{2}[4 + 2 + 2(3.2)] = 0.25 \times 12.4 = 3.1$.
💡 Hint 2
The exact value is $\displaystyle\int_0^1 \dfrac{4}{1 + x^2}\,\text{d}x = 4\big[\arctan x\big]_0^1 = 4 \cdot \dfrac{\pi}{4} = \pi \approx 3.14159$.
Compare your trapezium estimates with $\pi$ — they should approach $\pi$ as $h$ shrinks.
The curve is concave-down near $x = 0$ (where $y'' = \dfrac{8(3x^2 - 1)}{(1 + x^2)^3} < 0$ for $x < \dfrac{1}{\sqrt{3}}$) and the function takes its largest values there, so the trapezium rule underestimates the true value overall.
👁 Reveal Solution
$h = 1$ (1 strip): $x = 0, 1$; $y = 4, 2$.
$\text{T}_1 = \dfrac{1}{2}[4 + 2] = \mathbf{3.000}$
$h = 0.5$ (2 strips): $x = 0, 0.5, 1$; $y = 4, 3.2, 2$.
$\text{T}_2 = \dfrac{0.5}{2}[4 + 2 + 2(3.2)] = 0.25 \times 12.4 = \mathbf{3.100}$
$h = 0.25$ (4 strips): $x = 0, 0.25, 0.5, 0.75, 1$; $y = 4, 3.7647, 3.2, 2.56, 2$.
$\text{T}_4 = \dfrac{0.25}{2}[4 + 2 + 2(3.7647 + 3.2 + 2.56)] = 0.125 \times [6 + 19.0494] = 0.125 \times 25.0494 \approx \mathbf{3.131}$
$h = 0.125$ (8 strips): $x = 0, 0.125, 0.25, \ldots, 1$; $y = 4, 3.9385, 3.7647, 3.5068, 3.2, 2.8767, 2.56, 2.2655, 2$.
Sum of middle ordinates: $3.9385 + 3.7647 + 3.5068 + 3.2 + 2.8767 + 2.56 + 2.2655 = 22.1122$.
$\text{T}_8 = \dfrac{0.125}{2}[4 + 2 + 2(22.1122)] = 0.0625 \times [6 + 44.2244] = 0.0625 \times 50.2244 \approx \mathbf{3.139}$
The exact value is $\displaystyle\int_0^1 \dfrac{4}{1+x^2}\,\text{d}x = 4\big[\arctan x\big]_0^1 = 4 \cdot \dfrac{\pi}{4} = \pi$.
The trapezium estimates approach $\pi$ from below (the curve is concave-down where the function is largest, near $x = 0$):
$3.000 \to 3.100 \to 3.131 \to 3.139 \to \pi \approx 3.14159$.
Each halving of $h$ roughly halves the error. From $\text{T}_8 \approx 3.139$ (3 d.p.), we can confidently state:
$\boxed{\displaystyle\int_0^1 \dfrac{4}{1+x^2}\,\text{d}x \approx 3.14 \text{ (to 2 decimal places, justified by convergence)}}$
Interpretation: since the exact value is $\pi$, this is a numerical method for computing $\pi$ itself — the connection between integration and the famous constant.
Comprehensive application — full modeling workflow: formulate, separate, partial fractions, integrate, apply IC, solve for $t$.
An underground storage tank is being filled with liquid. Initially the tank is empty. At time $t$ hours after filling begins, the volume of liquid is $V\,\text{m}^3$ and the depth of liquid is $h\,\text{m}$. It is given that $V = \dfrac{4}{3}h^3$. The liquid is poured in at a rate of $20\,\text{m}^3$ per hour, but owing to leakage, liquid is lost at a rate proportional to $h^2$. When $h = 1$, $\dfrac{dh}{dt} = 4.95$.
(i) Show that $h$ satisfies the differential equation $\dfrac{dh}{dt} = \dfrac{5}{h^2} - \dfrac{1}{20}$.
(ii) Verify that $\dfrac{20h^2}{100 - h^2} \equiv -20 + \dfrac{2000}{(10 - h)(10 + h)}$.
(iii) Hence solve the differential equation in part (i), obtaining an expression for $t$ in terms of $h$.
💡 Hint 1
For part (i): let the leakage rate be $kh^2$. Then $\dfrac{\text{d}V}{\text{d}t} = 20 - kh^2$.
Since $V = \dfrac{4}{3}h^3$, we have $\dfrac{\text{d}V}{\text{d}h} = 4h^2$, and by the chain rule $\dfrac{\text{d}V}{\text{d}t} = 4h^2 \dfrac{\text{d}h}{\text{d}t}$.
So $4h^2 \dfrac{\text{d}h}{\text{d}t} = 20 - kh^2$. Use $h = 1, \dfrac{\text{d}h}{\text{d}t} = 4.95$ to find $k$.
💡 Hint 2
For part (ii): combine the RHS over a common denominator $(10 - h)(10 + h) = 100 - h^2$.
For part (iii): separate as $\dfrac{20h^2}{100 - h^2}\,\text{d}h = \text{d}t$, then use the identity from part (ii) and partial fractions $\dfrac{2000}{(10-h)(10+h)} = \dfrac{A}{10-h} + \dfrac{B}{10+h}$. Multiply through by the full denominator $(10-h)(10+h)$ to get $2000 \equiv A(10+h) + B(10-h)$, then substitute $h = 10$ and $h = -10$ to find $A$ and $B$. Integrate, then apply $h = 0$ at $t = 0$.
👁 Reveal Solution
Let the leakage rate be $kh^2$ for some $k > 0$. Then $\dfrac{\text{d}V}{\text{d}t} = 20 - kh^2$.
$V = \dfrac{4}{3}h^3 \Rightarrow \dfrac{\text{d}V}{\text{d}h} = 4h^2$.
Chain rule: $\dfrac{\text{d}V}{\text{d}t} = \dfrac{\text{d}V}{\text{d}h}\cdot\dfrac{\text{d}h}{\text{d}t} = 4h^2\,\dfrac{\text{d}h}{\text{d}t}$.
So $4h^2\,\dfrac{\text{d}h}{\text{d}t} = 20 - kh^2$, giving $\dfrac{\text{d}h}{\text{d}t} = \dfrac{20 - kh^2}{4h^2} = \dfrac{5}{h^2} - \dfrac{k}{4}$.
Apply $h = 1, \dfrac{\text{d}h}{\text{d}t} = 4.95$: $4.95 = 5 - \dfrac{k}{4}$, so $\dfrac{k}{4} = 0.05$, giving $k = 0.2 = \dfrac{1}{5}$.
Therefore $\dfrac{k}{4} = \dfrac{1}{20}$, and $\;\boxed{\dfrac{\text{d}h}{\text{d}t} = \dfrac{5}{h^2} - \dfrac{1}{20} = \dfrac{100 - h^2}{20h^2}}$
RHS $= -20 + \dfrac{2000}{(10 - h)(10 + h)} = -20 + \dfrac{2000}{100 - h^2}$.
Combine over common denominator $100 - h^2$:
$= \dfrac{-20(100 - h^2) + 2000}{100 - h^2} = \dfrac{-2000 + 20h^2 + 2000}{100 - h^2} = \dfrac{20h^2}{100 - h^2}$
$=$ LHS. ✓
From $\dfrac{\text{d}h}{\text{d}t} = \dfrac{100 - h^2}{20h^2}$, separate: $\dfrac{20h^2}{100 - h^2}\,\text{d}h = \text{d}t$.
Using part (ii): $\left(-20 + \dfrac{2000}{(10 - h)(10 + h)}\right)\text{d}h = \text{d}t$.
Partial fractions: $\dfrac{2000}{(10 - h)(10 + h)} = \dfrac{A}{10 - h} + \dfrac{B}{10 + h}$.
$2000 = A(10 + h) + B(10 - h)$. Set $h = 10$: $2000 = 20A \Rightarrow A = 100$. Set $h = -10$: $2000 = 20B \Rightarrow B = 100$.
So $\dfrac{2000}{(10-h)(10+h)} = \dfrac{100}{10-h} + \dfrac{100}{10+h}$.
Integrate: $\displaystyle\int\!\left(-20 + \dfrac{100}{10-h} + \dfrac{100}{10+h}\right)\text{d}h = \displaystyle\int \text{d}t$.
$-20h - 100\ln|10 - h| + 100\ln|10 + h| = t + C$.
$-20h + 100\ln\!\left|\dfrac{10 + h}{10 - h}\right| = t + C$.
Apply $h = 0$ at $t = 0$: $0 + 100\ln 1 = 0 + C \Rightarrow C = 0$.
$\boxed{t = -20h + 100\ln\!\left(\dfrac{10 + h}{10 - h}\right)}$
Interpretation: as $h \to 10$, the logarithm $\to \infty$, so $t \to \infty$. The depth asymptotically approaches $h = 10$ m (the equilibrium where inflow = leakage) but never quite reaches it — a classic example of a logistic-style approach to a limiting value.
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