Question: A point moves along the curve $y = (2x-1)\sqrt{4x+1}$. Given $\dfrac{dx}{dt} = 0.3$ unit/s at $x = 6$, find $\dfrac{dy}{dt}$ at that instant.

Step 1. Differentiate $y$ w.r.t. $x$ (product rule, with $u = 2x-1$, $v = (4x+1)^{1/2}$):

• $u' = 2$, $v' = \dfrac{2}{\sqrt{4x+1}}$

$$\dfrac{dy}{dx} = 2\sqrt{4x+1} + (2x-1) \cdot \dfrac{2}{\sqrt{4x+1}} = \dfrac{2(4x+1) + 2(2x-1)}{\sqrt{4x+1}} = \dfrac{12x}{\sqrt{4x+1}}$$

Step 2. Evaluate at $x = 6$: $\dfrac{dy}{dx} = \dfrac{72}{\sqrt{25}} = \dfrac{72}{5} = 14.4$.

Step 3. Chain rule:

$$\dfrac{dy}{dt} = \dfrac{dy}{dx} \cdot \dfrac{dx}{dt} = 14.4 \times 0.3 = 4.32 \text{ unit/s}$$

Answer: $\dfrac{dy}{dt} = 4.32$ unit/s.

Tip: For composite-function rate problems, compute $\dfrac{dy}{dx}$ symbolically first, evaluate at the given $x$, then multiply by $\dfrac{dx}{dt}$.

Question: Water leaks from an inverted cone of height $4$ m and base radius $2$ m at the constant rate $0.1$ m³/min. Find the rate at which the water level drops when the depth is $2$ m.

Step 1. Similar triangles: at depth $h$, the water surface has radius $r = \dfrac{h}{2}$.

Step 2. Volume of water:

$$V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi \left(\dfrac{h}{2}\right)^2 h = \dfrac{\pi h^3}{12}$$

Step 3. Differentiate:

$$\dfrac{dV}{dh} = \dfrac{\pi h^2}{4}$$

Step 4. Chain rule $\dfrac{dV}{dt} = \dfrac{dV}{dh} \cdot \dfrac{dh}{dt}$:

$$-0.1 = \dfrac{\pi (2)^2}{4} \cdot \dfrac{dh}{dt} = \pi \cdot \dfrac{dh}{dt}$$ $$\dfrac{dh}{dt} = -\dfrac{0.1}{\pi} \approx -0.0318 \text{ m/min}$$

Answer: Water level drops at $\dfrac{0.1}{\pi} \approx 0.0318$ m/min.

Tip: ALWAYS use similar triangles to eliminate $r$ before differentiating. $V = \dfrac{1}{3}\pi r^2 h$ has two variables — you can't differentiate w.r.t. $h$ alone without substituting $r = f(h)$.

Question: A spherical balloon is inflated at the rate of $50$ cm³/s. Find the rate at which the radius is increasing when $r = 5$ cm.

Step 1. Volume: $V = \dfrac{4}{3}\pi r^3$.

Step 2. Differentiate:

$$\dfrac{dV}{dr} = 4\pi r^2$$

Step 3. Chain rule:

$$\dfrac{dV}{dt} = \dfrac{dV}{dr} \cdot \dfrac{dr}{dt} \Rightarrow 50 = 4\pi (5)^2 \cdot \dfrac{dr}{dt} = 100\pi \cdot \dfrac{dr}{dt}$$ $$\dfrac{dr}{dt} = \dfrac{50}{100\pi} = \dfrac{1}{2\pi} \approx 0.159 \text{ cm/s}$$

Answer: $\dfrac{dr}{dt} = \dfrac{1}{2\pi} \approx 0.159$ cm/s.

Tip: Sphere formulas: $V = \dfrac{4}{3}\pi r^3$, $A = 4\pi r^2$. Hemisphere volumes use $V = \dfrac{2}{3}\pi r^3$. Always check the question's wording — "diameter" vs "radius" trips students up.

Question: The side of a cube increases at $0.5$ cm/s. Find the rate at which the surface area is increasing when the side is $4$ cm.

Step 1. Surface area of cube with side $s$:

$$A = 6s^2$$

Step 2. Differentiate:

$$\dfrac{dA}{ds} = 12s$$

Step 3. Chain rule:

$$\dfrac{dA}{dt} = \dfrac{dA}{ds} \cdot \dfrac{ds}{dt} = 12s \cdot 0.5 = 6s$$

At $s = 4$: $\dfrac{dA}{dt} = 24$ cm²/s.

Answer: Surface area increases at $24$ cm²/s.

Tip: Standard 2D/prismatic formulas to memorise:

• Circle: $A = \pi r^2$, $C = 2\pi r$
• Cube side $s$: $V = s^3$, $A = 6s^2$
• Cylinder: $V = \pi r^2 h$, $A = 2\pi r^2 + 2\pi rh$
• Sector (angle $\theta$ in rad): $A = \tfrac{1}{2}r^2\theta$, arc $= r\theta$

Question: A reservoir's volume changes at $\dfrac{dV}{dt} = \dfrac{4}{t+2}$ m³/min. If $V = 10$ m³ when $t = 0$, find $V$ when $t = 4$.

Step 1. Integrate both sides w.r.t. $t$:

$$V(t) = \int \dfrac{4}{t+2}\,dt = 4\ln|t+2| + C$$

Step 2. Use initial condition: $V(0) = 10$:

$$10 = 4\ln 2 + C \Rightarrow C = 10 - 4\ln 2$$

Step 3. Evaluate at $t = 4$:

$$V(4) = 4\ln 6 + 10 - 4\ln 2 = 10 + 4\ln\dfrac{6}{2} = 10 + 4\ln 3$$

Numerically: $V(4) \approx 10 + 4(1.0986) \approx 14.39$ m³.

Answer: $V(4) = 10 + 4\ln 3 \approx 14.39$ m³.

Tip: When the rate is given as a function of time, you don't need a chain rule — just integrate directly w.r.t. $t$. The initial condition pins down the constant of integration.

Question: A car's distance from a fixed point is $y = \dfrac{5}{x-1} + 3$ where $x > 1$. Show that $y$ is a decreasing function of $x$.

Step 1. Differentiate (chain rule on $5(x-1)^{-1}$):

$$\dfrac{dy}{dx} = -5(x-1)^{-2} = -\dfrac{5}{(x-1)^2}$$

Step 2. For $x > 1$: $(x-1)^2 > 0$ (any non-zero number squared is positive). So $\dfrac{5}{(x-1)^2} > 0$, and:

$$\dfrac{dy}{dx} = -\dfrac{5}{(x-1)^2} < 0$$

Step 3. Since $\dfrac{dy}{dx} < 0$ for all $x > 1$, $y$ is a decreasing function of $x$. $\checkmark$

Tip: Decreasing $\Leftrightarrow \dfrac{dy}{dx} < 0$. The "argue without solving" version exploits $(x-a)^2 > 0$ for $x \neq a$, plus the explicit minus sign. Cleaner than a sign table.

Question: A bacteria population grows according to $N = 1000 e^{0.2t}$, where $t$ is in hours. Find the rate of growth when $N = 5000$.

Step 1. Differentiate:

$$\dfrac{dN}{dt} = 1000 \cdot 0.2 \cdot e^{0.2t} = 200 e^{0.2t}$$

Step 2. Note that $\dfrac{dN}{dt} = 0.2 \cdot 1000 e^{0.2t} = 0.2 N$.

Step 3. When $N = 5000$:

$$\dfrac{dN}{dt} = 0.2 \times 5000 = 1000 \text{ bacteria/hour}$$

Answer: Growth rate $= 1000$ bacteria/hour.

Tip: For exponential growth $N = N_0 e^{kt}$, the rate $\dfrac{dN}{dt} = kN$ is proportional to the current population This is the defining property of exponential growth — you don't always need to find $t$ first.