Rate of Change and Related Rates

Two facts drive the whole subtopic:

• Time rate via the chain rule: if $Q = Q(x)$ and $x = x(t)$, then $\dfrac{dQ}{dt} = \dfrac{dQ}{dx}\cdot\dfrac{dx}{dt}$. Differentiate the relation w.r.t. $t$, then substitute the known rate and the given instant.
• Rate of one quantity w.r.t. another: $\dfrac{dQ}{dP} = \dfrac{dQ/dt}{dP/dt} = \dfrac{dQ/dx}{dP/dx}$. This is a RATIO of derivatives, never $Q/P$.

The most tested instance is volume vs. surface area of a sphere: with $V = \tfrac43\pi r^3$ and $S = 4\pi r^2$, $\dfrac{dV}{dS} = \dfrac{dV/dr}{dS/dr} = \dfrac{4\pi r^2}{8\pi r} = \dfrac{r}{2}$.

Standard formulas to differentiate w.r.t. $t$:

• Circle: $A = \pi r^2 \Rightarrow \dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt}$.
• Sphere: $V = \tfrac43\pi r^3 \Rightarrow \dfrac{dV}{dt} = 4\pi r^2\,\dfrac{dr}{dt}$; $S = 4\pi r^2 \Rightarrow \dfrac{dS}{dt} = 8\pi r\,\dfrac{dr}{dt}$.
• Square (side $a$): area $A = a^2$, perimeter $P = 4a$, diagonal $d = a\sqrt2$ so $A = \tfrac{d^2}{2}$.

When only the AREA rate is given but the perimeter rate is wanted, chain through the side: $\dfrac{dP}{dt} = 4\dfrac{da}{dt}$ and $\dfrac{dA}{dt} = 2a\dfrac{da}{dt}$.

Formulas and the key reductions:

• Cone: $V = \tfrac13\pi r^2 h$; lateral (curved) surface $A = \pi r \ell = \pi r\sqrt{r^2 + h^2}$. If the semi-vertical angle fixes $r = kh$ (e.g. $r = h/2$), substitute FIRST so $V$ is a function of $h$ alone: $V = \tfrac{\pi h^3}{12}$ when $r = h/2$.
• Hemispherical bowl of radius $R$, water depth $x$: $V = \pi\left(Rx^2 - \dfrac{x^3}{3}\right)$, so $\dfrac{dV}{dt} = \pi(2Rx - x^2)\dfrac{dx}{dt}$.
• Cylinder (radius $R$ fixed): $V = \pi R^2 h \Rightarrow \dfrac{dV}{dt} = \pi R^2\dfrac{dh}{dt}$.
• Melting spherical shell (inner radius $R$ fixed, ice out to $R + r$): the outer radius is $R + r$, so $\dfrac{dV}{dt} = 4\pi(R + r)^2\dfrac{dr}{dt}$.

For a rod/ladder of fixed length $L$ with ends at distances $x$ (horizontal) and $y$ (vertical):

• Length constraint: $x^2 + y^2 = L^2$. Differentiate: $2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$, so $\dfrac{dy}{dt} = -\dfrac{x}{y}\dfrac{dx}{dt}$.
• String/kite variant: if the string length is $z$ and the height $h$ is fixed, $x^2 + h^2 = z^2$ gives $\dfrac{dz}{dt} = \dfrac{x}{z}\dfrac{dx}{dt}$.
• Angle variant: with $\sin\theta = \dfrac{y}{L}$, $\cos\theta\,\dfrac{d\theta}{dt} = \dfrac{1}{L}\dfrac{dy}{dt}$ — solve for $\dfrac{d\theta}{dt}$ using $\cos\theta = x/L$ at the instant.

For a point on $y = f(x)$ with $x = x(t)$:

• Coordinate rates: $\dfrac{dy}{dt} = f'(x)\dfrac{dx}{dt}$. Setting $\dfrac{dy}{dt} = k\dfrac{dx}{dt}$ gives $f'(x) = k$ — solve for the points.
• Distance from origin: $D = \sqrt{x^2 + y^2}$, so $\dfrac{dD}{dt} = \dfrac{x\,\dot x + y\,\dot y}{\sqrt{x^2 + y^2}}$.
• Area of a triangle with one moving vertex $(x, y)$: write the area by the coordinate formula $\Delta = \tfrac12|\cdots|$ as a function of the moving parameter, then differentiate.
• Implicit constraint (e.g. on a circle $x^2 + y^2 = 1$): differentiate the constraint, $2x\dot x + 2y\dot y = 0$, and solve for the wanted rate.

The differentiation ladder for motion:

• Velocity: $v = \dfrac{ds}{dt}$. The body is momentarily at rest where $v = 0$.
• Acceleration: $a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}$.
• Read the instant from the condition: 'stops' / 'at rest' $\Rightarrow v = 0$; 'acceleration zero' $\Rightarrow a = 0$; then evaluate the wanted quantity at that $t$.
• Planar motion $x = x(t),\, y = y(t)$: resultant acceleration $= \sqrt{\left(\dfrac{d^2x}{dt^2}\right)^2 + \left(\dfrac{d^2y}{dt^2}\right)^2}$.
• Coefficients from data: for $s = at^2 + bt + c$, $v = 2at + b$, $a_{\text{accel}} = 2a$; solve the given conditions as simultaneous equations.

When a rate is supplied and its accumulated quantity is asked:

• Marginal rate to total: if $\dfrac{dP}{dx} = g(x)$, the extra amount from $x = 0$ to $x = n$ is $\displaystyle\int_0^{n} g(x)\,dx$; add the base level $P_0$: total $= P_0 + \int_0^n g(x)\,dx$.
• Acceleration to velocity: if $a = a(t)$ starting from rest, $v(t) = \displaystyle\int_0^t a(\tau)\,d\tau$; evaluate at the instant the condition fixes (e.g. where $a = 0$).

Always carry the initial value / lower limit — the most common error is dropping the base amount.