Quick questions on Related rates and rates of change: VCE Math Methods Unit 4

Read the problem. Name the time-dependent variables (volume, radius, height, area, distance). Write the equation that relates them, drawn from geometry (cone, sphere, cylinder, triangle, circle), physics, or algebra.

Apply $\frac{d}{dt}$ to the entire equation. Each variable contributes its time derivative via the chain rule.

After differentiating, substitute the values at the specific moment the question describes. Note that the differentiation must happen before the substitution; you cannot plug in a value for $r$ first and then differentiate (because then $V$ would be a constant).

The equation from step 3 is linear in the unknown rate. Solve algebraically. State units.

Differentiate: $\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.

Volume: $V = \frac{4}{3} \pi r^3$, so $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.

If both vary: $V = \pi r^2 h$, $\frac{dV}{dt} = 2 \pi r h \frac{dr}{dt} + \pi r^2 \frac{dh}{dt}$ (product rule).

A conical tank with fixed total height $H$ and total radius $R$. As water fills to depth $h$ with surface radius $r$, the ratio $\frac{r}{h} = \frac{R}{H}$ holds. Eliminate $r$ to get $V = \frac{\pi R^2}{3 H^2} h^3$, a function of $h$ alone.

A ladder of length $L$ slides down a wall. Let $x$ be the horizontal distance from the wall to the foot of the ladder and $y$ the height of the top of the ladder.

A trough has cross-section of varying area as the water rises. Compute the cross-sectional area as a function of depth, then $V = A(h) \cdot $ (length), giving $\frac{dV}{dt} = $ (length) $\cdot A'(h) \cdot \frac{dh}{dt}$.

Let $x$ be the distance from the lamp to the person and $s$ the distance from the lamp to the tip of the shadow. By similar triangles (lamp / shadow tip and person / shadow tip), $\frac{5}{s} = \frac{1.8}{s - x}$, giving $5(s - x) = 1.8 s$, i.e. $3.2 s = 5 x$, so $s = \frac{25}{16} x$.

$\frac{ds}{dt} = \frac{25}{16} \frac{dx}{dt}$.

$\frac{dx}{dt} = 1.2$.

$\frac{ds}{dt} = \frac{25}{16} \cdot 1.2 = 1.875$ m/s.

If you replace $r$ with $10$ before differentiating, you implicitly treat $r$ as constant, and $\frac{dr}{dt}$ disappears from the equation. Always differentiate first, then substitute.