Related rates

A related-rates problem has four steps: draw and name the changing quantities, write the equation that links them at every instant, differentiate that equation with respect to t, and only then substitute the given values.

Substituting the numbers before differentiating freezes a variable that is really changing, and the rate comes out as 0.

Worked example: a cube's volume grows at 9 cm³/s. V = x³ gives dx/dt = 3/x². Its surface S = 6x² then changes at 12x·dx/dt = 36/x, which is 3.6 cm²/s when x = 10 cm.

Worked example: a rectangle's length x falls at 3 cm/min while its width y grows at 2 cm/min. At x = 10 and y = 6, the perimeter changes at 2(−3 + 2) = −2 cm/min and the area at 10 × 2 + 6 × (−3) = 2 cm²/min.

Worked example: water enters an inverted cone, with radius equal to half the depth, at 5 m³/h. V = πh³/12 gives dV/dt = (πh²/4)dh/dt, so at h = 4 m the level rises at 5/(4π) m/h, which is 35/88 m/h with π taken as 22/7.

Worked example: a man 2 m tall walks away from a 6 m lamp post at 5 km/h. Similar triangles give shadow length s = x/2 for his distance x, so the shadow lengthens at 5/2 km/h, whatever his position.