Given that 1/y + 1/x = 1/(12) and y decreases at a rate of 1 cm s⁻¹, find the rate of change of x…

Differentiate $\frac{1}{y} + \frac{1}{x} = \frac{1}{12}$ with respect to $t$: $-\frac{1}{y^2}\frac{dy}{dt} - \frac{1}{x^2}\frac{dx}{dt} = 0$ $\frac{dx}{dt} = -\frac{x^2}{y^2}\frac{dy}{dt}$ With $\frac{dy}{dt} = -1$ cm s$^{-1}$, $x = 5$ cm, $y = 1$ cm: $\frac{dx}{dt} = -\frac{25}{1}(-1) = 25$ Hence $x$ increases at $25$ cm s$^{-1}$.