Question

Consider the hyperbola y=100/x^2. Y decreases at a rate of 4 units/second on the x>0. What is x' at y=20?

Answer 1

Implicit differentiation yields\[y=\frac{100}{x^2}\implies \frac{\mathrm{d}y}{\mathrm{d}t}=-\frac{200}{x^3}\frac{\mathrm{d}x}{\mathrm{d}t}\]You're given that \(\dfrac{\mathrm{d}y}{\mathrm{d}t}=-4\text{ unit/sec}\) (negative because \(y\) is decreasing). You're interested in the value of \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) when \(y=20\), but you need to know what \(x\) is at this precise moment.
\[20=\frac{100}{x^3}\implies \color{red}{x}=\cdots\]So you have
\[-4=-\frac{200}{\color{red}x^3}\frac{\mathrm{d}x}{\mathrm{d}t}\implies\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{\color{red}x^3}{50}=\cdots\]