Question

in a simple physical apparatus, the distance x cm and y cm are related by the equation 1/x+1/y=1/9, if x is increasing at a rate of 8 cm/s calculate the rate at which y is changing when x=15 cm

Answer 1

related rate problem yes?

Answer 2

you have
\[\frac{1}{x}+\frac{1}{y}=\frac{1}{9}\]
both \(x\) and \(y\) are presumed to be functions of time therefore you can take the derivative with respect to time using the chain rule to get
\[-\frac{1}{x^2}x'-\frac{1}{y^2}y'=0\]

Answer 3

you are told that \(x'=8\) and want to solve for \(y'\) when \(x=15\) so replace \(x\) by 15, \(x'\) by 8 and solve for \(y'\)

Answer 4

oh of course you also need \(y\) but you can find \(y\) when \(x=15\) because you are given \(\frac{1}{x}+\frac{1}{y}=\frac{1}{9}\)

Answer 5

thanks