math
posted by mat .
the distance x cm and y cm are related by the eqt. 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. [ans:18]

(1/x^2)dx/dt  (1/y^2)dy/dt = 0
when x=15
1/15 + 1/y = 1/9
1/y = 2/45
y = 45/2
(1/225)(8)  (4/2025)(dy/dt) = 0
I will let you do the arithmetic 
dx/x^2  dy/y^2 = 0
so
dy/dt =  (y^2/x^2) dx/dt
when x = 15, find y
1/15 + 1/y = 1/9
1/y = 5/45  3/45 = 2/45
so y = 45/2
dy/dy =  (45^2/4/225)(8) = 18