MATH
posted by SYAZ on .
IN A PHYSICS EXPERIMENT,THE DISTANCE OF X AND Y CM HAS A CONNECTION TO EACH OTHER BY 1/X + 1/Y = 1/5.IF THE DISTANCE OF Y IS INCREASING AT THE RATE 16 CMS^(1),FIND THE RATE OF CHANGE OF THE DISTANCE OF X WHEN Y=9 CM.

So we have
1/x + 1/y = 1/5
differentiate with respect to t
(1/x^2)dx/dt + (1/y^2)dy/dt = 0
(1/x2)dx/dt = (1/y^2)dy/dt
dx/dt = (x^2/y^2)(dy/dt)
given:
when y = 9
1/x + 1/9 = 1/5
1/x = 4/45
x = 45/4
and dy/dt = ???? , not clear what you mean by
16 CM^(1)
what ever it is, sub into
dx/dt = ((45/4)^2 /81) (????)
= (25/16)(????)