If two resistors of `R_(1)` and `R_(2)` ohms are connected in parallel in an electric circuit to make an R ohm resistor, the value of R can be found from the equation,
`(1)/(R)=(1)/(R_(1))+(1)/(R_(2))`
If `R_(1)` is decreasing at the rate of 1 ohm/sec and `R_(2)` i increasing at the rate of 0.5 ohm/sec, at what rate is R changing when `R_(1)=75` ohm and `R_(2)=50` ohm ?

Explanation

`R_(eq)=(R_(1)R_(2))/(R_(1)+R_(2))=(50xx75)/(125)=30 Omega`
`(1)/(R_(eq))=(1)/(R_(1))+(1)/(R_(2))`
Differentiating,
`(1)/(R_(eq)^(2))(dR_(eq))/(dt)=(1)/(R_(1)^(2))(dR_(1))/(dt)+(1)/(R_(2)^(2))(dR_(2))/(dt)`
`:. " " (dR_(eq))/(dt)=((R_(eq))/(R_(1)))^(2
)(dR_(1))/(dt)+((R_(eq))/(R_(2)))^(2) (dR_(2))/(dt)`
`=((30)/(75))^(2)xx(-1)+((30)/(50))^(2) xx(1)/(2)`
`=(1)/(50)Omega//s`