SOLUTION: Suppose that we have two resistors connected in parallel with resistances R1 and R2 measured in ohms (Ω). The total resistance, R, is then given by:
1/R=1/R1+1/R2
Suppose th
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Question 1193543: Suppose that we have two resistors connected in parallel with resistances R1 and R2 measured in ohms (Ω). The total resistance, R, is then given by:
1/R=1/R1+1/R2
Suppose that R1 is increasing at a rate of 0.4Ω/min and R2 is decreasing at a rate of 0.7Ωmin. At what rate is R changing when R1 = 80Ω and R2=105Ω?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Suppose that we have two resistors connected in parallel with resistances R1 and R2 measured in ohms (Ω). The total resistance, R, is then given by:
1/R=1/R1 + 1/R2
Suppose that R1 is increasing at a rate of 0.4Ω/min and R2 is decreasing at a rate of 0.7Ωmin. At what rate is R changing when R1 = 80Ω and R2=105Ω?
====================
At t = 0, R = R1*R2/(R1+R2) = 80*105/185 = 2160/37 ohms
-----------------
Call R1 a and R2 b
---
1/r = 1/a + 1/b
r^-1 = a^-1 + b^-1
Differentiate
-(r^-2)dr/dt = -(a^-2)da/dt -(b^-2)db/dt
dr/dt = r^2*(da/dt/a^2 + db/dt/b^2)
dr/dt = (2160/37)^2*(0.4/6400 - 0.7/11025)
dr/dt = 0.123002191 ohms/minute
===========================
At t=0, R = 1680/37 ohms, not 2160/37. Oooops.
You can redo the arithmetic.
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