If R1 and R2 are increasing at rates of 0.3 Ω/s and 0.2 Ω/s, respectively, how fast is R changing when R1 = 70 Ω and R2 = 110 Ω? (Round your answer to three decimal places.)

Question

If two resistors with resistances R₁ and R₂ are connected in parallel, as in the figure below, then the total resistance R, measured in ohms (Ω), is given by

1/R=1/R1+1/R2

.

If R₁ and R₂ are increasing at rates of 0.3 Ω/s and 0.2 Ω/s, respectively, how fast is R changing when R₁ = 70 Ω and R₂ = 110 Ω? (Round your answer to three decimal places.)

  ________________ Ω/s

Answer 1

Hi Brigitte, I saw that you posted a number of questions that are all similar. I think you're going to find it kind of hard to learn through these text based posts so I can tutor you in calculus online to help you with the class. I will help you answer this question though:

With related rate questions, you want to take note of the original equation and also differentiate with respect to the common variable (time)to make the related rates equation:

1/R=1/R1+1/R2

=> (-1/R²)dR/dt = (-1/(R1)²)dR1/dt+(-1/(R2)²)dR2/dt

Our job is to calculate dR/dt. First we need to find the right value of R by using the first equation with the given values of R1 and R2:

1/R = 1/70+1/110 = 9/385 => R=385/9Ω. The problem told us the values of dR1/dt and dR2/dt so now we just plug in

-1/(385/9)²(dR/dt) = (-1/(70)²)*0.3+(-1/(110)²)0.2

solving this for dR/dt gives

dR/dt=461/3240 Ω/s