Math Problem Statement

If two resistors with resistances R1 and R2 are connected in parallel, as in the figure, 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 increasing when R1 = 80 Ω and R2 = 100 Ω?

Solution

To solve this problem, we’ll differentiate the formula for the total resistance RR with respect to time tt and then substitute the values given for R1R_1, R2R_2, and their rates of change.

Problem Analysis

  1. The formula for total resistance RR in parallel is: 1R=1R1+1R2\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}

  2. Given:

    • dR1dt=0.3Ω/s\frac{dR_1}{dt} = 0.3 \, \Omega/s
    • dR2dt=0.2Ω/s\frac{dR_2}{dt} = 0.2 \, \Omega/s
    • R1=80ΩR_1 = 80 \, \Omega
    • R2=100ΩR_2 = 100 \, \Omega

We need to find dRdt\frac{dR}{dt}, the rate at which RR is changing.

Step-by-Step Solution

  1. Differentiate both sides with respect to tt: ddt(1R)=ddt(1R1+1R2)\frac{d}{dt} \left( \frac{1}{R} \right) = \frac{d}{dt} \left( \frac{1}{R_1} + \frac{1}{R_2} \right)

  2. Using the chain rule: 1R2dRdt=1R12dR1dt1R22dR2dt-\frac{1}{R^2} \frac{dR}{dt} = -\frac{1}{R_1^2} \frac{dR_1}{dt} - \frac{1}{R_2^2} \frac{dR_2}{dt}

  3. Solving for dRdt\frac{dR}{dt}: dRdt=R2(1R12dR1dt+1R22dR2dt)\frac{dR}{dt} = R^2 \left( \frac{1}{R_1^2} \frac{dR_1}{dt} + \frac{1}{R_2^2} \frac{dR_2}{dt} \right)

  4. Calculate RR when R1=80R_1 = 80 and R2=100R_2 = 100: 1R=180+1100=5+4400=9400\frac{1}{R} = \frac{1}{80} + \frac{1}{100} = \frac{5 + 4}{400} = \frac{9}{400} So, R=400944.44ΩR = \frac{400}{9} \approx 44.44 \, \Omega

  5. Substitute values into the derivative expression:

    • R=4009R = \frac{400}{9}
    • R1=80R_1 = 80, R2=100R_2 = 100
    • dR1dt=0.3\frac{dR_1}{dt} = 0.3, dR2dt=0.2\frac{dR_2}{dt} = 0.2

    dRdt=(4009)2(18020.3+110020.2)\frac{dR}{dt} = \left( \frac{400}{9} \right)^2 \left( \frac{1}{80^2} \cdot 0.3 + \frac{1}{100^2} \cdot 0.2 \right)

  6. Calculate each term:

    • (4009)2=16000081\left( \frac{400}{9} \right)^2 = \frac{160000}{81}
    • 1802=16400\frac{1}{80^2} = \frac{1}{6400}
    • 11002=110000\frac{1}{100^2} = \frac{1}{10000}

    So, dRdt=16000081(0.36400+0.210000)\frac{dR}{dt} = \frac{160000}{81} \left( \frac{0.3}{6400} + \frac{0.2}{10000} \right)

  7. Evaluate inside the parentheses: 0.36400=4.6875×105\frac{0.3}{6400} = 4.6875 \times 10^{-5} 0.210000=2×105\frac{0.2}{10000} = 2 \times 10^{-5} 0.36400+0.210000=6.6875×105\frac{0.3}{6400} + \frac{0.2}{10000} = 6.6875 \times 10^{-5}

  8. Final calculation: dRdt=16000081×6.6875×105\frac{dR}{dt} = \frac{160000}{81} \times 6.6875 \times 10^{-5} dRdt0.132Ω/s\frac{dR}{dt} \approx 0.132 \, \Omega/s

Answer:

The rate at which RR is increasing is approximately 0.132Ω/s0.132 \, \Omega/s.

Would you like a more detailed breakdown of any step, or have any questions?

Related Questions:

  1. How would the rate of change of RR vary if R1R_1 and R2R_2 were decreasing?
  2. What happens to the total resistance RR if R1R_1 and R2R_2 approach infinity?
  3. How would this calculation differ if there were three resistors in parallel?
  4. Can this method be applied to resistors in series? Why or why not?
  5. What would the result be if R1R_1 and R2R_2 were equal?

Tip:

In problems with implicit differentiation, be careful with signs when differentiating expressions with 1R\frac{1}{R} or similar inverse terms.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Implicit Differentiation
Parallel Resistor Formula

Formulas

Parallel resistance formula: 1/R = 1/R1 + 1/R2
Derivative of 1/R: d(1/R)/dt = -1/R^2 * dR/dt
Rate of change formula for resistors in parallel

Theorems

Chain Rule for Differentiation

Suitable Grade Level

Undergraduate Engineering or Physics