Electricity The combined electrical resistance R of two...

Electricity The combined electrical resistance $R$ of two resistors $R_{1}$ and $R_{2},$ connected in parallel, is given by $\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$ where $R, R_{1},$ and $R_{2}$ are measured in ohms. $R_{1}$ and $R_{2}$ are increasing at rates of 1 and 1.5 ohms per second, respectively. At what rate is $R$ changing when $R_{1}=50$ ohms and $R_{2}=75$ ohms?

Electricity The combined electrical resistance $R$ of two resistors $R_{1}$ and $R_{2},$ connected in parallel, is given by

$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$

where $R, R_{1},$ and $R_{2}$ are measured in ohms. $R_{1}$ and $R_{2}$ are increasing at rates of 1 and 1.5 ohms per second, respectively. At what rate is $R$ changing when $R_{1}=50$ ohms and $R_{2}=75$ ohms?

Key Concepts

Parallel Resistor Combination

This concept involves calculating the total resistance of resistors connected in parallel. When resistors are arranged in this manner, the reciprocal of the total resistance is equal to the sum of the reciprocals of the individual resistances. This method is crucial in electrical circuit analysis for determining how parallel pathways affect current flow and resistance.

Implicit Differentiation

Implicit differentiation is a technique used in calculus to differentiate equations where the dependent variable is not isolated on one side. In problems involving electrical resistance, it enables the determination of how changes in one parameter affect another when the relationship between them is defined implicitly by an equation.

Related Rates

Related rates problems in calculus deal with finding the rate at which one quantity changes with respect to another. By differentiating both sides of an equation that relates multiple variables with respect to time, one can determine the instantaneous rate of change of one variable, given the rates of change of others. This is particularly useful in dynamics where multiple quantities vary simultaneously.

Transcript

00:01 We are given the equation 1 over r equal 1 over r1 plus 1 over r2, and we are told that r1 is increased at the rate of 1 um per second, and r2 at the rate of 1 .5 1 ums per second.

00:16 So that will mean that d r1 d t is equal to 1 per second, and d r2 d t is equal to 1 .5 ums.

00:33 Per second.

00:36 We want to find a greater change of r, so we want d r d t, when r1 is equal to 50 oms, and r2 is equal to 75 oms.

00:54 The equation relates the variables are r1 and r2, they are all functions of the time t, so we can look at our given equation, and take the derivative with respect to t.

01:11 So d d t of 1 over r.

01:15 Here we take the derivative of a sum, so it's the sum of the derivatives.

01:20 D d t of 1 over r1 plus d d d t of 1 over r 2.

01:30 All of these have the form 1 over x, so let's recall how the derivative of that will be computed.

01:37 We rewrite 1 over x as x to negative 1 and take the derivative using the power rule we get minus 1 x minus 1 minus 1 and that equals minus x to negative 2 which can be rewritten as minus 1 over x squared using that here we will have minus 1 over r squared times dr d d d d we are using the chain rule here because r is a function of d and similarly on this side we have minus 1 over r1 squared times the r1 d plus minus 1 over r2 squared times d r2 d t so on the right side we all these quantities, r1 is 50, dr1 dt, we know it equals 1.

02:46 We know the r2 dt equals 1 .5...

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