Problem 61 - Calculus with Concepts in Calculus

Question

When two resistors having resistances \(R_{1}\) and \(R_{2}\) are connected in parallel, the combined resistance \(R\) is given by \(R=R_{1} R_{2} /\left(R_{1}+R_{2}\right) .\) Show that $$ \frac{\partial^{2} R}{\partial R_{1}^{2}} \frac{\partial^{2} R}{\partial R_{2}^{2}}=\frac{4 R^{2}}{\left(R_{1}+R_{2}\right)^{4}} $$

Step-by-Step Solution

Answer

The result shows that \( \frac{\partial^{2} R}{\partial R_{1}^{2}} \frac{\partial^{2} R}{\partial R_{2}^{2}}=\frac{4 R^{2}}{(R_{1}+R_{2})^{4}} \) as required.

Step 1: Differentiate R with respect to R1

The given combined resistance is \( R = \frac{R_1 R_2}{R_1 + R_2} \). First, differentiate \( R \) with respect to \( R_1 \). Using the quotient rule \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \), set \( u = R_1 R_2 \) and \( v = R_1 + R_2 \). Thus, \( u = R_1R_2 \), so \( u' = R_2 \). Also, \( v = R_1 + R_2 \), so \( v' = 1 \). Therefore, \[ \frac{\partial R}{\partial R_1} = \frac{(R_2)(R_1 + R_2) - (R_1R_2)(1)}{(R_1 + R_2)^2} = \frac{R_2^2}{(R_1 + R_2)^2}. \]

Step 2: Differentiate R with respect to R2

Similarly, differentiate \( R \) with respect to \( R_2 \) again using the quotient rule. Set \( u = R_1R_2 \) and \( v = R_1 + R_2 \). Thus \( u' = R_1 \) and \( v' = 1 \). Therefore, \[ \frac{\partial R}{\partial R_2} = \frac{(R_1)(R_1 + R_2) - (R_1R_2)(1)}{(R_1 + R_2)^2} = \frac{R_1^2}{(R_1 + R_2)^2}. \]

Step 3: Find the second partial derivatives

Now, find the second partial derivative of \( R \) with respect to \( R_1 \). The first derivative was \( \frac{\partial R}{\partial R_1} = \frac{R_2^2}{(R_1 + R_2)^2} \). Differentiate again: \[ \frac{\partial^2 R}{\partial R_1^2} = \frac{d}{dR_1} \left( \frac{R_2^2}{(R_1 + R_2)^2} \right) = \frac{0 \cdot (R_1 + R_2)^2 - (-2R_2^2)(1)}{(R_1 + R_2)^4} = \frac{2R_2^2}{(R_1 + R_2)^3}. \]

Step 4: Find the second partial derivatives

Similarly, find the second partial derivative with respect to \( R_2 \). The first derivative was \( \frac{\partial R}{\partial R_2} = \frac{R_1^2}{(R_1 + R_2)^2} \). Differentiate again: \[ \frac{\partial^2 R}{\partial R_2^2} = \frac{d}{dR_2} \left( \frac{R_1^2}{(R_1 + R_2)^2} \right) = \frac{0 \cdot (R_1 + R_2)^2 - (-2R_1^2)(1)}{(R_1 + R_2)^4} = \frac{2R_1^2}{(R_1 + R_2)^3}. \]

Step 5: Verify the product of second derivatives

Compute the product: \[ \frac{\partial^2 R}{\partial R_1^2} \cdot \frac{\partial^2 R}{\partial R_2^2} = \frac{2R_2^2}{(R_1 + R_2)^3} \cdot \frac{2R_1^2}{(R_1 + R_2)^3} = \frac{4R_1^2 R_2^2}{(R_1 + R_2)^6}. \]

Step 6: Simplify and Compare

The left side of the identity states \( \frac{4R^2}{(R_1 + R_2)^4} \) with \( R = \frac{R_1 R_2}{R_1 + R_2} \). Substitute \( R \) into the identity: \[ \frac{4 \left(\frac{R_1 R_2}{R_1 + R_2}\right)^2}{(R_1 + R_2)^4} = \frac{4R_1^2 R_2^2}{(R_1 + R_2)^6}. \] This matches the result from Step 5, confirming the identity.

Key Concepts

Partial Derivatives

Partial derivatives are a crucial concept in multivariable calculus, where we deal with functions of several variables. When we take a partial derivative, we're focusing on how the function changes as we alter just one of those variables, keeping all others constant. In the exercise concerning resistor values, we are interested in examining how the combined resistance \( R \) changes with each individual resistor \( R_1 \) and \( R_2 \).

To find the partial derivatives with respect to \( R_1 \) and \( R_2 \), we apply standard differentiation techniques, holding other variables constant. This effectively reduces a multivariable function into a single-variable scenario, allowing us to see the effect of one specific variable on the function.

By computing these derivatives, we gain insights into relationships in the problem, like how changing one resistor affects the overall resistance, which can be very valuable in practical electrical engineering applications.

Quotient Rule

The quotient rule is a method used in calculus to differentiate expressions that are ratios of two functions. Given two functions, \( u(x) \) and \( v(x) \), the quotient rule formula is:

\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}.\]

This rule is particularly useful when our function of interest, like the combined resistance \( R = \frac{R_1 R_2}{R_1 + R_2} \), is presented as a fraction where both the numerator and denominator depend on the variables we're interested in.

In our problem, the quotient rule helps us derive the first partial derivatives of \( R \) with respect to \( R_1 \) and \( R_2 \). By identifying the parts of the function as \( u = R_1 R_2 \) and \( v = R_1 + R_2 \), and then applying the quotient rule, we can systematically find how \( R \) changes with each resistor independently, providing a clear path to the second derivatives as well.

Second Derivative Test

The second derivative test helps determine the concavity of a function, telling us whether a critical point is a local minimum or maximum. In our exercise, we find the second partial derivatives of the function to better understand how changes in \( R_1 \) and \( R_2 \) affect the shape of the resistance function \( R \).

When we compute second partial derivatives \( \frac{\partial^2 R}{\partial R_1^2} \) and \( \frac{\partial^2 R}{\partial R_2^2} \), we are essentially looking into the curvature along each coordinate direction. These second derivatives help us verify the relationship stated in the problem. By multiplying these individual second derivatives, we confirm the identity related to the curvature of the function. This verification assures us that the theory aligns with practical applications, especially when examining electric circuits and how resistance is modified by component values.