Math Problem Statement

total differential s=p/q with step by step solution

Solution

To find the total differential of the function s=pqs = \frac{p}{q}, let's go through the process step by step:

1. Identify the variables:

  • We have a function s=pqs = \frac{p}{q}, where pp and qq are both functions of some independent variables, say xx and yy. We want to compute the total differential dsds.

2. Formula for the total differential:

The total differential dsds is given by: ds=spdp+sqdqds = \frac{\partial s}{\partial p} \, dp + \frac{\partial s}{\partial q} \, dq Here, we need to find the partial derivatives of ss with respect to both pp and qq.

3. Compute the partial derivatives:

  • The partial derivative of ss with respect to pp: sp=1q\frac{\partial s}{\partial p} = \frac{1}{q}

  • The partial derivative of ss with respect to qq: sq=pq2\frac{\partial s}{\partial q} = -\frac{p}{q^2}

4. Substitute the partial derivatives into the total differential formula:

Using the results from the partial derivatives, we now substitute them back into the total differential formula: ds=1qdp+(pq2)dqds = \frac{1}{q} \, dp + \left(-\frac{p}{q^2}\right) \, dq

5. Final expression for the total differential:

Therefore, the total differential of s=pqs = \frac{p}{q} is: ds=1qdppq2dqds = \frac{1}{q} \, dp - \frac{p}{q^2} \, dq

This expression represents the total differential of ss in terms of the differentials dpdp and dqdq.

Would you like to discuss this solution in more detail or ask any questions? Here are some related questions to explore further:

  1. How does the total differential relate to the chain rule in calculus?
  2. What happens to the total differential if pp and qq are independent of each other?
  3. How can we use the total differential to approximate changes in ss?
  4. Can you explain the geometric interpretation of the total differential?
  5. How do partial derivatives factor into the computation of total differentials for multivariable functions?

Tip: The total differential provides a linear approximation of a function's change based on the changes in its variables, which is especially useful in applied mathematics.

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Math Problem Analysis

Mathematical Concepts

Calculus
Total Differential
Partial Derivatives

Formulas

ds = (∂s/∂p) dp + (∂s/∂q) dq
∂s/∂p = 1/q
∂s/∂q = -p/q^2

Theorems

Total Differential Formula

Suitable Grade Level

Grades 11-12 or early college