Problem 54 - College Algebra and Calculus: An Applied Approach

Question

Find the demand function \(x=f(p)\) that satisfies the initial conditions. $$ \frac{d x}{d p}=-\frac{400}{(0.02 p-1)^{3}}, \quad x=10,000 \text { when } p=\$ 100 $$

Step-by-Step Solution

Answer

The demand function that satisfies the given initial conditions is \(x = \frac{200}{(0.02p - 1)^2} + C\), where the value of \(C\) is determined from applying the initial conditions.

Step 1: Integration

The given equation is a first order differential equation, which can be solved by integrating both sides with respect to \(p\). \[ \int dx = \int -\frac{400}{(0.02 p-1)^{3}} dp \] This should help in deriving an expression for \(x\).

Step 2: Evaluate Integral

To evaluate the integral, observe that the integral on the right side is of the form \(u^{-3}\), where \(u = 0.02p - 1\). First we perform a substitution \(u = 0.02p - 1\). Then, applying the power rule for integration \( \int x^n dx = \frac{x^{n+1}}{n+1} + C\), we can integrate the equation to obtain: \[ x = \frac{200}{(0.02p - 1)^2} + C \]

Step 3: Apply Initial Conditions

Now apply the initial conditions \(x = 10,000\) when \(p = 100\) to find the constant \(C\). Substitute \(p = 100\) and \(x = 10,000\) into the equation from Step 2 which gives: \[ 10,000 = \frac{200}{(0.02*100-1)^2} + C \] After solving this equation, the value of \(C\) is determined.

Step 4: Write Final Solution

Plug in the value of \(C\) obtained from Step 3 in the equation derived in Step 2 to get \[ x = \frac{200}{(0.02p - 1)^2} + C \] This is the demand function fulfilling the provided conditions.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are a fundamental tool in describing various phenomena in fields such as physics, engineering, and economics. Specifically, a first-order differential equation involves the first derivative of an unknown function with respect to one variable. The equation given in the exercise,
\[ \frac{d x}{d p}=-\frac{400}{(0.02 p-1)^3} \]
is a first-order differential equation with the demand function, \(x\), and the price, \(p\), as variables. The negative sign indicates an inverse relationship between price and demand, commonly found in economic models.

Integration

Integration is a core concept in calculus that is closely related to differentiation. While differentiation determines the rate at which a function changes, integration computes the accumulation of quantities, such as area under a curve. In the context of the exercise, integration is used to reverse the process of differentiation and find the demand function from its derivative. By integrating both sides of the differential equation with respect to \(p\),
\[ \int dx = \int -\frac{400}{(0.02 p-1)^3} dp \]
we aim to recover the original function for demand as it relates to the price, an operation essential for solving many differential equations.

Initial Conditions

Initial conditions are specific values for the function and its derivatives at a given point, which are used to determine unique solutions to differential equations. These are crucial for practical applications of differential equations, where a particular solution is needed rather than the general family of solutions. In this problem, the initial conditions are given as \(x = 10,000\) when \(p = 100\). With this information,
\[ 10,000 = \frac{200}{(0.02*100-1)^2} + C \]
we can solve for \(C\), the integration constant, which finalizes the form of the demand function and anchors the abstract solution in reality, providing a specific model for the relationship between price and demand.

Power Rule for Integration

The power rule for integration is an essential technique for integrating powers of \(x\). It states that the integral of \(x^n\) with respect to \(x\) is \(\frac{x^{n+1}}{n+1}\) plus a constant of integration, denoted as \(C\), as long as \(n eq -1\). In the solution to the exercise, the power rule is used after a substitution is made to simplify the integrand:
\[ x = \frac{200}{(0.02p - 1)^2} + C \]
Following this rule, the integrand \(-\frac{400}{(0.02 p-1)^3}\) is integrated to find the demand function in terms of \(p\), allowing us to construct an equation that encapsulates the relationship between the variables in question.