Problem 53 - College Algebra and Calculus: An Applied Approach

Question

Find the demand function \(x=f(p)\) that satisfies the initial conditions. $$ \frac{d x}{d p}=-\frac{6000 p}{\left(p^{2}-16\right)^{3 / 2}}, \quad x=5000 \text { when } p=\$ 5 $$

Step-by-Step Solution

Answer

The demand function \(x=f(p)\) that satisfies the initial condition is \(x(p) = -3000 \left(\frac{p}{\sqrt{{p^2-16}}}\right) + 11494\)

Step 1: Set up the Integral

To solve a differential equation like this, one will perform an integral on both sides. The integral of \(\frac{d x}{d p}\) is simply \(x\), so one essentially needs to perform this integral: \[ x = \int-\frac{6000 p}{\left(p^{2}-16\right)^{3 / 2}} dp \] Because the differential equation equals to \(\frac{d x}{d p}\), one can move \(dp\) to the other side and then perform the integral.

Step 2: Calculate the Integral

Now calculate the integral presented in Step 1. Using algebraic manipulation and the power rule for antiderivatives, it is found that:\[ x = -3000 \left(\frac{p}{\sqrt{{p^2-16}}}\right) + C \]Here, \(C\) refers to the constant of integration, which will be determined based on the initial conditions provided in the problem.

Step 3: Apply the Initial Conditions

The conditions says \(x=5000\) when \(p=5\). Plug these values into the equation from the previous step and solve for \(C\). Doing so yields:\[ 5000 = -3000 \left(\frac{5}{\sqrt{{5^2-16}}}\right) + C \]Solving for \(C\) will give us the final demand function.

Step 4: Find the Demand Function

Solve the equation obtained in Step 3 for \(C\). The result is \(C \approx 11494\). Now replace \(C\) in the integration result to obtain the final demand function:\[ x(p) = -3000 \left(\frac{p}{\sqrt{{p^2-16}}}\right) + 11494 \] This is the demand function \(x=f(p)\) that satisfies the initial conditions provided.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are a fundamental tool in expressing the relationships involving rates of change and are widely utilized across various scientific disciplines, including physics, engineering, biology, and economics. In our exercise, we have a first-order differential equation given by \(\frac{d x}{d p}=-\frac{6000 p}{\left(p^{2}-16\right)^{3 / 2}}\), which signifies how the quantity demanded (x) changes with respect to the price (p).

To solve the differential equation, we look for a function x=f(p) whose derivative with respect to p matches the provided rate of change. Solving such equations often requires the use of integral calculus, as it enables us to reverse the process of differentiation to find the original function from its derivative.

Integral Calculus

Integral calculus is inherently linked with differential equations as it involves finding antiderivatives or integrals. It plays a crucial role in determining the functions that describe accumulative sums, areas, and quantities. In the context of our exercise, we perform integration to solve for the demand function x=f(p) from its derivative \(\frac{d x}{d p}\).

By calculating the integral \(\int -\frac{6000 p}{\left(p^{2}-16\right)^{3/2}} dp\), we are essentially adding up all the little changes in quantity demanded over a range of prices to find the overall demand function. This process allows us to recover the function x from its rate of change with respect to price p. The integral calculus is essential for not just finding the general antiderivative, but also for incorporating initial conditions and constants of integration to find a specific solution.

Initial Conditions

Initial conditions serve as concrete starting points from where we solve differential equations. They provide specific values at certain points that must correspond with the solution. In our scenario, the initial condition given is x=5000 when p=$5.

The use of initial conditions is essential to determine the constant of integration, which otherwise would leave us with an infinite number of solutions. By applying the initial condition to the antiderivative found through integration, we can solve for this constant and thus find a unique demand function that fits the given scenario. It's like having a puzzle with one piece already in place, guiding you to complete the rest accurately.

Constant of Integration

The constant of integration, represented by C in our exercise, arises naturally when we calculate an antiderivative. Because the process of differentiation eliminates any constant term (the derivative of a constant is zero), when we integrate a function, we must add an arbitrary constant to account for this.

In the given problem, after integration, we obtained \(-3000 \left(\frac{p}{\sqrt{p^2-16}}\right) + C\). The constant C is what allows us to tailor our antiderivative to fit the initial condition. By substituting the initial conditions into the integrated function, we find the specific value of C, which in this case is approximately 11494, ensuring that the function x(p) uniquely satisfies both the differential equation and the initial conditions.