Problem 51 - College Algebra and Calculus: An Applied Approach

Question

Find the supply function \(x=f(p)\) that satisfies the initial conditions. $$ \frac{d x}{d p}=p \sqrt{p^{2}-25}, \quad x=600 \text { when } p=\$ 13 $$

Step-by-Step Solution

Answer

The solution will be a function \(x=p\), it involves integration technique, substitution method and solving for constant using initial condition

Step 1: Determine the integral.

The integral equation of \(\frac{d x}{d p}=p\sqrt{p^{2}-25}\) should be determining first, which is \[ \int p\sqrt{p^2-25} dp \]

Step 2: Integrate using substitution method

Let's denote \(u=p^2-25\), the differential \(du = 2p \, dp\), therefore, the integration equation will be \[ \frac{1}{2}\int u^{1/2} du \]. Which has a solution of \[ \frac{2}{3}u^{3/2}\]

Step 3: Back substitute

Substitute \(u\) back into the integral, we have \[ \frac{2}{3}(p^2-25)^{3/2} + C \]

Step 4: Solve for the constant using the initial condition

When \(p=13\) and \(x=600\), we can solve for the constant \(C\). Replace p and x with 13 and 600 in the equation, we get: \[ 600 = \frac{2}{3}(13^2-25)^{3/2} + C \] Solve for \(C\)

Step 5: Final supply function

Replace \(C\) in supply function equation, thus obtaining the final supply function

Key Concepts

Integration Techniques

The integration of functions is one of the central operations in calculus. It allows us to find quantities like areas, volumes, and the accumulation of quantities over time. Understanding integration techniques is crucial for solving problems where antiderivatives are needed, such as in the given exercise.

In our example, the integration technique used is substitution, also known as 'u-substitution'. This method is highly effective for integrating compounds of functions or functions that are products of a function and its derivative. Substitution simplifies the integral into a more familiar form by introducing a new variable, making the integration process more manageable.

Substitution Method:

When looking at the function to be integrated, you observe if it can be broken down into another function and its derivative. In the step by step solution, after identifying such a pattern, the term \(u=p^2-25\) was introduced, with its differential \(du = 2p \, dp\). This substitution transforms the complex original function into a simpler form that is more straightforward to integrate, which results in an integral in terms of \(u\) that is easier to solve.

Differential Calculus

Differential calculus is primarily concerned with the concept of change. It provides us tools to analyze and calculate rates at which quantities change. The central tool in differential calculus is the derivative. The derivative of a function at a point is the rate at which the function's value changes with respect to changes in its input. Derivatives can be represented symbolically as \(\frac{dx}{dp}\), which in our exercise denotes the rate at which the supply function x changes with the price p.

In the given problem, we are presented with a differential equation \(\frac{dx}{dp}=p\sqrt{p^2-25}\). Differential equations involve derivatives and are powerful in describing various phenomena such as movement, growth, decay, and in this case, the behavior of a supply function in economics. Our goal in such cases is to find the original function before it was differentiated, also known as finding the antiderivative or solving the differential equation.

Initial Conditions in Calculus

Initial conditions are specific values assigned at the starting point of the problem that are used to determine the constants in the general solutions of differential equations. In calculus, these conditions play a vital role in specifying a unique solution to a differential equation which otherwise would have infinitely many solutions.

In the context of the exercise, we have the initial condition \(x=600\) when \(p=\$13\). Using this piece of information, we apply the condition to the general solution obtained from the integration process to compute the constant \(C\). It is this step that fine-tunes our general solution and makes it specific to the problem at hand. Without the initial condition, we would not be able to finalize the supply function, and consequently, could not utilize it for any practical application or analysis within the realm of the supply chain.