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Find an equation defining a function \(f\) given that (a) the slope of the tangent line to the graph of \(f\) at any point \(P(x, y)\) on the graph is given by $$ \frac{d y}{d x}=\frac{3 x^{2}}{2 y} $$ and (b) the graph of \(f\) passes through the point \((1,3)\).

Short Answer

Expert verified
The equation defining the function is \(f(x) = \sqrt{x^3 + 8}\).

Step by step solution

01

Integrate the given derivative

We are given the derivative as \(\frac{dy}{dx} = \frac{3x^2}{2y}\). To find the equation defining the function, we need to integrate this expression. First, let's write it in differential form: \(dy = \frac{3x^2}{2y} dx\) Now, let's separate the variables: \(2y dy = 3x^2 dx\) Now we integrate both sides: \(\int 2y dy = \int 3x^2 dx\)
02

Perform the integration

Next, we integrate both sides with respect to their corresponding variables: \(y^2 = x^3 + C_1\) To make our equation in the function form \(y = f(x)\), let's take the square root of both sides: \(y = \sqrt{x^3 + C_1}\), where \(C_1\) is the constant of integration.
03

Find the constant of integration using the given point

We are given that the function passes through the point \((1,3)\), which means we can substitute the values \(x=1\) and \(y=3\) into the equation to find the constant of integration. Plugging the coordinates into the function equation, we get: \(3 = \sqrt{1^3 + C_1}\) Squaring both sides: \(9 = 1 + C_1\) Now, we can find the value of \(C_1\): \(C_1 = 8\)
04

Write the final equation for the function

Substituting the value of \(C_1\) back into the equation, we get the equation for the function: \(y = \sqrt{x^3 + 8}\) So, the equation defining the function is: \(f(x) = \sqrt{x^3 + 8}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrating Differential Equations
Integrating differential equations is a foundational technique in calculus used to find a function when its derivative is known. It involves reversing the process of differentiation, which can be interpreted as finding the area under the curve of a given function's rate of change. In the context of our exercise, we are given the slope of the tangent line at any point on the function's graph, which is the derivative, \(\frac{dy}{dx} = \frac{3x^2}{2y}\).

To integrate this differential equation, we first rewrite it in a way that separates the variables (more on this in the next section). After separation, we're left with integrals on both sides of the equation, one with respect to \(x\) and the other with respect to \(y\).

Upon integration, we obtain an expression involving the function \(y\) and the integration constant \(C_1\). Integration constants are crucial as they represent the 'family' of possible functions that satisfy the original differential equation. The next steps involve finding the specific constant for our function by using the given initial condition.
Tangent Line Slope
The slope of the tangent line to a curve at any given point represents the instantaneous rate of change of the function at that point. It is effectively the derivative of the function at that specific point on its graph. Differential equations often involve these slopes, as seen in the exercise, where the differential equation relates the slope of the tangent line at any point \((x, y)\) on the graph to the variables themselves.

Mathematically, if \(f(x)\) is a function, then the slope of the tangent line at \(x\) is given by the derivative \(f'(x)\). In this problem, the slope given is \(\frac{3x^2}{2y}\), so finding \(f(x)\) requires solving this specific differential equation where the slope, inherently connected to the concept of the derivative, varies with both \(x\) and \(y\).
Separation of Variables
The separation of variables is a technique for solving differential equations where we manipulate the equation to isolate one variable and its differential on one side, and the other variable and its differential on the other side. Here's how this process looked in our exercise:
\[2y dy = 3x^2 dx\]

By separating \(x\) terms on one side and \(y\) terms on the other, we set up the equation to integrate both sides independently. This creates two separate integrals that can be solved using standard calculus techniques. The goal of separation of variables is to simplify a complex differential equation into a form where we can easily integrate to find the relationship between \(x\) and \(y\).
Integration Constants
When we integrate a function, we often add an arbitrary constant, known as the integration constant. This is because indefinite integration is the reverse of differentiation, and when differentiating a constant, the result is zero—hence, the constant 'disappears' during differentiation. For our exercise, after integrating the separated variables, we introduced an integration constant \(C_1\).

The integration constant plays a vital part in the solution of differential equations. It represents an infinite number of possible functions that satisfy the differential equation. However, we can find the exact value of this constant using initial or boundary conditions. In our exercise, we used the point \((1,3)\) to solve for \(C_1\), thus pinpointing the specific function that not only satisfies the differential equation but also passes through the given point on the graph.

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