Problem 4 - Differential Equations and Linear Algebra

Question

Determine the differential equation giving the slope of the tangent line at the point \((x, y)\) for the given family of curves. $$y=c / x$$

Step-by-Step Solution

Answer

The differential equation that gives the slope of the tangent line at any point \((x, y)\) on the curve \(y = \frac{c}{x}\) is: \[\frac{dy}{dx} = -\frac{y}{x}.\]

Step 1: Differentiate the function with respect to x

Differentiate the function $$y = \frac{c}{x}$$ with respect to x using the power rule: \[y = c \cdot x^{-1}.\] By differentiating both sides with respect to x, we obtain: \[\frac{dy}{dx} = -1 \cdot c \cdot x^{-2}.\]

Step 2: Solve for c in terms of x and y

To eliminate \(c\) from the equation, we need to rewrite the equation in terms of \(x\) and \(y\). From the original equation, we can solve for \(c\): \[c = xy.\]

Step 3: Substitute c in the derivative equation

Substitute the expression for \(c\) from step 2 into the equation from step 1: \[\frac{dy}{dx} = -1 \cdot (xy) \cdot x^{-2}.\]

Step 4: Simplify the expression

Simplify the expression we obtained in step 3: \[\frac{dy}{dx} = -1 \cdot (xy) \cdot x^{-2} = -\frac{xy}{x^2}.\] Now, cancel out \(x\) from the numerator and denominator: \[\frac{dy}{dx} = -\frac{y}{x}.\] So, the differential equation that gives the slope of the tangent line at any point \((x, y)\) on the curve is: \[\frac{dy}{dx} = -\frac{y}{x}.\]

Key Concepts

Understanding the Slope of the Tangent Line

In the realm of calculus, finding the slope of the tangent line to a curve at a given point is a fundamental concept. Imagine the curve as a hill and the tangent line as a straight path touching the hill only at one point without cutting into it. This line provides an approximation of the curve's steepness right at that point. The slope of this line is what we call the derivative of the function at that point.

When tackling problems involving the slope of the tangent line, we aim to calculate the derivative of the function that describes the curve. Once we obtain this derivative, it not only tells us the slope at a particular point but also the rate of change of the function's value at that point. Using the derivative, we can then write a differential equation that relates all possible slopes of tangent lines to their corresponding points on the curve.

Power Rule Differentiation Made Simple

Power rule differentiation is a quick and reliable tool for finding derivatives of functions where the variable, often denoted as x, is raised to a power. The rule states: to differentiate an expression of the form \(x^n\), you multiply the exponent \(n\) by the coefficient of \(x\), and then subtract one from the exponent. For instance, if you have \(x^3\), the derivative will be \(3x^2\).

This rule is immensely useful in solving differential equations like the one in our exercise. We utilized the power rule to differentiate \(y = c \/ x\), which is the same as \(y = cx^{-1}\), resulting in \(dy/dx = -1 \/ c \/ x^{-2}\). The power rule simplifies what could be a complex process, making calculus more accessible to students.

The Beauty of Separable Differential Equations

Separable differential equations are a class of ordinary differential equations that can be expressed and solved by separating the variables. This means we can move all the terms involving one variable to one side of the equation and all the terms involving the other variable to the opposite side.

Our textbook example illustrates this beautifully. After finding the derivative, we have a differential equation that can be rearranged to get \(dx/dy = -x/y\), which means all x terms can be placed on one side, and all y terms on the opposite side. This sets the stage for integrating each side separately to find the general solution of the equation. Students often find separable differential equations more intuitive because they can physically manipulate the variables, simplifying the path to the solution.