91Ó°ÊÓ

In the context of differential equations, the slope of the graph refers to the value of the derivative at a specific point. To calculate this, we use the given differential equation: \(y' = t^2 - y^2\).

When asked to find the slope at a specific point, such as \(t=2\) in our exercise, it means determining the value of \(y'\) at that point. Here’s how it’s done step-by-step:

• Identify the specific values for \(t\) and \(y\). In our problem, we're given \(t = 2\) and \(y = 3\).
• Substitute these values into the differential equation.
• Solve for \(y'\) to find the slope.

For example, substituting \(t = 2\) and \(y = 3\):

\(y' = 2^2 - 3^2\)

Calculate the arithmetic next:

\(y' = 4 - 9 = -5\)

Therefore, the slope of the graph at \(t = 2\) is \(-5\).

The substitution method is a technique used in solving differential equations, particularly when trying to find specific values like a slope at a given point. This method involves replacing variables in the equation with given values to simplify calculations.

In this exercise, to determine \(y'\) at the point where \(t = 2\) and \(y = 3\), we perform the substitution directly:

• First, locate the equations and initial points. We know our differential equation is \(y' = t^2 - y^2\) and the point of interest is \( (2, 3) \).
• Substitute \( t = 2 \) into the equation. Also, substitute \( y = 3 \).
• Simplify to find \(y'\).

Thus, the substitution in action goes as follows:

\(y' = 2^2 - 3^2\)

This calculates to:

\(y' = 4 - 9 = -5\)

This data yields that \(y'\), or the slope at the given point, is \(-5\). By following these steps, one can handle similar problems efficiently and correctly.

Derivatives represent the rate at which a function changes. For instance, if \(y\) is a function of \(t\), then \(y'\) (the derivative of \(y\)) would denote the rate of change of \(y\) with respect to \(t\).

In this exercise, our differential equation \(y' = t^2 - y^2\) connects the derivative of \(y\) (i.e., the slope) with the values of \(t\) and \(y\) themselves. Here’s how we used derivatives:

• The equation \(y' = t^2 - y^2\) directly gives the derivative of \(y\) in terms of \(t\) and \(y\).
• Given a specific point, substituting the values of \(t\) and \(y\) into the equation allows us to calculate the derivative—\the slope at a particular point.

The derivative \(y'\) tells us how steep the graph is at any given point. In our example, substituting \(t = 2\) and \(y = 3\) in the equation gives us:

\(y' = 2^2 - 3^2 = 4 - 9 = -5\)

This means that at \(t = 2\), the graph’s slope is \(-5\), indicating that the graph is steeply decreasing. Understanding and utilizing derivatives in this way is fundamental to mastering differential equations.