Question

Consider the differential equation \(y^{\prime}(t)=t^{2}-3 y^{2}\) and the solution curve that passes through the point \((3,1) .\) What is the slope of the curve at (3,1)\(?\)

Short answer

Answer: The slope of the solution curve at the point (3, 1) is 6.

Step-by-step solution

Identify the given information

The differential equation is given by \(y'(t) = t^2 - 3y^2\). We are also given the point \((3, 1)\) on the solution curve, where \(t=3\) and \(y=1\).

Substitute the values of \(t\) and \(y\) into the differential equation

We have the differential equation \(y'(t) = t^2 - 3y^2\). At the point \((3,1)\), we can substitute \(t = 3\) and \(y = 1\) to find the slope at this specific point: \(y'(3) = (3)^2 -3(1)^2\).

Evaluate and simplify to find the slope

Now, we will solve for the slope. Given the expression for \(y'(3)\), let's simplify: \(y'(3) = 3^2 - 3(1)^2 = 9 - 3 = 6\).

State the final answer

The slope of the solution curve at the point \((3, 1)\) for the given differential equation is \(6\).

Key concepts

Solution Curve

A solution curve in a differential equation is a graphical representation of all solutions that satisfy a given differential equation, typically expressed in terms of variables. In simple terms, it's the path that connects all the points \((t, y)\) that satisfy the differential equation. Think of it as a story told through graphing. The curves are drawn based on solving the equation (often involving derivatives), and they can show an infinite number of potential solutions, plotting the relation between the variables as influenced by the equation's terms.

Slope of the Curve

The slope of a curve at a particular point gives you a sense of its steepness, which tells how rapidly or slowly the values change. In the context of differential equations, the slope is computed as a derivative. Here, the derivative \(y'(t)\) represents the rate of change of \(y\) with respect to \(t\).For the exercise given, the slope of the solution curve at the point \((3,1)\) is found by substituting the specific values of \(t\) and \(y\) into the differential equation. After substituting these values, you perform the necessary calculations to get a numerical slope value. At the given point, the slope is \(6\), indicating the steepness of the curve at that exact spot.

Substitution Method

The substitution method is a technique often used in calculus and algebra to solve equations. It involves replacing variables with known values to simplify the equation, making it easier to solve.In this differential equation example, you substitute the provided point \((3,1)\) directly into the equation \(y'(t) = t^2 - 3y^2\). By doing this, you're evaluating the specific scenario at \(t = 3\) and \(y = 1\). This method can be a straightforward way to find unknown quantities by transforming the original equation into simpler terms.

Evaluate and Simplify Expressions

Evaluating and simplifying expressions are steps you take to make mathematical statements easier to understand and work with. When you evaluate, you calculate the value of a particular expression by substituting values into it.In the exercise, after substituting \(t = 3\) and \(y = 1\) into the differential equation \(y'(t) = t^2 - 3y^2\), you evaluate it by performing arithmetic: \(3^2 - 3(1)^2 = 9 - 3\). This simplifies the expression to \(6\), giving you the slope of the curve at the specific point. Simplification is crucial because it reduces complexity, making the result cleaner and more straightforward to interpret.