Question

Find the function \(y=f(t)\) passing through the point (0,10) with the given first derivative. Use a graphing utility to graph the solution. $$ \frac{d y}{d t}=\frac{1}{2} t $$

Short answer

The function passing through the point (0,10) is \( y = \frac{t^2}{4} + 10 \)

Step-by-step solution

- Integrate the derivative

The integral of the derivative \( \frac{1}{2} t \) yields the general solution, which is the function \( y = \frac{t^2}{4} + C \) , where \( C \) is the constant of integration.

- Use known point to find the constant

To find the value of \( C \), substitute the point (0,10) into the equation: \( 10 = \frac{(0)^2}{4} + C \), which leads to \( C = 10 \). Therefore, the function \( y \) that passes through the point (0,10) is \( y=f(t)=\frac{t^2}{4} + 10 \).

- Graph the function

Graphing the function helps visualizing the solution. Create a graph where \( t \) is on the x-axis and \( y \) on the y-axis. Plot the function \( y=f(t)=\frac{t^2}{4} + 10 \).

Key concepts

Antiderivative Calculus

Antiderivatives are at the heart of one of the most fundamental operations in calculus: integration. But what exactly is an antiderivative? It's a function that reverses the process of differentiation. When you have a derivative, finding an antiderivative means discovering a function whose rate of change (derivative) matches the given derivative.

For example, in our exercise, we were given the first derivative of a function, \( \frac{d y}{d t} = \frac{1}{2} t \). The task was to find the function \( y = f(t) \) whose derivative is \( \frac{1}{2} t \). Through integration, we found that antiderivative to be \( \frac{t^2}{4} \), as every time we differentiate \( \frac{t^2}{4} \), we'll end up with \( \frac{1}{2} t \).

In essence, antiderivative calculus involves using integration to navigate from the rate of change back to the original function, which describes the cumulative change. This concept is pivotal in solving many real-world problems where we have information about the change and need to determine the quantity that is changing.

Integration Constant

When integrating a derivative to find an antiderivative, we encounter a unique component known as the integration constant, denoted by \( C \). This constant arises because integration is the reverse process of differentiation, and when we differentiate a constant, it vanishes. Hence, when we integrate, we must consider all possible constants that could have been lost in the differentiation process.

This integration constant is pivotal because it allows for the existence of an infinite number of antiderivatives for a given derivative. Each value of \( C \) gives us a different antiderivative. In our exercise, we integrated \( \frac{d y}{d t} = \frac{1}{2} t \) to find the general antiderivative \( y = \frac{t^2}{4} + C \). To determine the specific \( C \) for our problem, we used the known condition that the function passes through the point (0,10), leading to \( C = 10 \). This shows the importance of the integration constant in finding precise solutions that fulfill additional conditions or constraints.

Graphing Functions

Graphing functions is a visual tool in mathematics that helps us understand the behavior and properties of functions. By representing functions on a coordinate system, we can explore their shape, identify where they increase or decrease, and even determine potential points of inflection.

In our exercise, after determining the antiderivative and the integration constant, graphing the specific function \( y = f(t) = \frac{t^2}{4} + 10 \) added a visual dimension to our analysis. It allowed us to see how the function behaves for different values of \( t \). To do this, we placed \( t \) on the x-axis and \( y \) on the y-axis of the graph. By plotting points and connecting them smoothly, the parabolic nature of the graph becomes evident, showcasing how it opens upwards and passes through the point (0,10).

This graphical representation is not just an abstract exercise; it serves a practical purpose by allowing us to predict outcomes, understand relationships between variables, and better communicate mathematical concepts, particularly in educational and collaborative settings.