Question

Finding a Particular Solution In Exercises \(15-18\) , 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 y = \( \frac{1}{4} t^2 + 10 \) is the solution that satisfies the given conditions.

Step-by-step solution

Finding Antiderivative

The first step is to find the antiderivative. To do this, integrate the derivative, which in this case is \( \frac{1}{2} t \). Integrating gives \( \int \frac{1}{2} t dt = \frac{1}{4} t^2 + C \). Now the equation is \( y = \frac{1}{4} t^2 + C \), where C is the constant of integration that we will solve for.

Determine the Constant of Integration

Use the given point (0,10) to solve for 'C'. Since y=10 when t=0, substitute these values into the equation to find: \( 10 = \frac{1}{4} * 0^2 + C \) which simplifies to \( C = 10 \). So, the equation of our solution function is \( y = \frac{1}{4} t^2 + 10 \).

Graphing the Function

The final step is to graph the function. Plotting the graph of the function \( y = \frac{1}{4} t^2 + 10 \) on a graphing utility should give a parabola that opens upwards.

Key concepts

Antiderivative

When it comes to understanding calculus, the antiderivative is a fundamental concept. In simple terms, finding an antiderivative means discovering a function whose derivative is the given function. Think of it as reverse engineering the process of differentiation.

For instance, in our exercise, we're given the derivative \( \frac{d y}{d t}=\frac{1}{2} t \) and asked to find the original function, or antiderivative. To do this, we perform integration, which, in essence, is the inverse operation of taking a derivative. The integration of \( \frac{1}{2} t \) with respect to \( t \) gives us \( \frac{1}{4} t^2 \) because in general, when you integrate \( t^n \) your result is \( \frac{t^{n+1}}{n+1} \) (where \( neq -1 \) ).

This is only a portion of our final solution since we're not accounting for all the possible functions that differentiate back to \( \frac{1}{2} t \)—yet that's where the constant of integration comes into play! It's essential to remember the significance of antiderivatives in calculus as they're a critical step in solving many types of problems, including finding the area under a curve and solving differential equations.

Constant of Integration

Every time you integrate, you must remember to include the constant of integration, symbolized as \( C \). This represents the infinite number of antiderivatives that exist for any given function since adding any constant to a function doesn't change its derivative.

Let's relate this concept to our exercise. After integrating \( \frac{1}{2} t \) to get \( \frac{1}{4} t^2 \) as mentioned earlier, we append \( C \) to account for all possible vertical translations of the graph. This rectifies the potential infinity of parallel curves we could draw, each representing an antiderivative of \( \frac{1}{2} t \).

In the next step, we use the given information - the function passes through the point \( (0,10) \) - to find the specific value of \( C \) that will give us our particular solution to the problem. By plugging in the coordinates, we solve for \( C \) and determine that \( C = 10 \). So, while there are many potential antiderivative functions of \( \frac{1}{2} t \), the presence of \( C \) allows us to pinpoint the exact one that corresponds to our given conditions, illustrating the invaluable role the constant of integration plays in calculus.

Graphing Functions

Graphing functions is an effective way to visualize mathematical concepts and often helps in understanding the behavior of different functions, particularly in calculus. A graph can show properties such as rate of change, extrema, and the areas under and between curves. It translates algebraic expressions into a visual form, aiding in comprehension and allowing us to identify key characteristics at a glance.

In the context of our exercise, after determining the particular solution \( y = \frac{1}{4} t^2 + 10 \), we move to graph this quadratic function. Using a graphing utility, this specific equation will produce a parabola that opens upwards. The vertex of this parabola, due to the positive coefficient of \( t^2 \) and the value of \( C \) (which we solved to be 10), will be at the point \( (0,10) \).

The visualization confirms whether our solution curve indeed passes through the given point, validating our algebraic work. It's a form of reinforcement learning - by both solving the problem algebraically and seeing it graphically, students can gain a deeper understanding and retain the knowledge more effectively. Hence, graphing functions is not just a tool for solution validation, but also an impactful educational device.