Problem 75 - Calculus and its applications

Question

Solve for f(t). $$ f^{\prime}(t)=\sqrt{t}+\frac{1}{\sqrt{t}}, \quad f(4)=0 $$

Step-by-Step Solution

Answer

The function is \( f(t) = \frac{2}{3} t^{3/2} + 2t^{1/2} - \frac{28}{3} \).

Step 1: Identify the Problem Type

This problem involves finding a function \( f(t) \) where the derivative \( f'(t) \) is given, along with an initial condition. This requires integrating \( f'(t) \) to find \( f(t) \).

Step 2: Set Up the Integral

To find \( f(t) \), integrate the function \( f'(t) = \sqrt{t} + \frac{1}{\sqrt{t}} \). This means setting up the integral: \[ f(t) = \int \left( \sqrt{t} + \frac{1}{\sqrt{t}} \right) \, dt.\]

Step 3: Integrate Each Term

Separate the integral into two parts and integrate each term individually:\[\int \sqrt{t} \, dt + \int \frac{1}{\sqrt{t}} \, dt.\]The integral of \( \sqrt{t} = t^{1/2} \) is given by the power rule: \[\int t^{1/2} \, dt = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2}.\]The integral of \( \frac{1}{\sqrt{t}} = t^{-1/2} \):\[\int t^{-1/2} \, dt = \frac{t^{1/2}}{1/2} = 2t^{1/2}.\]

Step 4: Combine the Results of Integration

Add the results of the integration together:\[ f(t) = \frac{2}{3} t^{3/2} + 2t^{1/2} + C,\]where \( C \) is the constant of integration.

Step 5: Solve for the Constant using Initial Condition

Use the initial condition \( f(4) = 0 \) to solve for \( C \). Substitute \( t = 4 \) into the function:\[0 = \frac{2}{3} (4)^{3/2} + 2(4)^{1/2} + C.\]Calculate:- \( 4^{3/2} = 8 \) so \( \frac{2}{3} \cdot 8 = \frac{16}{3} \).- \( 4^{1/2} = 2 \) so \( 2 \cdot 2 = 4 \).Thus,\[ 0 = \frac{16}{3} + 4 + C.\]Simplify and solve for \( C \):\[ C = 0 - \frac{16}{3} - 4 = -\frac{28}{3}.\]

Step 6: Write the Final Solution for f(t)

Substitute \( C \) back into the expression for \( f(t) \):\[ f(t) = \frac{2}{3} t^{3/2} + 2t^{1/2} - \frac{28}{3}.\]

Key Concepts

Integration Techniques

Integration techniques are the strategies used to solve integrals, which are foundational in integral calculus. In our exercise, the problem was to integrate the given function to find another function, specifically integrating \( f'(t) \) to find \( f(t) \). To do this, we split the function into simpler parts: \( \sqrt{t} \) and \( \frac{1}{\sqrt{t}} \).
One common technique used here is the **power rule for integration**, which is essential in handling polynomials and terms that can be expressed as powers of \( t \).

• For \( \sqrt{t} \), expressed as \( t^{1/2} \), the power rule tells us the integral is \( \int t^{n} \, dt = \frac{t^{n+1}}{n+1} + C \).
• Similarly, \( \frac{1}{\sqrt{t}} \) is rewritten as \( t^{-1/2} \), allowing us to apply the power rule again.

These steps simplify the process, ensuring each component of the function is integrated correctly.

Initial Value Problem

An initial value problem involves finding a function not only based on its differential equation but also considering a given specific value of the function at a point. This condition, called the "initial condition," anchors the function to a specific value, providing a unique solution.
In our exercise, the initial condition is \( f(4) = 0 \), indicating that when \( t = 4 \) the value of \( f(t) \) should be zero. To abide by the initial condition:

• First, integrate the given derivative to get the general family of solutions \( f(t) \) with an unknown constant \( C \).
• Then, use the initial condition to solve for \( C \). This involves substituting \( t = 4 \) into the integrated function and setting the equation equal to zero, as follow.

This process ensures that the final solution satisfies the given initial condition and accurately represents the behavior of the original differential equation at that point.

Antiderivatives

The antiderivative is the reverse operation of differentiation in calculus, giving us back the original function from its derivative. It's crucial in obtaining the function \( f(t) \) from \( f'(t) \). In simple terms, if the derivative represents the rate of change, the antiderivative represents the accumulated sum or total value over an interval.

1. The first step is recognizing the form of the integrand, here it is the sum of \( \sqrt{t} \) and \( \frac{1}{\sqrt{t}} \).
2. Then, apply integration techniques to each part, effectively reversing the differentiation.

Antiderivatives essentially provide a template for all possible solutions, differentiated by a constant term \( C \), representing that indefinite integral nature. This constant is later determined through initial values or conditions. Thus, finding the antiderivative allows us to understand the original behavior described by the derivative.