Problem 76 - Calculus and its applications

Question

Solve for f(t). $$ f^{\prime}(t)=t^{\sqrt{3}}, \quad f(0)=8 $$

Step-by-Step Solution

Answer

\( f(t) = \frac{t^{\sqrt{3} + 1}}{\sqrt{3} + 1} + 8 \).

Step 1: Introduce the Problem

We need to find the function \( f(t) \) given that \( f'(t) = t^{\sqrt{3}} \) and \( f(0) = 8 \). To do this, we'll integrate \( f'(t) \) to find \( f(t) \).

Step 2: Integrate f'(t)

To find \( f(t) \), integrate the derivative \( f'(t) = t^{\sqrt{3}} \). The formula for integrating a power of \( t \) is: \( \int t^a \, dt = \frac{t^{a+1}}{a+1} + C \). Set \( a = \sqrt{3} \) and integrate:

Step 3: Apply Integration Formula

Perform the integration: \[\int t^{\sqrt{3}} \, dt = \frac{t^{\sqrt{3} + 1}}{\sqrt{3} + 1} + C,\]where \( C \) is the constant of integration.

Step 4: Solve for the Constant C

To find \( C \), use the given initial condition \( f(0) = 8 \). Substitute \( t = 0 \) and \( f(0) = 8 \) into the expression for \( f(t) \):\[8 = \frac{0^{\sqrt{3} + 1}}{\sqrt{3} + 1} + C \]Simplifying gives us:\[C = 8.\]

Step 5: Combine Results for f(t)

Substitute \( C = 8 \) back into the expression for \( f(t) \):\[f(t) = \frac{t^{\sqrt{3} + 1}}{\sqrt{3} + 1} + 8.\]

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are essential in modelling situations where rates of change are important, such as in physics, engineering, and other sciences. In this exercise, we are given a first-order differential equation, which means it involves the first derivative of the function. The equation is \( f'(t) = t^{\sqrt{3}} \). This tells us that the rate of change of the function \( f(t) \) with respect to \( t \) is given by \( t^{\sqrt{3}} \). To solve a differential equation means to find the function \( f(t) \) that satisfies this equation.

• First-order differential equations involve only the first derivative.
• The solution to a differential equation is a function.
• Differential equations can describe a wide range of real-world phenomena.

Understanding how these equations work is foundational for solving many problems in calculus.

Initial Value Problem

An initial value problem is a type of differential equation that comes with an initial condition. This additional piece of information is crucial because it enables us to find a specific solution, rather than a generic one. In our exercise, the initial condition is given by \( f(0) = 8 \).
This means that when \( t = 0 \), the value of the function \( f(t) \) is 8. This initial condition helps us determine the constant of integration when we find the general solution of the differential equation. The process typically involves:

• Solving the differential equation to find the general solution, which includes an arbitrary constant \( C \).
• Substituting the initial conditions to find the exact value of \( C \).

Once \( C \) is determined, we have an explicit solution to the initial value problem. This solution gives us the specific behavior of \( f(t) \) that satisfies both the differential equation and the initial condition.

Integration Techniques

Integration techniques are methods used to find integrals of functions. Integral calculus focuses on determining the antiderivative or integral of a function. In this problem, there is the need to integrate \( f'(t) = t^{\sqrt{3}} \) to find the original function \( f(t) \). The step we use here is power rule integration, which is one of the fundamental integration techniques. The power rule for integration states that \( \int t^a \, dt = \frac{t^{a+1}}{a+1} + C \) for \( a eq -1 \). Applying this to our function involves setting \( a = \sqrt{3} \).
This yields the integral:

• \( \int t^{\sqrt{3}} \, dt = \frac{t^{\sqrt{3} + 1}}{\sqrt{3} + 1} + C \)

The constant \( C \) represents all possible vertical shifts of the function. Once integration is performed, substituting the initial condition \( f(0) = 8 \) allows us to find the specific value of \( C \). Thus, by using integration techniques, we solve the differential equation and gain insights into the specific problem setup.