Question

Consider the function \(f(t)=2 t^{4}\). Evaluate the following integrals: (a) \(\int_{0}^{10} f(t) \hat{\delta}(t-3) d t\) (b) \(\int_{0}^{10} f(t) \dot{\bar{\delta}}(t-3) d t\) (c) \(\int_{0}^{10} f(t) \ddot{\bar{\delta}}(t-3) d t\) (d) \(\int_{0}^{10} \ddot{f}(t) \hat{\delta}(t-3) d t\)

Short answer

Question: Evaluate the following integrals containing the Dirac Delta function and its derivatives: a) Evaluate \(\int_{0}^{10} f(t) \hat{\delta}(t-3) dt\) b) Evaluate \(\int_{0}^{10} f(t) \dot{\hat{\delta}}(t-3) dt\) c) Evaluate \(\int_{0}^{10} f(t) \ddot{\hat{\delta}}(t-3) dt\) d) Evaluate \(\int_{0}^{10} \ddot{f}(t) \hat{\delta}(t-3) dt\) Given function: \(f(t) = 2t^4\) Answer: a) The integral evaluates to 162. b) The integral evaluates to -216. c) The integral evaluates to 216. d) The integral evaluates to 216.

Step-by-step solution

a) Evaluate \(\int_{0}^{10} f(t) \hat{\delta}(t-3) dt\)

To evaluate the integral, we can use the property of Dirac Delta function, which is \(\int_{-\infty}^{\infty} f(t) \hat{\delta}(t-a) dt = f(a)\). We have the integral \(\int_{0}^{10} 2t^4 \hat{\delta}(t-3) dt\), and using the property mentioned, the result is just \(f(3)\). So, we can calculate \(f(3) = 2(3^4) = 2(81) = 162\). The integral evaluates to 162.

b) Evaluate \(\int_{0}^{10} f(t) \dot{\hat{\delta}}(t-3) dt\)

The integral involves the derivative of the Dirac Delta function, which has the property: \(\int_{-\infty}^{\infty} f(t) \dot{\hat{\delta}}(t-a) dt = -f'(a)\). First, we need to find the derivative of \(f(t)\). \(f'(t) = \frac{d}{dt}(2t^4) = 8t^3\). So, the integral evaluates to \(-\int_{0}^{10} 8t^3 \dot{\hat{\delta}}(t-3) dt = -f'(3)\), which is \(-8(3^3) = -216\). The integral evaluates to -216.

c) Evaluate \(\int_{0}^{10} f(t) \ddot{\hat{\delta}}(t-3) dt\)

The integral involves the second derivative of the Dirac Delta function, which has the property: \(\int_{-\infty}^{\infty} f(t) \ddot{\hat{\delta}}(t-a) dt = f''(a)\). To calculate this integral, first, we need to find the second derivative of \(f(t)\). \(f''(t) = \frac{d^2}{dt^2}(2t^4) = 24t^2\). So, the integral evaluates to \(\int_{0}^{10} 24t^2 \ddot{\hat{\delta}}(t-3) dt = f''(3)\), which is \(24(3^2) = 216\). The integral evaluates to 216.

d) Evaluate \(\int_{0}^{10} \ddot{f}(t) \hat{\delta}(t-3) dt\)

In this integral, we have the second derivative of \(f(t)\) and the Dirac Delta function. We've already calculated the second derivative as: \(\ddot{f}(t) = 24t^2\). Now, we can apply the main property of the Dirac Delta function again, and the integral evaluates to \(\ddot{f}(3)\), which is \(24(3^2) = 216\). The integral evaluates to 216.

Key concepts

Integral Calculus and the Dirac Delta Function

Integral calculus is a mathematical tool that helps in finding the total accumulation of quantities. One fascinating application of integral calculus is working with the Dirac Delta function. This function is quite unique as it is not a function in the traditional sense, but more of a tool used in calculus to 'pick out' values of another function at a specific point.

When dealing with the Dirac Delta function, \(\delta(t-a)\), and a function \(f(t)\), the integral \(\int_{-\infty}^{\infty} f(t) \delta(t-a) dt = f(a)\). This integral essentially evaluates \(f(t)\) at \(t=a\). This is particularly helpful when solving problems in various areas of physics and engineering.

Understanding this key property can make complex calculations much simpler. It transforms challenging integrals into straightforward function evaluations. For someone analyzing step (a) in our exercise, the integral \(\int_{0}^{10} 2t^4 \hat{\delta}(t-3) dt\) used this principle to directly solve the problem by determining \(f(3) = 162\).

• The Dirac Delta function simplifies integrals by reducing them to function evaluations at a single point.
• Integral calculus becomes more accessible with such properties, especially for students and professionals working in fields that require signal processing or systems analysis.

The Role of Derivatives in Dirac Delta Integrals

Derivatives are a fundamental aspect of calculus, representing how functions change. In the context of the Dirac Delta function, derivatives play a crucial role and can often transform an integral into a quick evaluation.

Suppose we have the derivative of the Dirac Delta function, denoted \(\dot{\delta}(t-a)\). The integration involving this derivative changes the problem: \(\int_{-\infty}^{\infty} f(t) \dot{\delta}(t-a) dt = -f'(a)\). This expression tells us that instead of integrating \(f(t)\), we integrate its rate of change, then evaluate at \(a\). If we look at step (b) in our exercise, the integral \(\int_{0}^{10} 2t^4 \dot{\hat{\delta}}(t-3) dt\) simplifies to \(-f'(3)\), resulting in \(-216\).

The second derivative of the Dirac Delta function, denoted \(\ddot{\delta}(t-a)\), follows a similar pattern. The property becomes: \(\int_{-\infty}^{\infty} f(t) \ddot{\delta}(t-a) dt = f''(a)\), requiring a second derivative of \(f(t)\) before evaluation at \(a\). In our exercise's step (c), this means \(f''(3) = 216\).

• Derivatives used with Dirac Delta functions simplify integrals by leveraging change rates of functions.
• This concept streamlines analysis and computation, an essential practice in real-world engineering mathematics.

The Application of Engineering Mathematics

Engineering mathematics integrates various areas of mathematics to solve practical engineering problems. The Dirac Delta function, combined with integral and differential calculus, is a cornerstone in this field. Engineering often requires dealing with impulses, systems' resets, or changes that happen instantaneously, which is why the Dirac Delta is so important.

The use of the Dirac Delta function in engineering is prominent in areas like signal processing and control systems. Engineers utilize it to model impulse responses in systems. For instance, when designing systems that need to respond to sudden changes, mastering the use of the Dirac Delta function can prove quite advantageous.

In the exercise presented, applying the Dirac Delta function allows engineers to model scenarios easily and handle changes at specific points without complex calculations. The common patterns of using derivatives with the Dirac Delta reflect the dynamic nature of engineering systems — whether second derivatives for acceleration in mechanical systems or the rate of change in electrical circuits.

• The integration of Dirac Delta functions is crucial for analyzing signals and systems in engineering.
• These mathematical tools simplify complex changes into manageable computations, making them indispensable for engineers.