Professional Courses
Industry-relevant training in Business, Technology, and Design to help professionals and graduates upskill for real-world careers.
Categories
Interactive Games
Fun, engaging games to boost memory, math fluency, typing speed, and English skills—perfect for learners of all ages.
Typing
Memory
Math
English Adventures
Knowledge
Enroll to start learning
You’ve not yet enrolled in this course. Please enroll for free to listen to audio lessons, classroom podcasts and take practice test.
Interactive Audio Lesson
Listen to a student-teacher conversation explaining the topic in a relatable way.
Introduction to Laplace Transform of Derivatives
Unlock Audio Lesson
Today, we're discussing the Laplace Transform and how it applies to derivatives. Who can remind us what the Laplace Transform of a function is?
It's the integral of e^{-st} times the function, from 0 to infinity!
Correct! Now, when we differentiate a function and then take its Laplace Transform, we use a specific formula. Does anyone remember that formula?
Is it L{f'(t)} = sF(s) - f(0)?
Exactly! Let's break that down. The term 'sF(s)' represents the transformed variable, while 'f(0)' is the initial value of the function. This ties back to the idea that the Laplace Transform converts differentiation into algebraic terms, which are easier to work with.
So, we can solve differential equations more easily using this transform?
Yes! That’s one of its primary applications. Great job, everyone!
Proof of the First Derivative Transform
Unlock Audio Lesson
Now, let's move on to the proof of the first derivative's Laplace Transform using integration by parts. Can anyone remind us how integration by parts works?
It’s u dv = uv - ∫v du!
Exactly! We can set u = f(t) and dv = e^{-st} dt. Applying this gives us the foundation for our proof. What happens as t approaches infinity if our function is of exponential order?
The term e^{-st} f(t) goes to zero!
That's right! So, we can simplify our integral greatly. What do we arrive at?
We end up with sF(s) - f(0)!
Perfect! This proof is essential as it lays the groundwork for understanding how transformations work for higher derivatives as well. Let’s revisit the second derivative in our next session.
Laplace Transform of Higher Derivatives
Unlock Audio Lesson
Now, let's talk about the second derivative. Can someone tell me the formula for its Laplace Transform?
It’s L{f''(t)} = s^2F(s) - sf(0) - f'(0)!
Correct! By applying the first result, we can derive this by taking the Laplace Transform of L{f'(t)}. How do we handle more than two derivatives?
We use the general formula! L{f(n)(t)} = s^n F(s) - Σ from k=0 to n-1 of s^{n-1-k} f^(k)(0).
Excellent! This formula summarizes the process. What’s great is that it enables us to solve complex IVPs efficiently. Any questions about this?
How is this useful in real life?
It’s particularly vital in engineering for solving circuit equations and modeling systems. This connection is vital as it shows the practical application of our mathematical work!
Introduction & Overview
Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.
Quick Overview
Standard
In this section, we explore how the Laplace Transform can be applied to first derivatives of functions. We provide the mathematical proof of the transformation, alongside its application to differential equations, highlighting its significance in engineering and physical sciences.
Detailed
Detailed Summary
In Section 1.1.4, we focus on the Laplace Transform of derivatives, particularly on the first derivative. The equation for the Laplace Transform of a function is introduced, defined as:
$$L \{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) dt$$
This integral requires functions to be piecewise continuous and of exponential order. The Laplace Transform converts derivatives into algebraic equations which are significantly easier to solve than their differential counterparts.
For the first derivative, it is shown that:
$$L \{f'(t)\} = sF(s) - f(0)$$
The proof utilizes integration by parts, highlighting essential integration techniques and limits. As the variable t approaches infinity, the term involving f(t) approaches zero which simplifies the equation significantly.
Further, we derive the Laplace Transform of the second derivative:
$$L \{f''(t)\} = s^2F(s) - sf(0) - f'(0)$$
This formula is derived by applying the Laplace Transform again to the first derivative. We also generalize this process for higher-order derivatives:
$$L \{f^{(n)}(t)\} = s^n F(s) - \sum_{k=0}^{n-1} s^{n-1-k} f^{(k)}(0)$$
These transformations play a vital role in solving initial value problems (IVPs) in differential equations, enabling easier manipulation for solutions in fields such as engineering and physics, particularly in control systems, circuits, and mechanics.
Audio Book
Dive deep into the subject with an immersive audiobook experience.
Introduction to the Laplace Transform of the First Derivative
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
Let f(t) be a function such that f′ (t) exists and is continuous on [0, ∞). Then:
L{f′ (t)}=sF(s)−f(0)
Detailed Explanation
This chunk introduces the Laplace Transform of the first derivative of a function. The notation L{f′(t)} denotes the Laplace Transform of the derivative of f(t). According to the formula, the Laplace Transform of f′(t) is equal to s times the Laplace Transform of f(t), denoted as F(s), minus the value of the function f(t) at t=0, which is f(0). This means that when taking the Laplace Transform of the derivative, we not only account for how the function changes (the derivative) but we also need to adjust by the original function's value at the starting point.
Examples & Analogies
Think of the function f(t) as the position of a car over time. Its derivative, f′(t), is the speed of the car. The Laplace Transform of f′(t) helps us analyze the speed in a transformed space where we can solve problems more easily, similar to how knowing both the speed and the initial position of the car helps us understand its journey.
Derivation of the Laplace Transform of the First Derivative
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
Proof:
Using integration by parts:
L{f′ (t)}=∫e−stf′ (t)dt=[e−stf(t)] +s∫e−stf(t)dt
Detailed Explanation
This chunk describes the proof process for the Laplace Transform of the first derivative using integration by parts. Integration by parts is a technique from calculus that allows us to rewrite integrals in a different form. Here we start with the integral of e raised to a power times f′(t). After applying integration by parts, we obtain two parts: the boundary term [e−stf(t)] evaluated from 0 to infinity and another integral involving f(t). The boundary term approaches zero as t approaches infinity, so we can simplify the expression to arrive at L{f′(t)} = sF(s) − f(0).
Examples & Analogies
Imagine you're tracking the growth of a plant over time. If you know its height at various moments (the integrals) and how much it grows at those moments (the derivative), using integration by parts allows you to get a complete picture of how height relates to growth over time, ultimately simplifying the process of understanding the plant's overall height at the end.
Conclusion of the Proof
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
As t→∞, e−stf(t)→0 (since f(t) is of exponential order). So:
L{f′ (t)}=sF(s)−f(0)
Detailed Explanation
This chunk concludes the proof by discussing the behavior of the term e−stf(t) as time approaches infinity. Since f(t) is of exponential order, the product e−stf(t) goes to zero when we take the limit as t approaches infinity. This is crucial because it allows us to disregard this term when calculating the Laplace Transform, leading us to the final result, confirming the relationship found earlier: L{f'(t)} = sF(s) - f(0).
Examples & Analogies
Consider a balloon that is slowly deflating as time passes. Initially, it might have a certain volume (similar to f(0)), but as time goes on, and the balloon gradually deflates, its volume approaches zero. Just as we can ignore the negligible volume of a fully deflated balloon when calculating its final state, we disregard the term e−stf(t) in our calculations as t grows very large, simplifying our analysis.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
-
Laplace Transform of the First Derivative: A transformation that converts the first derivative of a function into algebraic form.
-
Integration by Parts: A method of integration used to find the Laplace Transform of derivatives, based on the product rule of differentiation.
-
Initial Value Problems (IVPs): Differential equations along with specified starting values that can be solved effectively using the Laplace Transform.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
-
Example of Laplace Transform of a first derivative: Given f(t) = t, L{f'(t)} = L{1} = s.
-
Example of applying the Laplace Transform to a second derivative, deriving the necessary steps to arrive at the final expression.
Memory Aids
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
🎵 Rhymes Time
-
To find the first derivative's name, sF(s) minus f(0's fame.
📖 Fascinating Stories
-
In a world of calculus, every function feared differentiation until they discovered the magic of the Laplace Transform, turning their rates of change into friendly algebraic forms.
🧠 Other Memory Gems
-
Remember as 'Fools Strive Effectively' for L{f'(t)} = sF(s) - f(0).
🎯 Super Acronyms
LFD for 'Laplace First Derivative' to recall L{f'(t)}.
Flash Cards
Review key concepts with flashcards.
Glossary of Terms
Review the Definitions for terms.
-
Term: Laplace Transform
Definition:
An integral transform that converts a function of time into a function of a complex variable, facilitating easier solutions to differential equations.
-
Term: First Derivative
Definition:
The rate of change of a function at a given point, representing the slope of the function at that point.
-
Term: Integration by Parts
Definition:
An integration technique derived from the product rule of differentiation, used to integrate products of functions.
-
Term: Exponential Order
Definition:
A condition where a function grows at a rate no faster than an exponential function as t approaches infinity.
-
Term: Initial Value Problem (IVP)
Definition:
A differential equation along with specified values at a given point, typically at t=0.