Math / Calculus

QuestionFind the inverse Laplace transform $f(t)$ of $F(s)=\frac{e^{-12 s}}{s^{2}}$ . Then sketch the graph of $f$ . $f(t)=$ $\square$ Choose the correct graph below. A. B. © C. D.

Studdy Solution

STEP 1

What is this asking? We need to find the inverse Laplace transform of a given function and then pick the correct graph that represents it! Watch out! Don't forget about the exponential term $e^{-12s}$ , it shifts the function in time!

STEP 2

1. Inverse Transform
2. Graph Selection

STEP 3

Alright, let's start by looking at our Laplace transform: $F(s) = \frac{e^{-12s}}{s^2}$ .
We need to bring this back to the time domain!

STEP 4

Remember the time-shifting property of the Laplace transform?
It says that if $\mathcal{L}\{f(t)\} = F(s)$ , then $\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s)$ , where $u(t-a)$ is the unit step function.
This is super important because we have that $e^{-12s}$ term, which means a time shift is involved!

STEP 5

Let's focus on $\frac{1}{s^2}$ first.
We know that the inverse Laplace transform of $\frac{1}{s^2}$ is $t$ .
We can write this as $\mathcal{L}^{-1}\{\frac{1}{s^2}\} = t$ .

STEP 6

Now, bringing back the exponential term $e^{-12s}$ , we can apply the time-shifting property.
Here, our $a$ is 12, and our $F(s)$ is $\frac{1}{s^2}$ .
So, our $f(t)$ is $t$ , and $f(t-a)$ becomes $f(t-12) = (t-12)$ .

STEP 7

Multiplying by the unit step function $u(t-12)$ , we get our final inverse Laplace transform: $f(t) = (t-12)u(t-12)$ .
This means the function is 0 before $t=12$ and then increases linearly with a slope of 1 afterwards!

STEP 8

Our inverse transform $f(t) = (t-12)u(t-12)$ tells us a lot about the graph.
It's zero until $t$ reaches 12, and then it starts increasing linearly.

STEP 9

Looking at the provided graphs, we can immediately eliminate graphs B and C.
Graph B decreases, and graph C is a parabola, neither of which match our function.

STEP 10

Graphs A and D both show lines starting at the origin and increasing linearly.
However, our function doesn't start increasing until $t=12$ !
So, the correct graph must start at $t=12$ .

STEP 11

Between graphs A and D, only graph A starts increasing at $t=12$ .
Graph D starts increasing right away from $t=0$ .
Therefore, graph A is the correct representation of our function!

STEP 12

$f(t) = (t-12)u(t-12)$ and the correct graph is A.

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QuestionFind the inverse Laplace transform $f(t)$ of $F(s)=\frac{e^{-12 s}}{s^{2}}$ . Then sketch the graph of $f$ . $f(t)=$ $\square$ Choose the correct graph below. A. B. © C. D.

Studdy Solution

STEP 1

What is this asking? We need to find the inverse Laplace transform of a given function and then pick the correct graph that represents it! Watch out! Don't forget about the exponential term $e^{-12s}$ , it shifts the function in time!

STEP 2

1. Inverse Transform
2. Graph Selection

STEP 3

Alright, let's start by looking at our Laplace transform: $F(s) = \frac{e^{-12s}}{s^2}$ .
We need to bring this back to the time domain!

STEP 4

STEP 5

Let's focus on $\frac{1}{s^2}$ first.
We know that the inverse Laplace transform of $\frac{1}{s^2}$ is $t$ .
We can write this as $\mathcal{L}^{-1}\{\frac{1}{s^2}\} = t$ .

STEP 6

Now, bringing back the exponential term $e^{-12s}$ , we can apply the time-shifting property.
Here, our $a$ is 12, and our $F(s)$ is $\frac{1}{s^2}$ .
So, our $f(t)$ is $t$ , and $f(t-a)$ becomes $f(t-12) = (t-12)$ .

STEP 7

Multiplying by the unit step function $u(t-12)$ , we get our final inverse Laplace transform: $f(t) = (t-12)u(t-12)$ .
This means the function is 0 before $t=12$ and then increases linearly with a slope of 1 afterwards!

STEP 8

Our inverse transform $f(t) = (t-12)u(t-12)$ tells us a lot about the graph.
It's zero until $t$ reaches 12, and then it starts increasing linearly.

STEP 9

Looking at the provided graphs, we can immediately eliminate graphs B and C.
Graph B decreases, and graph C is a parabola, neither of which match our function.

STEP 10

Graphs A and D both show lines starting at the origin and increasing linearly.
However, our function doesn't start increasing until $t=12$ !
So, the correct graph must start at $t=12$ .

STEP 11

Between graphs A and D, only graph A starts increasing at $t=12$ .
Graph D starts increasing right away from $t=0$ .
Therefore, graph A is the correct representation of our function!

STEP 12

$f(t) = (t-12)u(t-12)$ and the correct graph is A.

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QuestionFind the inverse Laplace transform f(t)f(t)f(t) of F(s)=e−12ss2F(s)=\frac{e^{-12 s}}{s^{2}}F(s)=s2e−12s​. Then sketch the graph of fff. f(t)=f(t)=f(t)= □\square□ Choose the correct graph below. A. B. © C. D.

Studdy Solution

STEP 1

What is this asking? We need to find the *inverse Laplace transform* of a given function and then pick the correct graph that represents it! Watch out! Don't forget about the exponential term e−12se^{-12s}e−12s, it shifts the function in time!

STEP 2

1. Inverse Transform2. Graph Selection

STEP 3

Alright, let's **start** by looking at our Laplace transform: F(s)=e−12ss2F(s) = \frac{e^{-12s}}{s^2}F(s)=s2e−12s​.We need to bring this back to the time domain!

STEP 4

STEP 5

Let's **focus** on 1s2\frac{1}{s^2}s21​ first.We know that the inverse Laplace transform of 1s2\frac{1}{s^2}s21​ is ttt.We can write this as L−1{1s2}=t\mathcal{L}^{-1}\{\frac{1}{s^2}\} = tL−1{s21​}=t.

STEP 6

Now, bringing back the exponential term e−12se^{-12s}e−12s, we can apply the time-shifting property.Here, our aaa is **12**, and our F(s)F(s)F(s) is 1s2\frac{1}{s^2}s21​.So, our f(t)f(t)f(t) is ttt, and f(t−a)f(t-a)f(t−a) becomes f(t−12)=(t−12)f(t-12) = (t-12)f(t−12)=(t−12).

STEP 7

Multiplying by the unit step function u(t−12)u(t-12)u(t−12), we get our **final inverse Laplace transform**: f(t)=(t−12)u(t−12)f(t) = (t-12)u(t-12)f(t)=(t−12)u(t−12).This means the function is **0** before t=12t=12t=12 and then increases linearly with a slope of **1** afterwards!

STEP 8

Our inverse transform f(t)=(t−12)u(t−12)f(t) = (t-12)u(t-12)f(t)=(t−12)u(t−12) tells us a lot about the graph.It's **zero** until ttt reaches **12**, and then it starts increasing linearly.

STEP 9

Looking at the provided graphs, we can immediately **eliminate** graphs B and C.Graph B decreases, and graph C is a parabola, neither of which match our function.

STEP 10

Graphs A and D both show lines starting at the origin and increasing linearly.However, our function *doesn't start* increasing until t=12t=12t=12!So, the correct graph must start at t=12t=12t=12.

STEP 11

Between graphs A and D, only graph A starts increasing at t=12t=12t=12.Graph D starts increasing right away from t=0t=0t=0.Therefore, graph A is the **correct representation** of our function!

STEP 12

f(t)=(t−12)u(t−12)f(t) = (t-12)u(t-12)f(t)=(t−12)u(t−12) and the correct graph is **A**.

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QuestionFind the inverse Laplace transform $f(t)$ of $F(s)=\frac{e^{-12 s}}{s^{2}}$ . Then sketch the graph of $f$ . $f(t)=$ $\square$ Choose the correct graph below. A. B. © C. D.

What is this asking? We need to find the inverse Laplace transform of a given function and then pick the correct graph that represents it! Watch out! Don't forget about the exponential term $e^{-12s}$ , it shifts the function in time!

1. Inverse Transform
2. Graph Selection

Alright, let's start by looking at our Laplace transform: $F(s) = \frac{e^{-12s}}{s^2}$ .
We need to bring this back to the time domain!

Let's focus on $\frac{1}{s^2}$ first.
We know that the inverse Laplace transform of $\frac{1}{s^2}$ is $t$ .
We can write this as $\mathcal{L}^{-1}\{\frac{1}{s^2}\} = t$ .

Now, bringing back the exponential term $e^{-12s}$ , we can apply the time-shifting property.
Here, our $a$ is 12, and our $F(s)$ is $\frac{1}{s^2}$ .
So, our $f(t)$ is $t$ , and $f(t-a)$ becomes $f(t-12) = (t-12)$ .

Multiplying by the unit step function $u(t-12)$ , we get our final inverse Laplace transform: $f(t) = (t-12)u(t-12)$ .
This means the function is 0 before $t=12$ and then increases linearly with a slope of 1 afterwards!

Our inverse transform $f(t) = (t-12)u(t-12)$ tells us a lot about the graph.
It's zero until $t$ reaches 12, and then it starts increasing linearly.

Looking at the provided graphs, we can immediately eliminate graphs B and C.
Graph B decreases, and graph C is a parabola, neither of which match our function.

Graphs A and D both show lines starting at the origin and increasing linearly.
However, our function doesn't start increasing until $t=12$ !
So, the correct graph must start at $t=12$ .

Between graphs A and D, only graph A starts increasing at $t=12$ .
Graph D starts increasing right away from $t=0$ .
Therefore, graph A is the correct representation of our function!

$f(t) = (t-12)u(t-12)$ and the correct graph is A.