I have a problem that reads; Find f(t) if f '(t)=t, and f(0)=3

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- Jul 1st 2008, 08:49 PM #1

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- Jul 1st 2008, 08:52 PM #2
Note that $\displaystyle f(t)=\int f'(t)\,dt$

Integrating, we get

$\displaystyle \int t \,dt = \frac{1}{2}t^2+C$

However, you're given an initial condition. Use it to find the value of C.

$\displaystyle \because f(0)=3, \ 3=\frac{1}{2}(0)^2+C\implies \color{blue}\boxed{C=3}$

Thus, $\displaystyle \color{blue}\boxed{f(t)=\frac{1}{2}t^2+3}$

Hope that this makes sense!

--Chris

- Jul 1st 2008, 08:54 PM #3

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- Jul 1st 2008, 08:57 PM #4

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- Jul 1st 2008, 09:02 PM #5
Haha, maybe this will be fun

Let $\displaystyle f'(t)=at+b$

So then

$\displaystyle \int{f'(t)}dt=f(t)=\frac{at^2}{2}+bt+C$

Now seeing that

$\displaystyle f(0)=3\Rightarrow{C=3}$

So we have that

$\displaystyle f(t)=\frac{at^2}{2}+bt+3$

...you guys should thank me for doing all the heavy lifting around here.

- Jul 1st 2008, 10:41 PM #6

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That is not the question asked, and is a pointless extension why not $\displaystyle f'(x)=P_n(x)$ where $\displaystyle P_n(x)$ is an $\displaystyle n$-th order polynomial in x?

Also, the idea is to give the minimum help to allow the original poster to complete the solution themselves.

RonL

- Jul 1st 2008, 10:52 PM #7