Question

Find\[
y = f(x)
\]Such that \[
\int_0^tf'(x)~dy = 2t^2
\]

Answer 1

I want to see what you guys come up with for this one.

Answer 2

I know one solution, I'm not sure how many there are.

Answer 3

Why did you delete your reply?

Answer 4

found my mistake

Answer 5

Oh

Answer 6

\[\int\limits_{0}^{t}(f'(x))^2 dx=G(t)-G(0) \\ \text{ where } G(t)-G(0)=2t^2 \\ G'(t)=4t \\ \text{ but } G'(x)=(f'(x))^2 \\ g(x)=(f'(x))^2 \\ 4x=(f'(x))^2 \\ 2 \sqrt{x}=f'(x)\]

Answer 7

still something wrong

Answer 8

Yep, something weird is going on here, lol

Answer 9

but this isn't too far off actually

Answer 10

it is a constant multiple off

Answer 11

oops check the wrong function

Answer 12

lol

Answer 13

I wonder what's bad about what I wrote.

Answer 14

oh wait it does work

Answer 15

sorry lol

Answer 16

\[f'(x)=2 \sqrt{x} \\ =2x^\frac{1}{2} \\ f(x)=2 \frac{x^\frac{3}{2}}{\frac{3}{2}}+C =2 \frac{2}{3} x^\frac{3}{2}+C=\frac{4}{3}x^\frac{3}{2}+C\]

Answer 17

Checking:
\[f(x)=\frac{4}{3}x^\frac{3}{2}+C \\ f'(x)=2x^\frac{1}{2} \\ (f'(x))^2=4x \\ \int\limits_{0}^{t}4x dx=\frac{4x^2}{2}|_0^t=2x^2|_0^t=2(t^2-0^2)=2t^2\]

Answer 18

Did you get a different function that also works?

Answer 19

Something not quite right is happening here. I'm not sure why though.

Answer 20

why my solution?

Answer 21

with*

Answer 22

Well, consider \[
y=\frac{4}{3}x^{3/2}
\]for a moment

Answer 23

\[
x = \frac 34 y^{2/3}
\]When we plug this into the derivative, we get: \[
2\sqrt{x} = 2\left(\frac 34y^{2/3}\right)^{1/2} = \sqrt{3}y^{1/3}
\]

Answer 24

The anti derivative of that doesn't look like it is correct

Answer 25

\[y=\frac{4}{3}x^\frac{3}{2} \\ \frac{3}{4}y=x^\frac{3}{2} \\ (\frac{3}{4} y)^\frac{2}{3}=x \\\]

Answer 26

Sure

Answer 27

Oh, I know what the issue is!

Answer 28

what?

Answer 29

You made a tiny mistake

Answer 30

Remember that we were integrating \(y\in [0,t]\)

Answer 31

But you changed it to \(x\in [0,t]\), which is not the same region of integration.

Answer 32

omg

Answer 33

I didn't think about that

Answer 34

Let me try again

Answer 35

I forgot the limits were in terms of y :(

Answer 36

Well, so far you have been very clever. Even I didn't completely understand why it wasn't working.

Answer 37

\[\int\limits\limits_{f^{-1}(0)}^{f^{-1}(t)}(f'(x))^2 dx=G(f^{-1}(t))-G(f^{-1}(0)) \\ \text{ where } G(f^{-1}(t))-G(f^{-1}(0))=2t^2 \\ [G(f^{-1}(t))]'=4t \\ (f^{-1})'(t)g(t)=4t \\ \frac{1}{f'(f^{-1}(t))} g(t)=4t \\ g(t)=4t f'(f^{-1}(t)) \text{ hmmm.... } \text{ we are also given } G'(x)=(f'(x))^2 \\ g(x)=(f'(x))^2 \\ g(t)=(f'(t))^2 \\ (f'(t))^2=4t f'(f^{-1}(t))\]
so far I have this...

Answer 38

I assume f was one to one for the function I'm looking for

Answer 39

not successful yet
still thinking

Answer 40

oh no I messed up lol

Answer 41

To make it easier, maybe let \(x=g(y)\).

Answer 42

To keep your inverses less hard to type

Answer 43

\[\int\limits\limits\limits_{f^{-1}(0)}^{f^{-1}(t)}(f'(x))^2 dx=G(f^{-1}(t))-G(f^{-1}(0)) \\ \text{ where } G(f^{-1}(t))-G(f^{-1}(0))=2t^2 \\ [G(f^{-1}(t))]'=4t \\ (f^{-1})'(t)g(f^{-1}(t))=4t \\ \frac{1}{f'(f^{-1}(t))} g((f^{-1}(t))=4t \\ g(f^{-1}(t))=4t f'(f^{-1}(t)) \text{ hmmm.... } \text{ we are also given } G'(x)=(f'(x))^2 \\ g(x)=(f'(x))^2 \\ \]
I think I corrected it
\[g(x)=4f(x)f'(x)\]
\[4f(x)f'(x)=(f'(x))^2\]

Answer 44

\[4 f(x)=f'(x) \text{ assume } f'(x) \neq 0\]

Answer 45

\[4 f=\frac{df}{dx} \\ 4 dx=\frac{df}{f} \\ 4x=\ln|f(x)|+C\]

Answer 46

\[4x+K=\ln|f(x)| \\ e^{4x+K}=f(x) \\ f(x)=ce^{4x}\]

Answer 47

I have to check that

Answer 48

Why, I don't understand some of your work above.

Answer 49

I mean, what happened to the square around \(f'\)?

Answer 50

divided both sides by f'

Answer 51

How did you get \(g=4ff'\)?

Answer 52

http://www.wolframalpha.com/input/?i=integrate%2816+e%5E%288t%29%2Ct%3D-inf..1%2F4+ln%28t%29%29 seems to work out

Answer 53

ok let me go through it

Answer 54

do you the line I said hmmm... in?

Answer 55

You are correct

Answer 56

I just was confused about the \(g=4ff'\) part.

Answer 57

\[\int\limits\limits\limits\limits_{f^{-1}(0)}^{f^{-1}(t)}(f'(x))^2 dx=G(f^{-1}(t))-G(f^{-1}(0)) \\ \text{ where } G(f^{-1}(t))-G(f^{-1}(0))=2t^2 \\ [G(f^{-1}(t))]'=4t \\ (f^{-1})'(t)g(f^{-1}(t))=4t \\ \frac{1}{f'(f^{-1}(t))} g((f^{-1}(t))=4t \\ g(f^{-1}(t))=4t f'(f^{-1}(t)) \text{ hmmm.... } \text{ we are also given } G'(x)=(f'(x))^2 \\ g(x)=(f'(x))^2 \\ \]
I replaced f inverse of t with x there
x=f inverse of t
means f(x)=t

Answer 58

So you let \(t=f(x)\)?

Answer 59

oh yes lol

Answer 60

I could have just said that

Answer 61

Oh, good thinking, I did something a bit different.

Answer 62

did you get the same function though?

Answer 63

\[
\frac{d}{dt}\int_{f^{-1}(0)}^{f^{-1}(t)}[f'(x)]^2dx = \bigg[ f'(f^{-1}(t))\bigg]^2\frac{d}{dt}f^{-1}(t)
\]Then I factored out a term: \[
\bigg[ f'(f^{-1}(t))\bigg]\bigg[ f'(f^{-1}(t))\frac{d}{dt}f^{-1}(t)\bigg]
\]Then here I undo the chain rule: \[
\bigg[ f'(f^{-1}(t))\bigg]\bigg[ f(f^{-1}(t))\bigg]' = \bigg[ f'(f^{-1}(t))\bigg](t)' = f'(f^{-1}(t))
\]So I had \[
f'(f^{-1}(t))=4t
\]Then I plug in \(t=f(x)\)\[
f'(x) = 4f(x)
\]

So yes, you get correct answer.

Answer 64

The correct answer is \[
y=Ce^{4x}
\]I believe.

Answer 65

yes yes

Answer 66

but you were wondering if someone could get a different solution

Answer 67

so I wonder if we could have approached it without the assumption it was one to one

Answer 68

and made a different assumption on f

Answer 69

What do you mean, one to one?

Answer 70

injective

Answer 71

You mean that \(f\) is injective?

Answer 72

yeah

Answer 73

i has to be 1 to 1 (injective) to have an inverse function

Answer 74

f

Answer 75

What if we let it equal \(f(t)\)? What is the most simplified form?

Answer 76

what equal f(t)?

Answer 77

Find\[
y = f(x)
\]Such that \[
\int_0^tf'(x)~dy = g(t)
\]

Answer 78

\[
f'(f^{-1}(t))=g'(t))
\]Do we stop here?

Answer 79

Or maybe: \[
f'(x) = g'(f(x))
\]

Answer 80

For our case, \(g(t) = 2t^2\).

Answer 81

And \(g'(f(x)) = 4f(x)\).

Answer 82

I like this question by the way. Where did you get it?

Answer 83

I made it up.

Answer 84

lol
Also I will probably try to see if I can come up with different solution that works...
But I do have to go for tonight unfortunately.
P.S. So you do think there are other different solutions maybe right?

Answer 85

I wonder what the identity is... if we let \(g(t)=f(t)\), then: \[
f'(x) = f'(f(x))
\]How to solve.... Maybe multiply both by \(f'(x)\): \[
[f'(x)]^2 = f'(f(x))f'(x) = [f(f(x))]'
\]Woah!

Answer 86

@freckles I think that your method got every possible solution. I don't think you did operations which would remove answer.

Answer 87

So the function f(x) doesn't have to be a function with a inverse function to write the limits as I did if that makes sense?

Answer 88

I mean what if f(x) has an inverse relation instead
that wouldn't make a difference in the way I wrote the limits?

Answer 89

Hmmm, good point. To be honest, I don't think you had to ever change the integration.

Answer 90

\[
\frac{d}{dt}\int_0^tf'(x)~dy = \frac{d}{dt}\int_0^tf'(f^{-1}(y))~dy = f'(f^{-1}(t))
\]

Answer 91

that is the only thing I was concerned about

Answer 92

lol I used present tense and past tense in the same sentence
I'm glad there are no language snobs around

Answer 93

It's okay because it is a different clause.

Answer 94

Like "I know what you did"

Answer 95

ah true
lol

Answer 96

anyways thanks so much for asking a question I could answer
it was a bit tricky which I like :)

Answer 97

goodnight