Find a function f that has a continuous derivative

[rjs123]

Homework Statement

Find a function f that has a continuous derivative on (0, ∞) and that has both of the following properties:

i. The graph of f goes through the point (1, 1)
ii. The length L of the curve from (1, 1) to any point (x, f(x)) is given by the formula L = lnx + f(x) - 1

Homework Equations

Recall that the arc length from (a, f(a)) to (b, f(b)) is given by the formula

 

[micromass]

So, what did you try already??

 

[Char. Limit]

I think the fundamental theorem of calculus will be useful here... since you know the arc length, you can set it equal to the integral.

 

[rjs123]

So, what did you try already??

I'm not really sure where to start...but I'm assuming I pick a random integer from 1 to infinity to represent b. Since I'm given the equation for f(x)'s arclength from 1 to any point to infinity: L = lnx + f(x) - 1


...should i set it equal to the equation for finding arclength in general? I've never seen this type of problem before so I'm a little puzzled.

 

[micromass]

Well, you can do what Char limit proposed. The length from 1 to x is given by ln(x)+f(x)-1. But on the other hand, the lenth is also given by

\int_1^x{\sqrt{1+(f^\prime(x))^2}dx}

So, setting them equal gives you

ln(x)+f(x)-1=\int_1^x{\sqrt{1+(f^\prime(x))^2}dx}

Now derive both sides...

 

[rjs123]

Well, you can do what Char limit proposed. The length from 1 to x is given by ln(x)+f(x)-1. But on the other hand, the lenth is also given by

\int_1^x{\sqrt{1+(f^\prime(x))^2}dx}

So, setting them equal gives you

ln(x)+f(x)-1=\int_1^x{\sqrt{1+(f^\prime(x))^2}dx}

Now derive both sides...

Let x = 3

ln(3)+f(3)-1=\int_1^3{\sqrt{1+(f^\prime(x))^2}dx}

ln(3)+f(3)-1=\int_1^3{{1+(f^\prime(x))}dx}

ln(3)+f(3)-1={{x+(f(x))}} from 3 to 1


This is what I have so far...am i on the right path?

 

[micromass]

Let x = 3

ln(3)+f(3)-1=\int_1^3{\sqrt{1+(f^\prime(x))^2}dx}

ln(3)+f(3)-1=\int_1^3{{1+(f^\prime(x))}dx}

I really don't know what you did there. You can't eliminate the square root like that. It is NOT true that \sqrt{a+b}=\sqrt{a}+\sqrt{b}!

Instead, you have to take the derivative of both sides and apply the fundamental theorem of calculus...

 

[rjs123]

I really don't know what you did there. You can't eliminate the square root like that. It is NOT true that \sqrt{a+b}=\sqrt{a}+\sqrt{b}!

Instead, you have to take the derivative of both sides and apply the fundamental theorem of calculus...

dont you mean integral of both sides, and i thought the left side should be left alone...no? I'm stuck on integrating the right side...any help would be appreciated.

 

[micromass]

No, I really mean to take the derivative of both sides! This will make the integral disappear...

 

[rjs123]

ln(x)+f(x)-1=\int_1^x{\sqrt{1+(f^\prime(x))^2}dx}


becomes this correct?


1/x+f^\prime(x)={\sqrt{1+(f^\prime(x))^2}

 

[Char. Limit]

Yes, good. Now square both sides and solve for f'(x). Then integrate.

 

[rjs123]

Yes, good. Now square both sides and solve for f'(x). Then integrate.

ln(x)+f(x)-1=\int_1^x{\sqrt{1+(f^\prime(x))^2}dx}


1/x+f^\prime(x)={\sqrt{1+(f^\prime(x))^2}

next steps:

(1/x+f^\prime(x))^2={1+(f^\prime(x))^2}

1/x^2+f^\prime(x)^2={1+f^\prime(x)^2}

I haven't had calc in a while...if solving for f'(x) wouldn't they cancel each other out here?

 

[Char. Limit]

You squared the left side wrong.

(a+b)^2 \neq a^2 + b^2

(a+b)^2 = a^2 + 2 a b + b^2

 

[rjs123]

You squared the left side wrong.

(a+b)^2 \neq a^2 + b^2

(a+b)^2 = a^2 + 2 a b + b^2

ln(x)+f(x)-1=\int_1^x{\sqrt{1+(f^\prime(x))^2}dx}


1/x+f^\prime(x)={\sqrt{1+(f^\prime(x))^2}

next steps:

(1/x+f^\prime(x))^2={1+(f^\prime(x))^2}

1/x^2+ 2f'(x)/x + f'(x)^2=1+(f^\prime(x))^2

1/x^2+ 2f'(x)/x=1


simple mistake...im now here

multiply by x?

 

[Char. Limit]

Multiplying by x will help, yes. Then you need to subtract 1/x.

 

[rjs123]

Multiplying by x will help, yes. Then you need to subtract 1/x.

(a+b)^2 \neq a^2 + b^2

(a+b)^2 = a^2 + 2 a b + b^2[/QUOTE]

ln(x)+f(x)-1=\int_1^x{\sqrt{1+(f^\prime(x))^2}dx}


1/x+f^\prime(x)={\sqrt{1+(f^\prime(x))^2}

next steps:

(1/x+f^\prime(x))^2={1+(f^\prime(x))^2}

1/x^2+ 2f'(x)/x + f'(x)^2=1+(f^\prime(x))^2

1/x^2+ 2f'(x)/x=1

2f'(x) = 1 - 1/x

f'(x) = (1 - 1/x)/2

whats a good simplification of the right side for easier integration...f'(x) = 1/2 - 2/x?

f(x) = 1/2x - 2lnx + C ...final answer? i'll double check this right now

 

[Char. Limit]

1/x^2+ 2f'(x)/x=1

2f'(x) = 1 - 1/x

There is a mistake here. After multiplying both sides by x, you should get 2f'(x) = x - 1/x. After that, I would treat them separately.

 

[rjs123]

There is a mistake here. After multiplying both sides by x, you should get 2f'(x) = x - 1/x. After that, I would treat them separately.

f'(x) = x/2 - 2/x

After integrating

f(x) = 1/4x^2 - 2lnx + C


is there a way to give you reputation or something for your help...i appreciate it.

 

[Char. Limit]

I don't know if there's a way to do that, and it's not really necessary. Just thanks are good.

Although you should have integrated 1/(2x), not 2/x...

 

[rjs123]

I don't know if there's a way to do that, and it's not really necessary. Just thanks are good.

Although you should have integrated 1/(2x), not 2/x...

f(x) = 1/4x^2 - 1/2ln(2x) + C


is the C necessary here?

 

[rjs123]

it appears

f(x) = 1/4x^2 - 1/2ln(2x)

does not intersect f(x)...im guessing to set f(x) to 1...then find C...by plugging in 1 for x?

1 = 1/4(1)^2 - 1/2ln(2) + C

1 = 1/4 - 1/2ln(2)

3/4 + 1/2ln(2) = C

f(x) = 1/4x^2 - 1/2ln(2x) + 3/4 + 1/2ln(2)