A Function So Cool That It Wears Shades

Calculus Level 3

There exists a unique, positive-valued, non-constant, continuous and differentiable function $y = f(x)$ such that

(i) over any specified interval, the area between $f(x)$ and the $x$ -axis is equal to the arclength of the curve, and

(ii) $f(0) = 1$ .

If $S = \displaystyle\int_{-1}^{2} f(x) dx$ , then find $\lfloor 1000S \rfloor$ .

The answer is 4802.

2 solutions

Brian Charlesworth
Oct 20, 2014

We require that $\displaystyle \int y dx =\displaystyle \int \sqrt{1 + (y')^{2}} dx$ over any interval. As the interval is arbitrary, the integrands must be equal. Thus

$y = \sqrt{1 + (y')^{2}}\\ \Rightarrow y^{2} = 1 + (\frac{dy}{dx})^2 \\ \Rightarrow \sqrt{y^{2} - 1} = \frac{dy}{dx}\\ \Rightarrow \frac{dy}{\sqrt{y^{2} - 1}} = dx \\ \Rightarrow \cosh^{-1}(y) = x + c \\ \Rightarrow y = \cosh(x + c)$ .

Now for $x = 0$ we have $y = \cosh(c) = 1 \Rightarrow c = 0$ .

Thus $y = f(x) = \cosh(x)$ , the integral of which from $x = -1$ to $x = 2$ is

$\sinh(2) - \sinh(-1) = \sinh(2) + \sinh(1) = 4.8020616...$ ,

giving us $\lfloor 1000S \rfloor = \boxed{4802}$ . $\square$

I did in this way :

$\int { \frac { dy }{ \sqrt { { y }^{ 2 }-1 } } } =\int { dx } \\ \Rightarrow \ln { (y\quad +\quad \sqrt { { y }^{ 2 }-1 } ) } \quad =\quad x\quad +C\\ \because \quad f\left( 0 \right) =1\\ \Rightarrow \quad C=0\\ \Rightarrow \quad y\quad +\quad \sqrt { { y }^{ 2 }-1 } \quad =\quad { e }^{ x }\quad \longrightarrow \quad (1)\\$ .

Now Rationalising The equation (1)

$(y\quad +\quad \sqrt { { y }^{ 2 }-1 } )(\frac { y\quad -\quad \sqrt { { y }^{ 2 }-1 } }{ y\quad -\quad \sqrt { { y }^{ 2 }-1 } } )\quad =\quad \quad { e }^{ x }\\ \Rightarrow \quad \frac { 1 }{ y\quad -\quad \sqrt { { y }^{ 2 }-1 } } =\quad { e }^{ x }\\ \\ \Rightarrow \quad y\quad -\quad \sqrt { { y }^{ 2 }-1 } \quad =\quad { e }^{ -x }\quad \longrightarrow (2)\\ \\ \therefore \quad (1)\quad +\quad (2)\quad \\ \\ \Rightarrow \quad y\quad =\quad \frac { { e }^{ x }+\quad { e }^{ -x }\quad }{ 2 } \\ \quad \int _{ -1 }^{ 2 }{ f(x)dx } \quad =\int _{ -1 }^{ 2 }{ \frac { { e }^{ x }+\quad { e }^{ -x }\quad }{ 2 } dx } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad { e }\quad +{ \quad e }^{ 2 }\quad -\quad \frac { 1 }{ e } \quad -\quad \frac { 1 }{ { e }^{ 2 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \approx \quad 4.802$ .

And @brian charlesworth Sir is you used Hyperbolic function in your solution ... I don't study about them yet. But I feel Glad if you give Some little idea about them related to your solution .. Thanks

Deepanshu Gupta - 6 years, 7 months ago

Well, we have that $\cosh(u) = \frac{e^{u} + e^{-u}}{2}$ and $\sinh(u) = \frac{e^{u} - e^{-u}}{2}$ .

From this we can show that

$\cosh^{2}(u) - \sinh^{2}(u) = 1$

$\Longrightarrow \cosh^{2}(u) - 1 = \sinh^{2}(u)$ .

We can also show that $\frac{d}{du}(\cosh(u)) = \sinh(u)$ . So in my integral, if we make the substitution $y = \cosh(u)$ then

$\frac{dy}{\sqrt{y^{2} - 1}} = du$ ,

the integral of which is just $u = \cosh^{-1}(y) + C$ .

Brian Charlesworth - 6 years, 7 months ago

Now I understand something about hyperbolic functions. Thanks a lot sir...!! And now I get too excited to learn more about them . So now I Google them ...not wanted to disturb you more.. ;)

Deepanshu Gupta - 6 years, 7 months ago

@Deepanshu Gupta – You're welcome. :)

Brian Charlesworth - 6 years, 7 months ago

In the 4th line, if before integrating we multiply both sides by 'y' and then integrate both sides then we get the answer as 1.732, even without actually finding the function f{x}. What's wrong in this method?????

Rahul Badenkal - 6 years, 7 months ago

Sorry, but I'm not sure quite what you mean. If we multiply by y then the process of separating variables has been nullified; we wouldn't then be able to go ahead and integrate both sides. Could you please provide more details so that I can better evaluate your method?

Brian Charlesworth - 6 years, 7 months ago

$We\quad have\quad \frac { dy }{ \sqrt { { y }^{ 2 }-1 } } =\quad dx\\ Multiplying\quad both\quad sides\quad by\quad 'y'\quad we\quad get\\ \frac { ydy }{ \sqrt { { y }^{ 2 }-1 } } =\quad ydx\\ Now\quad integrate\quad both\quad sides\quad from\quad the\quad limits\quad -1\quad to\quad 2.\\ \int _{ -1 }^{ 2 }{ \frac { ydy }{ \sqrt { { y }^{ 2 }-1 } } } =\quad \int _{ -1 }^{ 2 }{ \quad ydx } \\ but\quad \int _{ -1 }^{ 2 }{ \quad ydx } \quad =\quad \int _{ -1 }^{ 2 }{ \quad f(x)dx } \quad =\quad S\\ \int _{ -1 }^{ 2 }{ \frac { ydy }{ \sqrt { { y }^{ 2 }-1 } } } =\quad \sqrt { { y }^{ 2 }-1 } \quad (\quad from\quad the\quad limits\quad -1\quad to\quad 2\quad )\\ \int _{ -1 }^{ 2 }{ \frac { ydy }{ \sqrt { { y }^{ 2 }-1 } } } =\quad \sqrt { 3 } \\ \Rightarrow S\quad =\quad \sqrt { 3 } =\quad 1.732\\ or\\ 1000S\quad =\quad 1732\\ \quad$

Rahul Badenkal - 6 years, 7 months ago

@Rahul Badenkal – That's interesting; your method looks good at first glance, but I think that the problem is in the assumption that we can just multiply both sides by $y$ prior to integration. For the method of separation of variables to work from a theoretical standpoint, we must have one side of the equation only in terms of $y$ and $dy$ , and the other side only in terms of $x$ and $dx$ before we can integrate each side separately.

So while your method looks fine from an "algebraic" standpoint, it does not meet the theoretical requirements for the separation of variables method. You've raised an interesting question; I had to think about this for a while before realizing what the error was. :)

Here is an overview of the theory behind the separation of variables method, in case you're interested.

Brian Charlesworth - 6 years, 7 months ago

@Brian Charlesworth – I think @Rahul Badenkal 's method is giving a different result because the integrand $f (y)= \dfrac{y}{\sqrt{y^2-1}}$ has singularities at $y=\pm 1$

Ishan Singh - 5 years, 8 months ago

@Rahul Badenkal – When x moves from -1 to 3 why would we assume y to move between the same limits

Vamsi Manthena - 6 years, 4 months ago

@Vamsi Manthena – It shouldn't. I think that was the mistake I made.Thanks for pointing it out.

Rahul Badenkal - 6 years ago

can u please tell how did you got the arc length of the curve as $\int \sqrt{1 + (y')^{2}} dx$

Vighnesh Raut - 6 years, 5 months ago

by pythgoras theorem you can have element arc length as sqrt((dx)^2 +(dy)^2)=sqrt(1+(dy/dx)^2)dx. now just add these little elements within a given interval by reimann sum or simply integrate it.

Priyesh Pandey - 6 years, 3 months ago

Romeo Gomez
Jun 30, 2015

The first condition is $\int_a^x{f(t)}dt=\int_a^x{\sqrt{1+[f'(t)]^2}}dt,$ then using the first fundamental theorem of calculus we have that the first condition is $\boxed{f(x)=\sqrt{1+[f'(x)]^2}},$ squaring both sides and doing some algebra we get this differential equation $[f'(x)]^2+1=[f(x)]^2,$ and then taking the derivative of both sides $2f'(x)f''(x)=2f(x)f'(x)$ $\boxed{f''(x)=f(x)}.$ This is a homogeneous equation with constant coefficients and the solution is $\boxed{f(x)=C_1e^x+C_2e^{-x}}.$ At this point we need to find the constants $C_1, C_2$ . Substituing the solution in the differential equation and using the second condition of the problem we get this system of equations $C_1C_2=\frac{1}{4}$ $C_1+C_2=1$ and is easy to find thet the solution is $C_1=C_2=\frac{1}{2},$ hence $\boxed{f(x)=\frac{1}{2}(e^x+e^{-x})}$ so $S=\int_{-1}^2{f(x)}dx=4.8021...$ then $[1000S]=\boxed{4802}.$

@Brian Charlesworth thanks for this particular function because to me has a very beautiful property.

Romeo Gomez - 5 years, 11 months ago

Nice solution; thanks for posting it. :)

Brian Charlesworth - 5 years, 11 months ago

Thanks to you :D

Romeo Gomez - 5 years, 11 months ago

A Function So Cool That It Wears Shades

The answer is 4802.

2 solutions

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