SMT 2013 Calculus #8:

The function $f(x)$ is defined for all $x\ge 0$ and is always nonnegative. It has the additional property that if any line is drawn from the origin with any positive slope $m$, it intersects the graph $y=f(x)$ at precisely one point, which is $\frac{1}{\sqrt{m}}$ units from the origin. Let a be the unique real number for which f takes on its maximum value at $x=a$ (you may assume that such an a exists). Find $\int_{0}^{a}f(x) \, dx$.

My work so far:

I tried interpreting the question in the following way:

"if any line is drawn from the origin with any slope $m$, it intersects the graph $y = f(x)$ at precisely one point, which is $\frac{1}{\sqrt{m}}$ units from the origin"

This can be parameterized as $$\begin{cases} y = mx \\ x^2 + y^2 = \frac{1}{m}\end{cases}$$

Then to find the maximum value $x = a$, you simply maximize the function $y = \sqrt{\frac{m}{1+m^2}}$ (which is the same as maximizing the function $\frac{m}{1+m^2}$).

$\left(\frac{m}{1+m^2}\right)' = 0 \implies m = 1 \implies$ the maximum point is $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$, so $ a=\frac{1}{\sqrt{2}}$.

After this point, I don't know how to compute the integral $$\int_0^\frac{1}{\sqrt{2}}{f(x)}\, dx.$$ (I tried using polar coordinates, but then I got confused.)

Edit 2: I have found an algebraic solution! It involved solving $x$ in terms $y$ (too tired to type out the entire solution). But if anyone can explain step-by-step a solution using polar coordinates, that would be greatly appreciated :)


2 Answers 2


First Approach

If $y=mx$ and $x^2+y^2=\frac1m$, then $x^2+y^2=\frac xy$, which implies $$ yx^2-x+y^3=0\tag{1} $$ Taking the implicit derivative of $(1)$: $$ y'=\frac{1-2xy}{x^2+3y^2}\tag{2} $$ $y'=0$ means $1=2xy$; then applying $y=mx$, we get $$ 1=2xy=2mx^2\tag{3} $$ applying $x^2+y^2=\frac1m$ gives $$ 1+2my^2=2mx^2+2my^2=2m\tag{4} $$ using $y=mx$ yields $$ m^2=\frac{y^2}{x^2}=2m-1\tag{5} $$ therefore, $$ m=1\tag{6} $$ so we have $a=\frac1{\sqrt2}$.

Applying the quadratic formula to $(1)$: $$ x=\frac{1-\sqrt{1-4y^4}}{2y}\tag{7} $$ Integration by parts and substituting $z^2=1-4y^4$ gives $$ \begin{align} \int_0^{1/\sqrt2}y\,\mathrm{d}x &=\frac12-\int_0^{1/\sqrt2}x\,\mathrm{d}y\\ &=\frac12-\int_0^{1/\sqrt2}\frac{1-\sqrt{1-4y^4}}{2y}\,\mathrm{d}y\\ &=\frac12-\frac18\int_0^{1/\sqrt2}\frac{1-\sqrt{1-4y^4}}{4y^4}\,\mathrm{d}4y^4\\ &=\frac12-\frac18\int_0^1\frac{1-z}{1-z^2}2z\,\mathrm{d}z\\ &=\frac12-\frac14\int_0^1\frac{z}{1+z}\,\mathrm{d}z\\ &=\frac12-\frac14\int_0^1\left(1-\frac1{1+z}\right)\,\mathrm{d}z\\ &=\frac14(1+\log(2))\tag{8} \end{align} $$

Polar Coordinate Approach

By virtue of the work done above, we know that the curve extends from $m=1$ to $m=\infty$, that is $\theta=\frac\pi4$ to $\theta=\frac\pi2$. There is also the triangle below that is not bounded by the curve.

region of interest

Note that $r^2=x^2+y^2=\frac1m$ and $\tan(\theta)=\frac yx=m$. Therefore, $$ r=\frac1{\sqrt{\tan(\theta)}}\tag{9} $$ The region is given by $(9)$ for $\frac\pi4\le\theta\le\frac\pi2$, and for $0\le\theta\le\frac\pi4$ the region is a right triangle with legs of $\frac1{\sqrt2}$. Thus the area is $$ \begin{align} \color{#C000FF}{\frac12\cdot\frac1{\sqrt2}\cdot\frac1{\sqrt2}}+\color{#00A000}{\int_{\pi/4}^{\pi/2}\frac12r^2\,\mathrm{d}\theta} &=\frac14+\frac12\int_{\pi/4}^{\pi/2}\frac{\mathrm{d}\theta}{\tan(\theta)}\\ &=\frac14+\frac12\left[\vphantom{\int}\log(\sin(\theta))\right]_{\pi/4}^{\pi/2}\\ &=\frac14(1+\log(2))\tag{10} \end{align} $$

  • $\begingroup$ Hi, excellent solution :) I just have one question: can you explain what happens when $y\,dx$ becomes $x\,dy$? Where does the $\frac{1}{2}$ come from? $\endgroup$ Jan 24, 2015 at 15:59
  • $\begingroup$ That is the $[\vphantom{\int}xy]_0^{1/\sqrt2}$ term from the integration by parts. $\endgroup$
    – robjohn
    Jan 24, 2015 at 18:53
  • $\begingroup$ Can you explain how you know "the curve extends from $m = 1$ to $m = \infty$? $\endgroup$ Jan 25, 2015 at 2:19
  • 1
    $\begingroup$ The point at the right end of the curve (at the maximum) is the point for $m=1$, the points to the left of that will be for $m\gt1$. $\endgroup$
    – robjohn
    Jan 25, 2015 at 3:12
  • $\begingroup$ Wow thank you so much for putting in all this effort to explain everything so clearly :) $\endgroup$ Jan 25, 2015 at 6:19

here in one way to solve this problem. the parametric equation of the graph is $$x = \dfrac{1}{\sqrt{m(1 + m^2)}}, y = \sqrt{\dfrac{m}{1+m^2}}$$ note that $y$ reaches a maximum at $m = 1.$

$y \ dx = \sqrt{\dfrac{m}{1+m^2}} \left( -\dfrac{1}{2}[m(1 + m^2)]^{-3/2}(1 + 3m^2)\right) \ dm = -\dfrac{1}{2} \dfrac{1 + 3m^2}{m(1 + m^2)^2} \ dm$

so the area is $$ \dfrac{1}{2} \int_1^\infty \dfrac{1 + 3m^2}{m(1 + m^2)^2} \ dm$$

  • $\begingroup$ Unfortunately, your integral comes to $\frac{3\pi+7}{48}$ which is not the same as mine. I will recheck my work. $\endgroup$
    – robjohn
    Jan 24, 2015 at 17:41
  • $\begingroup$ @robjohn, i have corrected the integrant. now i have a bigger problem; my integral does not converge at $m = 0.$ $\endgroup$
    – abel
    Jan 24, 2015 at 17:46
  • $\begingroup$ There are two solutions. One has $y\sim x^{1/3}$ and the other has $y\sim\frac1{2x}$. Since only the first actually satisfies the equations, I chose that one. $\endgroup$
    – robjohn
    Jan 24, 2015 at 17:56
  • $\begingroup$ @robjohn, when i change the limits of the integration i get the same result as you. $\endgroup$
    – abel
    Jan 24, 2015 at 19:29
  • $\begingroup$ (+1) Indeed. I missed the fact that your limits were not the same as the limits I used below. Nice that you caught that. $\endgroup$
    – robjohn
    Jan 25, 2015 at 0:22

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