Brain Teaser - Area enclosed by a parabola and a line
09-18-2015, 08:20 PM
Post: #29
 Thomas Klemm Senior Member Posts: 1,685 Joined: Dec 2013
RE: Brain Teaser - Area enclosed by a parabola and a line
We start with a slightly different definition of $$h(x)=g(x)-f(x)$$ but now consider it as a function of two variables $$u$$ and $$x$$:
$h(u,x)=\frac{u-x}{2u}+u^2-x^2$
From this we define the indefinite integral:
$H(u,x)=\int h(u,x)\, dx$
For each $$u$$ we find the lower limit $$a(u)$$ and the upper limit $$b(u)$$ as roots of $$h(u,x)=0$$:
\begin{align*} a(u) &= -(u+\frac{1}{2u})\\ b(u) &= u \end{align*}
We evaluate the anti-derivative at these limits, calculate the derivative and set it equal to 0:
$\frac{\mathrm{d} }{\mathrm{d} u}(H(u,b)-H(u,a))=0$
To calculate the total differential we use partial differentials:
$\frac{\mathrm{d} }{\mathrm{d} u}H(u,x)=\frac{\partial H}{\partial u}+\frac{\partial H}{\partial x}\frac{\partial x}{\partial u}$
From the definition of $$H(u,x)$$ we find:
$\frac{\partial H(u,x)}{\partial x}=h(u,x)$
However this function is 0 when evaluated at the lower or the upper limit. Thus this term vanishes.
This leaves only the first term:
$\frac{\partial}{\partial u}H(u,x)=\frac{\partial}{\partial u}\int h(u,x)\,dx$
But we can switch the order of integration and differentiation:
\begin{align*} \frac{\partial}{\partial u}\int h(u,x)\,dx &=\int \frac{\partial}{\partial u}h(u,x)\,dx \\ &=\int\frac{\partial}{\partial u}(\frac{u-x}{2u}+u^2-x^2)\,dx \\ &=\int\frac{\partial}{\partial u}(\frac{1}{2}-\frac{x}{2u}+u^2-x^2)\,dx \\ &=\int\frac{x}{2u^2}+2u\,dx \\ &=\frac{x^2}{4u^2}+2ux \end{align*}
We evaluate this expression at the upper and lower limit and find the difference. But we can do that for both terms separately:
\begin{align*}
\frac{b^2-a^2}{4u^2} + 2u(b-a) &= (b-a)\left(\frac{b+a}{4u^2}+2u\right)\\
&= \left(2u+\frac{1}{2u}\right)\left(2u-\frac{1}{8u^3}\right) \\
&= \frac{4u^2+1}{2u} \cdot \frac{16u^4-1}{8u^3} \\
&= \frac{(4u^2+1)(16u^4-1)}{16u^4} \\
&= \frac{(4u^2+1)(4u^2+1)(4u^2-1)}{16u^4} \\
&= \frac{(4u^2+1)^2(4u^2-1)}{16u^4} \\
&= \frac{(4u^2+1)^2(2u+1)(2u-1)}{16u^4} = 0 \\
\end{align*}
We conclude that $$2u-1=0$$ which leads to the solution $$u=\tfrac{1}{2}$$.

Kind regards
Thomas
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 Messages In This Thread Brain Teaser - Area enclosed by a parabola and a line - CR Haeger - 09-11-2015, 04:57 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Gerson W. Barbosa - 09-13-2015, 12:17 AM RE: Brain Teaser - Area enclosed by a parabola and a line - CR Haeger - 09-13-2015, 01:19 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Claudio L. - 09-13-2015, 01:35 AM RE: Brain Teaser - Area enclosed by a parabola and a line - Gerson W. Barbosa - 09-13-2015, 02:40 AM RE: Brain Teaser - Area enclosed by a parabola and a line - Thomas Klemm - 09-13-2015, 04:23 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Gerson W. Barbosa - 09-15-2015, 03:29 AM RE: Brain Teaser - Area enclosed by a parabola and a line - fhub - 09-15-2015, 11:56 AM RE: Brain Teaser - Area enclosed by a parabola and a line - Gerson W. Barbosa - 09-15-2015, 01:30 PM RE: Brain Teaser - Area enclosed by a parabola and a line - fhub - 09-15-2015, 02:59 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Gerson W. Barbosa - 09-15-2015, 04:18 PM RE: Brain Teaser - Area enclosed by a parabola and a line - fhub - 09-15-2015, 05:00 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Gerson W. Barbosa - 09-15-2015, 05:08 PM RE: Brain Teaser - Area enclosed by a parabola and a line - fhub - 09-15-2015, 09:42 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Bunuel66 - 09-13-2015, 10:06 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Gerson W. Barbosa - 09-16-2015, 02:19 AM RE: Brain Teaser - Area enclosed by a parabola and a line - CR Haeger - 09-14-2015, 03:57 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Bunuel66 - 09-14-2015, 05:08 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Bunuel66 - 09-14-2015, 05:51 PM RE: Brain Teaser - Area enclosed by a parabola and a line - eried - 09-14-2015, 08:32 PM RE: Brain Teaser - Area enclosed by a parabola and a line - CR Haeger - 09-14-2015, 10:06 PM RE: Brain Teaser - Area enclosed by a parabola and a line - CR Haeger - 09-15-2015, 02:44 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Gerson W. Barbosa - 09-16-2015, 02:06 AM RE: Brain Teaser - Area enclosed by a parabola and a line - Thomas Klemm - 09-16-2015, 05:12 AM RE: Brain Teaser - Area enclosed by a parabola and a line - Gerson W. Barbosa - 09-16-2015, 03:09 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Thomas Klemm - 09-16-2015, 03:55 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Gerson W. Barbosa - 09-16-2015, 04:14 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Gerson W. Barbosa - 09-18-2015, 05:51 PM RE: Brain Teaser - Area enclosed by a parabola and a line - fhub - 09-21-2015, 05:53 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Gerson W. Barbosa - 09-21-2015, 06:40 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Gerson W. Barbosa - 09-21-2015, 09:48 PM RE: Brain Teaser - Area enclosed by a parabola and a line - Thomas Klemm - 09-18-2015 08:20 PM

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