# Word Problems,The illumination at a point is inversely proportional to the square

• November 2nd 2006, 09:03 PM
killasnake
Word Problems,The illumination at a point is inversely proportional to the square
I have two word problems that I do not know what to do. can some please explain how to I solve this.

Quote:

The illumination at a point is inversely proportional to the square of the distance of the point from the light source and directly proportional to the intensity of the light source. If two light sources are s feet apart and their intensities are I and J respectively, at what point between them will the sum of their illuminations be a minimum? Instruction: Give your answer in terms of s, I and J .

Solution:

Let x be the distance from I at which the sum of the illuminations be minimum. Then

x= ______
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Quote:

Find the equation of the line that is tangent to the ellipse http://img.photobucket.com/albums/v6...25750img13.gif in the first quadrant and forms with the coordinate axes the triangle with smallest possible area ( a and b are positive constants.)

The equation of the required line is:

____x + ____y + ______ = 0
• November 3rd 2006, 03:00 AM
earboth
2nd problem only
Quote:

Originally Posted by killasnake
I have two word problems that I do not know what to do. can some please explain how to I solve this...

Hello, killasnake,

as you may know an ellipse is constructed by mapping the circle
x² + y² = a²
perpendicularly to the x-axis with the ratio $r=\frac{b}{a}$.

So first I was looking for the triangle which is build by the coordinate axes and a tangent to the circle with the smallest area. It is the right isoscele triangle which area is a².

When you map the circle to the ellipse you are mapping this triangle too. The base of the triangle is constant, only the height of the triangle changes with the same ratio as all points of the coordinate plane.

Therefore the original smallest triangle has the area $A_{at circle} = a^2$, thus
the area of the smallest triangle at the ellipse is $A_{at ellipse}=\frac{b}{a} \cdot a^2 = b \cdot a$

you asked for the equation of the tangent:

$y=-\frac{b}{a} \cdot x + b \cdot \sqrt{2}$

EB

PS.: The preview doesn't work properly so I send you this reply without checking my text. There are of course some ugly mistakes but I hope you can use this post as a starter.
• November 3rd 2006, 03:51 AM
Soroban
Hello, killasnake!

I think I have the first one . . .

Quote:

The illumination at a point is inversely proportional to the square of the distance of the
point from the light source and directly proportional to the intensity of the light source.
If two light sources are s feet apart and their intensities are $I$ and $J$ respectively,
at what point between them will the sum of their illuminations be a minimum?
Instruction: Give your answer in terms of $\text{s, I, and J.}$

Let $x$ be the distance from $I$ at which the sum of the illuminations be minimum.

Then: . $x\:=\: \_\_\_$

We are told that: . $L \:=\:\frac{ki}{d^2}$ ... where $L$ = illumination, $i$ = intensity, $d$ = distance
Code:

                  P     I *-----------*---------* J       :    x    :  s-x  :

$P$ is the point we are seeking.

Let distance $PI = x$, then distance $PJ = s-x$

The illimunation from $I$ is: . $L_I\:=\:\frac{kI}{x^2}$
The illumination from $J$ is: . $L_J\:=\:\frac{kJ}{(s-x)^2}$
Their sum is: . $S \;= \;\frac{kI}{x^2} + \frac{kJ}{(s-x)^2}$

And we will minimize: . $S \;= \;kIx^{-2} + kJ(s-x)^{-2}$

Differentiate and equate to zero:

. . $S' \;=\;-2kIx^{-3} + 2kJ(s-x)^{-3} \;= \;0$

This simplifies to: . $J(s - x)^{-3} \;= \;Ix^{-3}\quad\Rightarrow\quad Jx^3\;=\;I(s-x)^3$

Take cube roots: . $x\sqrt[3]{J} \;= \;(s-x)\sqrt[3]{I}$

Solve for $x:\;\;x\sqrt[3]{J}\;=\;s\sqrt[3]{I} - x\sqrt[3]{I}$

. . . . . $x\sqrt[3]{I} + x\sqrt[3]{J} \;= \;s\sqrt[3]{I}$

. . . . $x\left(\sqrt[3]{I} + \sqrt[3]{J}\right) \;= \;s\sqrt[3]{I}$

. . Therefore: . . $\boxed{x \;=\;\frac{s\sqrt[3]{I}}{\sqrt[3]{I} + \sqrt[3]{J}}}$

• November 3rd 2006, 03:53 AM
Neutrino
Problem 1
Let f(x) be the sum of the illuminations.

f(x) = kI/x^2 + kJ/(s-x)^2

f '(x) = - 2kI/x^3 + 2kJ/(s-x)^3

f '(x) = 0 gives

I/x^3 = J/(s-x)^3

((s-x)/x)^3 = J/I

(s-x)/x = (J/I)^(1/3)

s-x = x(J/I)^(1/3)

x((J/I)^(1/3) + 1) = s

x = s/((J/I)^(1/3) + 1)

f(x) ---> infinity when x ---> 0 or s

Therefore f(x) must have a minimum for x.
• November 3rd 2006, 09:14 PM
killasnake
Thanks for the explanation everyone!