# The Radio Transmitter problem. help?

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• Dec 14th 2010, 04:09 PM
henryhighstudent
The Radio Transmitter problem. help?
The Radio Transmitter

In hilly areas, radio reception may be poor. Consider a situation where an FM transmitter is located at the point (−1,1) behind a hill modeled by the graph of y=x−x^2, and a radio receiver is on the opposite side of the hill. Assume that the x-axis represents ground level at the base of the hill.

1. What is the closest the radio can be to the hill so that reception is unobstructed?
2. Write the closest position x of the radio as a function of h if the transmitter is located at (−1,h).

i worked on this for about 2 hours and i came up with no solution...if you have any method of solving at all please feel free to share it with me :/
• Dec 14th 2010, 04:37 PM
skeeter
Quote:

Originally Posted by henryhighstudent
The Radio Transmitter

In hilly areas, radio reception may be poor. Consider a situation where an FM transmitter is located at the point (−1,1) behind a hill modeled by the graph of y=x−x^2, and a radio receiver is on the opposite side of the hill. Assume that the x-axis represents ground level at the base of the hill.

1. What is the closest the radio can be to the hill so that reception is unobstructed?
2. Write the closest position x of the radio as a function of h if the transmitter is located at (−1,h).

i worked on this for about 2 hours and i came up with no solution...if you have any method of solving at all please feel free to share it with me :/

Find the equation of a line tangent to the curve $\displaystyle y = x - x^2$ that passes through the point (-1,1), then find the line's x-intercept.

let $\displaystyle (a, a-a^2)$ be the point of tangency on the curve.

slope, $\displaystyle m = 1-2a$ ... point $\displaystyle (-1,1)$

$\displaystyle y - 1 = (1-2a)(x + 1)$

$\displaystyle y = (1-2a)(x+1) + 1$

at the point of tangency, the line equals the curve ...

$\displaystyle (1-2a)(a+1) + 1 = a - a^2$

solve for $\displaystyle a$ and you have the tangent line equation.

I'll tell you now that the value of $\displaystyle a$ is irrational.

Same drill with the second part, just generalize with the point (-1,h)
• Dec 14th 2010, 05:57 PM
Soroban
Hello, henryhighstudent!

Quote:

$\displaystyle \text{In hilly areas, radio reception may be poor. \: Consider a situation}$
$\displaystyle \text{where an FM transmitter is located at the point }(-1,1)$
$\displaystyle \text{behind a hill modeled by the graph of: }\,y \:=\:x-x^2$
$\displaystyle \text{and a radio is on the opposite side of the hill.}$
$\displaystyle \text{Assume that the }x\text{-axis represents ground level at the base of the hill.}$

$\displaystyle \text{(a) What is the closest the radio can be to the hill}$
. . . . $\displaystyle \text}so that reception is unobstructed?}$
Code:

      T      |       o      |   (-1,1)  o  |               | o               |      *               |  *    * o               | *        *    o               |*          *        o               |                          o     - - - - - * - - - - - - * - - - - - - - - o - -               O            1                R

The transmitter is at: .$\displaystyle T(-1,\,1)$
The radio is at $\displaystyle \,R.$

When $\displaystyle \,R$ is at the closest, line $\displaystyle \,T\!R$ is tangent to the parabola
. . but not at its vertex.

Let line $\displaystyle T\!R$ have slope $\displaystyle \,m.$

The equation of $\displaystyle T\!R$ is: .$\displaystyle y - 1 \:=\:m(x+1) \quad\Rightarrow\quad y \:=\:mx + (m+1)$ .[1]

Intersection of $\displaystyle T\!R$ and the parabola: .$\displaystyle mx + m + 1 \:=\:x - x^2$

$\displaystyle x^2 + (m\!-\!1)x + (m\!+\!1) \:=\:0 \quad\Rightarrow\quad x \;=\;\dfrac{-(m\!-\!1) \pm\sqrt{(m\!-\!1)^2 - 4(m\!+\!1)}}{2}$

If $\displaystyle T\!R$ is tangent to the parabola, there is one intersection point;
. . the discriminant is zero.

Hence, we have: .$\displaystyle (m-1)^2 - 4(m_1) \:=\:0 \quad\Rightarrow\quad m^2 - 6m - 3 \:=\:0$

. . and we have: .$\displaystyle m \:=\:\dfrac{6\pm\sqrt{36+12}}{2} \;=\;3 \pm 2\sqrt{3}$

We see that slope must be negative: .$\displaystyle m \:=\:3 - 2\sqrt{3}$

Substitute into [1].

The equation of $\displaystyle T\!R$ is: .$\displaystyle y \;=\;(3-2\sqrt{3})x + 4-2\sqrt{3}$

We want the $\displaystyle \,x$-intercept of $\displaystyle T\!R.$
. . $\displaystyle (3-2\sqrt{3})x + 4 - 2\sqrt{3} \:=\:0 \quad\Rightarrow\quad x \;=\;\dfrac{-4 + 2\sqrt{3}}{3-2\sqrt{3}}$

which rationalizes to: .$\displaystyle x \;=\;\dfrac{2\sqrt{3}}{3}$

HA! . . . Just noticed the wording of the question.

What is the closest the radio can be to the hill . . . ?

$\displaystyle \,x$ is the distance from $\displaystyle \,R$ to $\displaystyle \,O.$
The distance from $\displaystyle \,R$ to the hill is: .$\displaystyle \dfrac{2\sqrt{3}}{3} - 1$ units.