The Radio Transmitter problem. help?

**henryhighstudent** · December 14th 2010, 04:09 PM

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.

What is the closest the radio can be to the hill so that reception is unobstructed?
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 :/

**skeeter** · December 14th 2010, 04:37 PM

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.

What is the closest the radio can be to the hill so that reception is unobstructed?
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 $y = x - x^2$ that passes through the point (-1,1), then find the line's x-intercept.

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

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

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

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

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

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

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

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

Same drill with the second part, just generalize with the point (-1,h)

**Soroban** · December 14th 2010, 05:57 PM

Hello, henryhighstudent!

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

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

Code:

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

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

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

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

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

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

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

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

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

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

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

Substitute into [1].

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

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

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

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

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

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

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