A small radio transmitter broadcasts in a 27 mile radius. If you drive along a straight line from a city 35 miles north of the transmitter to a second city 31 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

evanborot Oct 22, 2019

#1**+2 **

Let the transmitter be located at (0,0)

The equation for its effective coverage area will be a circle centered at (0,0) with a radius of 27

So....the equation is

x^2 + y^2 = 27^2

x^2 + y^2 = 729

Let the point 35 miles north of the transmitter be (0, 35)

Let the point 31 miles east of the trasmitter be (31,0)

The line connecting these (the line of travel) will have a slope of [ 35 - 0] / [ 0 - 31] = -35/31

The equation of this line will be

y = (-35/31)x + 35

We could solve this algebraically, but a graph seems easier to deal with

See here : https://www.desmos.com/calculator/ksd2p5jdwl

The signal will be received betwen the points (8.221 , 25.718) and (26.523, 5.055)

The distance between these points is

√ [ ( 26.523 - 8.221)^2 + ( 25.718 - 5.055)^2 ] ≈ 27.6 miles (1)

The distance between (0, 35) and (31,0) = √ [ ( 31)^2 + ( 35)^2 ] ≈ 46.8 miles (2)

So....you will receive the signal about (1)/ (2) = 27.6 / 46.8 ≈ 59% of the drive

CPhill Oct 22, 2019

#4**+2 **

We need to find the x intersection points of

x^2 + y^2 = 729 and

y = (-35/31)x + 35

So....subbing the second equation into the first for y we have that

x^2 + [ (-35/31)x + 35]^2 = 729

x^2 + 1225x^2/961 -2450x/31 + 1225 = 729

[961 + 1225]x^2 / 961 - 2450x/31 + 496 = 0 multiply through by 961

2186 x^2 - 75950x + 476656 = 0

Using the quadratic formula we have that

75950 ±√ [ (75950^2 - 4(2186)(476656) ]

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2 * 2186

Evaluating this we get that x = 8.221 and x = 26.523

Putiing the first value into the linear equation we have that

y = (-35/31)(8.221) + 35 ≈ 25.718

Putting the second vallue into the linear eqation we have that

y = (-35/31)(26.523) + 35 ≈ 5.055

So....the intersection points of the circle and line of travel are ( 8.221, 25.718) and (25.523, 5.055)

And I just used the distance formula twice to solve the rest of the problem

CPhill Oct 22, 2019