Stuck on a question from Precalculus 2

[HRubss]

Homework Statement

"A small radio transmitter broadcasts in a 25 mile radius. If you drive along a straight line from a city 30 miles north of the transmitter to a second city 33 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?"

Homework Equations

x^2+y^2=r^2

The Attempt at a Solution

Attempted to the distance of the point that hit the circular radius but I don't know how since there is no points of the city

 

[fresh_42]

Have you tried to draw it?

 

[HRubss]

Yes I have.

 

[ehild]

Yes, you have to determine the intersections of the route of the car with the circle. How do you do it? What is the equation of the straight line?

 

[fresh_42]

The triangle (North,East,Origin) can be fully calculated. From there on you can regard other triangles.

 

[HRubss]

Yes, you have to determine the intersections of the route of the car with the circle. How do you do it? What is the equation of the straight line?

Y=mx+b? b=y-intercept so b=30, and m=slope so 30/35?

 

[fresh_42]

Y=mx+b? b=y-intercept so b=30, and m=slope so 30/35?

If at all then 33, not 35, but I think it's easier with the triangles.

 

[epenguin]

OK, OK, it will make no difference to this problem whether it's East or West.

 

[HRubss]

Okay guys, I got it. It took me a while to understand the concept because it combined both using the distance formula and the standard form with the quadratic formula!

 

[ehild]

It is correct. There was an other method, using the triangles in the figure below (as fresh_42 suggested) and applying the Law of Sines.

 

[epenguin]

It is correct?
You expect this to give you a quadratic equation and you have got one. But you expect this equation to have two positive roots, and the one you have given doesn't and can have only one.
Then what you have done with your square root seems wrong, I don't think there is a factor 25 there, and anyway calculation from the equation you give can't give two positive roots.

 

[ehild]

It is correct?

The results are correct. There is a sign mistake in the quadratic equation, -275, instead of +275, made during copying, I assume.

 

[ehild]

There is a very easy solution, even without trigonometry. The length of the whole path between A and B is $s=\sqrt{30^2+33^2}$. Using the area of the triangle AOB, 30*33=h*s. Knowing h, apply Pythagoras Theorem to get the length of the chord d: $d=2 \sqrt {r^2-h^2}$.

 

[epenguin]

<sub>Yes, using similar triangles and elementary circle geometry is the fastest way.
Absolutely whenever you can use similar triangles, use similar triangles!

I have been working on the different line, much delayed by mistakes. But it may be useful for the OP and others to realize for other cases, if he looks at his calculation it was not actually unnecessary to calculate all the x1, x2, y1, y2. It can be done just with equation coefficients and discriminants.

We are looking at the intersections between

y = x/1.1 + 30 , and x² + y² = 25.

Equating two expressions for y² gives us the quadratic

x²⁽(1 + 1/1.1²) + x (2×30/1.1) +30² - 25² = 0

I rewrite to make the leading coefficient 1

x² + (2×30×1.1/2.21) x + 5²×11×1.1²/2.21 = 0

I think this is the same as the OP got. (Numbers were factorised as much as possible in the hope that might simplify - but it doesn't.)

Now of greatest interest to us is not x1, x2 individually, but the difference or distance (x1 - x2) or more still its square (x1 - x2)².

Recall writing a quadratic

x² + b x + c = (x - x1)(x - x2) = x² - (x1 + x2)x + x1x2

The squared length we need

(x1 - x2)² = (x1 -x2)² - 4x1x2

= b² - 4c

= (2×30×1.1/2.21)² - 4×5²×11×1.1²/2.21 = 289.61 (corresponding to close to 17 miles)

For the square of the vertical length between the two points of interest we can similarly eliminate x terms between

x² + y² = 25, and x = 1.1y - 33

And proceed similarly to before. But actually we don't strictly need to workout the discriminant again, things are related.

In fact from the circle equation

(y1² - y2²) = (x1² - x2²)

And from that we can get for example

(y1 - y2 = (x1 - x2)(x1 + x2)/(y1 + y2)

The numerator terms we already have, and we can get the denominator from the y coefficient of the quadratic in y not sayy Josephet written but this time is
-(2×1.1×3372.21)</sub>

 

[vela]

Recall writing a quadratic

x² + b x + c = (x - x₁)(x - x₂) = x² - (x₁ + x₂)x + x₁x₂

The squared length we need

(x₁ - x₂)² = (x₁ -x₂)² - 4x₁x₂

= b² - 4c

Note that you can get the same result from the quadratic equation:
$$\Delta x = \frac{-b+\sqrt{b^2-4c}}2 - \frac{-b-\sqrt{b^2-4c}}2 = \sqrt{b^2-4c}$$

For the square of the vertical length between the two points of interest we can similarly eliminate x terms between

x² + y² = 25, and x = 1.1y - 33

A simpler way to find $\Delta y$ is to simply note that $\Delta y = m\Delta x$ where $m = -\frac{30}{33}$ since the two intersection points both lie on the line.

 

[epenguin]

The first is what the OP has done, wittingly or unwittingly, and is what I meant by "it was not actually unnecessary to calculate all the x1, x2, y1, y2. It can be done just with equation coefficients and discriminants." That is, this is equivalent to not using the full solution of the quadratic.

The second - doh! I felt this had to be simpler, but first did the quadratic treatment for y, was seeing some similar numbers to before with x under square root, and then wondered how to improve it, but missed this 'obvious'. Due to fatigue in getting anything right cause mistakes.

Just goes to show 'when you have the right answer it's not necessarily finished'.