Algebra, Hyperbola.

[amathproblemthatneedsolve]

1. The graph below shows: • the two existing curved roads are represented by a rectangular hyperbola with the equation:

y =4/x

One curved road passes through the point (2, 2) and the other passes through (– 2, – 2). • a new straight road, represented by a line passing through the points (0,4) and (– 16, 0).



a. Form a pair of equations and use them to find the coordinates of the points of intersection of the straight road with the two curved roads.

b. Interpret and present your solutions in the context of the situation.

2. Investigate how any straight roads passing through (0, 4) can be laid out so that there may be 0, 1 or 2 points of intersection with the two curved roads.

a. Show and describe each of the possible situations of the intersection, in context. You should include all essential details.

b. The gradient of the line that represents any straight road passing through (0, 4) can be denoted by m. Investigate the range of values of m for the situations where a straight road intersects the two curved roads at 0, 1 or 2 points.


Thanks. Once I get an idea of what to do I'll show all my OWN working.

 

[tkhunny]

Have you considered the equation of the line?

 

[amathproblemthatneedsolve]

@tkhunny What do you mean?

 

[Steven G]

To find where two curves intersect you simply set the equations equal to one another. You had to learn that somewhere. Well you have one of the equations, namely y=4/x. You need the other equation which is a line. tkhunny suggested that you find the equation of the line. Do you know how to do that?

 

[amathproblemthatneedsolve]

@Jomo Oh ok, The equation of the line that passes through the points
(0,4) and (-16,0)

is

y=1/4x+4? or y = 0.25x + 4


doing y=4/x
y=0.25x+4 I get the two curves to intersect at (-8-4 √5,2- √5) (-8+4 √5,2+ √5) This doesn't seem right to me?

 

[tkhunny]

This doesn't seem right to me?

1) Why is that important? Is it correct or isn't it? Did you make mistakes on purpose?

 

[Dr.Peterson]

@Jomo Oh ok, The equation of the line that passes through the points
(0,4) and (-16,0)

is

y=1/4x+4? or y = 0.25x + 4


doing y=4/x
y=0.25x+4 I get the two curves to intersect at (-8-4 √5,2- √5) (-8+4 √5,2+ √5) This doesn't seem right to me?

You didn't show all your work as you said you would (which we'd need if your answer were wrong); but if you evaluate these expressions, you'll find that they agree with the graph.

So, what doesn't seem right? Did you check the numbers in the equations and find they didn't satisfy them?

 

[amathproblemthatneedsolve]

What I meant was I didn't expect it to be a non-whole number result like that. So I've done (a)

 

[tkhunny]

What I meant was I didn't expect it to be a non-whole number result like that. So I've done (a)

No one can help you with (b). Interpret away!

 

[Dr.Peterson]

The remaining parts all involve creativity on your part (interpret, investigate), so if you want help, you'll have to show your ideas and let us know where we can help. It's definitely up to you to do the main thinking -- the point of the exercise is to give you experience doing this sort of open-ended thinking, so if we "closed the end" for you, we are doing you a disservice.

 

[Otis]

… b. Interpret and present your solutions …

I'm not sure what they mean above.

Interpret your solutions and then present them?

Interpret what they wrote in part (a) and then present your solutions?

I think "in context of the situation" just means they want answers containing phrases like "curved road" and "straight road", instead of "hyperbola" and "line".

You're self-studying, so the important parts are knowing how to set up equations and find the values you need to answer. That is, for you, I think the English in your answers is secondary.

?

 

[JeffM]

I'm not sure what they mean above.

Interpret your solutions and then present them?

Interpret what they wrote in part (a) and then present your solutions?

I think "in context of the situation" just means they want answers containing phrases like "curved road" and "straight road", instead of "hyperbola" and "line".

You're self-studying, so the important parts are knowing how to set up equations and find the values you need to answer. That is, for you, I think the English in your answers is secondary.

?

In a very friendly and respectful way, I am going to disagree somewhat with Otis (or perhaps just with how he said what he meant.)

If you are a pure mathematician, the correctness of your reasoning and mathematical mechanics is what your colleagues will care about. They will make their own interpretations of significance, meaning, etc. Similarly, if you are working on a practical problem and have no need to report your results to others, all that matters is the correctness of your reasoning and mathematical mechanics.

If, however, you are using math to determine practical answers in any kind of cooperative venture and must justify your answers, explaining them is as every bit as important an issue as getting the correct answer. Answers that will not be used are useless. So I would formulate things differently.

Setting things up correctly and working it out correctly is always primary, but being able to translate those answers back into the natural language that others understand is just as important in many (but not all) cases.

I ended up having a very profitable and fun career by being able to translate problems into mathematical form and then explaining the conclusions in comprehensible English. Getting things right and then persuading others that they are right is a valuable skill.

(I am not sure that Otis disagrees. I suspect that he was stressing the universal primacy of reasoning rather than dismissing explanation as unimportant.)

 

[amathproblemthatneedsolve]

Except for the case of a vertical line, our arbitrary line passing through the point (0,4) can be represented as y=mx+4 where m is an arbitrary constant representing the slope.

Now, supposing our line intersects with the rectangular hyperbola, that would occur specifically when given the same x value both equations give the same y value. We can find these occasions where the same x value gives the same result by setting the equations equal to one another, i.e as

mx+4=4/x

Supposing that x is nonzero, we can go ahead and multiply both sides by x and we arrive at mx^2+4x=4 and rearranging we arrive at
mx^2+4x=4

Now... by assuming that m≠0 we can use our handy quadratic formula to find what values of x this would be true. It just so happens that one of our values here is a variable, but that is okay. We get the values of x where this equality occurs to be:

Now, from here we can see that when m is equal to −1 the expression in the square root vanishes and so we are left with just the one x value of 2 where the intersection occurs corresponding to the point (2,2)

On the other hand, if m is greater than −1 (and is not zero) then the expression in the square root is positive and so we get two values, corresponding to the points of intersection

When m is less than −1 the expression in the square root would have been negative and so the result of the square root would have been complex and so there are no real values of x for which the corresponding y values would coincide, i.e. the (real) curves don't intersect.

Finally, when m is equal to zero, we weren't allowed to use the quadratic formula in the first place as it wasn't a quadratic, but rather a line. We find in that situation that the one and only point of intersection occurs when 4x−4=0 and that happens at (1,4).

There is one final case to consider after all of this, and that is the case of the line x=0 which corresponds to an "infinite" slope. This too is a line which will not intersect our hyperbola.

Summarizing, there are two lines which have one intersection which correspond to when the slope of the line is 0 at the point (1,4) or when the slope of the line is −1 at the point (2,2).

There are infinitely many lines which have two intersections which occur when the slope of the line is a number greater than −1 different than 0 and the intersections occur at

Except for the case of a vertical line, our arbitrary line passing through the point (0,4)(0,4) can be represented as y=mx+4y=mx+4 where mm is an arbitrary constant representing the slope.

Now, supposing our line intersects with the rectangular hyperbola, that would occur specifically when given the same xx value both equations give the same yy value. We can find these occasions where the same xx value gives the same result by setting the equations equal to one another, i.e as



mx+4=4xmx+4=4x


Supposing that xx is nonzero, we can go ahead and multiply both sides by xx and we arrive at mx2+4x=4mx2+4x=4 and rearranging we arrive at



mx2+4x−4=0mx2+4x−4=0


Now... by assuming that m≠0m≠0 we can use our handy quadratic formula for to find what values of xxthis would be true. It just so happens that one of our values here is a variable, but that is okay. We get the values of xx where this equality occurs to be:



−4±16+16m−−−−−−−−√2m−4±16+16m2m


Now, from here we can see that when mm is equal to −1−1 the expression in the square root vanishes and so we are left with just the one xx value of 22 where the intersection occurs corresponding to the point (2,2)(2,2)

On the other hand, if mm is greater than −1−1 (and is not zero) then the expression in the square root is positive and so we get two values, corresponding to the points of intersection (−4+16+16m√2m,2m−4+16+16m√)(−4+16+16m2m,2m−4+16+16m) and (−4−16+16m√2m,2m−4−16+16m√)(−4−16+16m2m,2m−4−16+16m) respectively.

When mm is less than −1−1 the expression in the square root would have been negative and so the result of the square root would have been complex and so there are no real values of xx for which the corresponding yy values would coincide, i.e. the (real) curves don't intersect.

Finally, when mm is equal to zero, we weren't allowed to use the quadratic formula in the first place as it wasn't a quadratic, but rather a line. We find in that situation that the one and only point of intersection occurs when 4x−4=04x−4=0 and that happens at (1,4)(1,4).

There is one final case to consider after all of this, and that is the case of the line x=0x=0, which corresponds to an "infinite" slope. This too is a line which will not intersect our hyperbola.

Summarizing, there are two lines which have one intersection which correspond to when the slope of the line is 00 at the point (1,4)(1,4) or when the slope of the line is −1−1 at the point (2,2)(2,2).

There are infinitely many lines which have two intersections which occur when the slope of the line is a number greater than −1−1 different than 00 and the intersections occur at (−4+16+16m√2m,2m−4+16+16m√)(−4+16+16m2m,2m−4+16+16m)and (−4−16+16m√2m,2m−4−16+16m√)(−4−16+16m2m,2m−4−16+16m)

There are infinitely many lines which have no intersections which occur when the slope of the line is a number less than −1 or when the line is x=0

 

[Otis]

… if you … have no need to report your results to others, all that matters is the correctness of your reasoning and mathematical mechanics … I suspect that [Otis] was stressing the universal primacy of reasoning rather than dismissing explanation as unimportant.

Yes, that's pretty much it. For someone studying on their own, I think once they understand the meaning of a word problem and its solution, variations in how they might choose to express their answer on paper isn't as important as the comprehension itself. I could have added that a person's technical English improves naturally with continued practice, as they adopt phrasing they see repeated from source to source. (Well, at least mine does).

?

 

[tkhunny]

I am delighted you thought of the uniqueness behavior associated with the vertical line. Good work.

 

[amathproblemthatneedsolve]

Any other ways of solving this problem?

 

[JeffM]

Basically, your way is the natural way to prove things though I think you made it harder than necessary. But you can get a qualitative answer by simply thinking.

Consider a vertical line through (0, 4). It obviously does not intersect 4/x because that function does not exist at x = 0.

Now consider a line with a positive slope through (0, 4). That line will eventually extend into the third quadrant and intersect f(x) in that quadrant. It will also intersect f(x) in the first quadrant. So m > 0 means a double intersection, one in the first quadrant and one in the third quadrant.

Now consider a line with zero slope through (0, 4), namely the equation y = 4. It lies only above the x-axis so it will never intersect with f(x) in the third quadrant. And it will only intersect with f(x) at x = 1. If 0 < x < 1, then 4/x > 4. If x < 0, then 4/x < 0 < 4.

It depends on your imagination whether you realize this by intuition, but a line through (0, 4) with negative slope can intersect f(x) two, one, or zero times, and any intersection will occur in the first quadrant. Basically such a line lies mainly in the second and fourth quadrants, where f(x) is not present. If the line in the first quadrant is everywhere below f(x), meaning that the line is very steep, it cannot intersect. If the slope is just right, it will be tangent to f(x) somewhere in the first quadrant. And if the slope is relatively mild, it will intersect twice, both times in the first quadrant.

Now you may need to prove all this intuition, you may want to confirm it, or you may need more quantitative specification.

The cases of x = 0 and y = 0 are relatively obvious. So let's do real work only on the cases where the line is y = 4 + mx.

[MATH]y = 4 + mx, \text { and } y = \dfrac{4}{x} \implies mx^2 + 4x - 4 = 0 \implies[/MATH]
[MATH]x = \dfrac{-\ 4 + \sqrt{16 + 16m}}{2m} \text { or } x = \dfrac{-\ 4 - \sqrt{16 + 16m}}{2m}.[/MATH]
[MATH]\therefore m > 0 \implies 16 + 16m > 16 > 0 \implies \sqrt{16 + 16m} > 4 > 0.[/MATH]
[MATH]\therefore -\ 4 + \sqrt{16 + 16m} > 0 \implies x > 0 \ \because \ m > 0.[/MATH]
[MATH]\therefore -\ 4 - \sqrt{16 + 16m} < 0 \implies x < 0 \ \because \ m > 0.[/MATH]
This is a proof that a positive slope entails two intersections, one in the first and one in the third quadrants. Moreover, it gives the x-coordinate, which lets you specify the y co-ordinates as well.

[MATH]m < - 1 \implies 16 + 16m < 0 \implies \text {no real intersection.}[/MATH]
[MATH]m = - 1 \implies 16 + 16m = 0 \implies \text {one real intersection at } x = -\ \dfrac{-\ 4}{2(-\ 1)} = 2 = y.[/MATH]
[MATH]\therefore -\ 1 < m < 0 \implies 0 < 16 + 16m < 16 > 0 \implies 0 < \sqrt{16 + 16m} < 4.[/MATH]
[MATH]\therefore -\ 4 + \sqrt{16 + 16m} < 0 \implies x > 0 \ \because \ m < 0.[/MATH]
[MATH]\therefore -\ 4 - \sqrt{16 + 16m} < 0 \implies x > 0 \ \because \ m < 0.[/MATH]
This is a proof that a slope less than - 1 entails no intersection; a slope of - 1 entails 1 intersection at (2, 2), and a negative slope > - 1 entails two intersections, both in the first quadrant.

 

[amathproblemthatneedsolve]

Just a question, Is the way I wrote this suitable to answer the questions:

b. Interpret and present your solutions in the context of the situation.

2. Investigate how any straight roads passing through (0, 4) can be laid out so that there may be 0, 1 or 2 points of intersection with the two curved roads.

a. Show and describe each of the possible situations of the intersection, in context. You should include all essential details.

b. The gradient of the line that represents any straight road passing through (0, 4) can be denoted by m. Investigate the range of values of m for the situations where a straight road intersects the two curved roads at 0, 1 or 2 points.


I'm pretty sure they are but just want to make sure they make sense. (if I wrote them IN the right order and is not confusing)

 

[JeffM]

Just a question, Is the way I wrote this suitable to answer the questions:

b. Interpret and present your solutions in the context of the situation.

2. Investigate how any straight roads passing through (0, 4) can be laid out so that there may be 0, 1 or 2 points of intersection with the two curved roads.

a. Show and describe each of the possible situations of the intersection, in context. You should include all essential details.

b. The gradient of the line that represents any straight road passing through (0, 4) can be denoted by m. Investigate the range of values of m for the situations where a straight road intersects the two curved roads at 0, 1 or 2 points.


I'm pretty sure they are but just want to make sure they make sense. (if I wrote them IN the right order and is not confusing)

I do not think your answer is complete as written. If I were grading it, I would give you partial credit.

The primary problem with what you wrote is that you do not indicate that there are two qualitatively different classes of double intersection. If the slope m is positive, the two roads represented by the different branches of the hyperbola will be joined by the road represented by the straight line (mathematically, there will be an intersection in the first quadrant and a second in the third quadrant). If m is not positive, the road represented by the straight line will not join the roads represented by the two branches of the parabola.

So you do not explain the results of your mathematical model in terms relevant to what is being modeled. And this is usually what people want. Furthermore, you do not explain that -1 < m < 0, the two intersections are with the curved road to the right, not the one to the left.

In short, you missed one mathematical conclusion, namely that all straight lines with positive slopes intersect both branches of the hyperbola but that straight lines through the given point with a non-positive slope will either never intersect the hyperbola at all or will intersect with only the branch in the first quadrant. And you did not explain what any of your mathematical conclusions meant in terms of the underlying engineering problem.

You did a quite nice job mathematically (although you missed the point about only positive slopes joining the two branches), but missing what the mathematics mean is a big oversight. I'd give half credit (but I am a tough grader).

 

[amathproblemthatneedsolve]

So to give what I write "a better grade" you'd include :

that there are two qualitatively different classes of a double intersection. If the slope m is positive, the two roads represented by the different branches of the hyperbola will be joined by the road represented by the straight line (mathematically, there will be an intersection in the first quadrant and a second in the third quadrant). If m is not positive, the road represented by the straight line will not join the roads represented by the two branches of the hyperbola.

that -1 < m < 0, the two intersections are with the curved road to the right, not the one to the left.

All straight lines with positive slopes intersect both branches of the hyperbola but that straight lines through the given point with a non-positive slope will either never intersect the hyperbola at all or will intersect with only the branch in the first quadrant. And you did not explain what any of your mathematical conclusions meant in terms of the underlying engineering problem.