stuck on angle between two lines

mikau · 2005-12-10 18:40:07

Find the angle between the slopes of the graphs y = x^3 and y = sqrt(x)

First we find the intersection(s) of the two functions.

x^3 = x^(1/2)

divide both sides by x^(1/2)

x^5/2 = 1

x = 1 took the 5/2th root of both sides.

If we find the derivatives of the original functions, we get:

3x^2 and -1/2sqrt(x)

Evaluated at the intersection (1), the slopes of the two lines are:

3 and -0.5

The angle between the slopes of two lines is given by:

tan A = (M1 - M2)/(1 - M1M2)

If we use 3 for M1 and -0.5 for M2 it evaluates to (5/2)/(5/2) which is 1.

So if the tangent of A equals 1, we know it must be either 45 or 225. 225 makes no sense so it must be 45. But in my book, the answer is 45 and 90. Ninety?! Where did that come from? If anything, I would have thought 45 or 135. (since two intersecting lines will form two pairs of suplementary angles) but 90 makes no sense at all.

What bugs me about the forumula tan A = (M1 - M2)/(1 - M1M2) is that it makes no specifications about which is M1 and which is M2. I had two slopes of 3 and -0.5. Depending on which I choose for M1 and M2 changes the sign of the outcome of the formula. I'll either get 1 or -1.

Dismaying.

Last edited by mikau (2005-12-10 18:40:59)

mikau · 2005-12-10 19:21:37

Ahh... now I see where 90 comes from. Its because if x^3 = sqrt x, x can equal 1 but it can also equal zero.

This brings up an interesting topic. When you have for instance a statement that:

x^2 = x^3

The manner in which you solve it gives different results. If you attempt to solve if with logarithms, or simply noting that if the bases are the same, the exponants are the same, then you will be left with the expression 2 = 3.

If you divide both sides by x^2 you get x = 1.

Or if x^2 = x^3

x^2 - x^3 = 0

x^2(x - 1) = 0

then x must equal either zero or 1.

Depending on how you solve it, you may get either an incorrect expression or only one of the solutions which may or may not be the correct one.

And again like I said, attemtpting to solve with logarithms:

x^m = x^n

m log x = n log x

You might be tempted to divide both sides by log x to find m = n but its not necesarily true. If x equals 1 the exponants do not have to be the same. So m may have a value of 3 and n a value of 2, and the rest of the problem, you would go about substituting 3 for 2 or 2 for 3. Disturbing.

Come to think of it, I think I can draw the following conclusion:

if x^n = x^m and m does not equal n then x = 1, will be a solution to the equation. x = -1 will be a solution if m and n are either odd numbers or a fraction of odd numbers, and x = 0 will be a solution if both m and n are not negative. (division by zero is undefined.)

Boy oh boy, someone needs to tell 1 and 0 to stop misbehaving. This sort of thing can be a major source of bugs in computer functions.

Last edited by mikau (2005-12-10 19:33:06)

irspow · 2005-12-11 14:34:31

I don't know how your book made the leap of faith to say 90° was definitely a solution. Maybe someone here can show such a proof.

I however see the intersection at zero as an indeterminant slope for the square root function. Just because the slope approaches infinity does not prove that the slope would indeed be completely vertical. I couldn't see how to apply L'Hopital's rule to this situation. Anything less than perfectly vertical would result in less than a 90° angle between the two lines.

As for the other intersection, well that is obvious. Simply take the derivative of both equations at x = 1. I would like to point out that the slope of √x at 1 is ½ and not -½.

The difference in angles from the origin is the angle between them:

arctan 3 - arctan ½ = 45°

but...

arctan 3 - arctan -½ = 98.13°

By the way, that is all you need to do for this problem. By definition m = tan dy.

mikau · 2005-12-11 14:50:56

Yeah I thought that saying the slope was vertical at zero looked kind of wierd. Oh well. I managed to answer the problem correctly. Or at least see where they were coming from.

mikau · 2005-12-11 14:53:47

Well the derivative of x^(1/2) is (1/2)/sqrt(x) set x = 0 and the slope is 0.5/0 which is a vertical line. The derivative is x^3 is 3x^2. When x equals zero it has a value of zero, or a slope of 0/1 which is a horizontal line. So I guess its true.

irspow · 2005-12-11 15:23:14

Uh.....sorry.

y = √x

dy/dx = 1÷2√x

dy/dx = 1/0 at x = 0

Either way it approaches ∞ and that is not a slope it is a concept!

Again, approaching infinity is not vertical, close but no cigar.

I would ask your teacher for such a proof if nobody here wants to prove it.

irspow · 2005-12-11 15:50:24

Besides, even if it were a vertical line how would you know if it were positive or negative in slope? Would the line be 90° above the x axis or -90° below it? These are questions that the best math minds are still trying to grasp. Perhaps you should contact the publisher of the book that you are using to prove that he/she is correct.

mikau · 2005-12-12 06:35:20

Are you trying to say that a line with a slope of x/0 is not vertical? Think of the rise over the run. It moves up x units every time it moves over zero units. Then its going straight up. If you prefer to avoid division by zero completely, you could say as the denominator approaches zero, the slope aproaches that of a vertical line, as a limit. If we use 1/infinity as the denominator, then if we call it a vertical line, our error will be infinitly small.

I don't know, I just read that the slope of a line is vertical when it is undefined (has a denominator of zero)

If you work backwards and try to find the slope of a vertical line, it will always result in division by zero.

Besides, what is the tangent of 90 degree's? Undefined. 1/0 on a unit circle.

irspow · 2005-12-13 15:18:38

I've been away. I am not disputing that it is generally accepted that the line is vertical when the slope is undefined, but you can not prove it. Say that a wall is has a slope of 3meters/1micrometer. Well, unless you have a very sensitive measuring device you will say that the slope is undefined and that the wall is perfectly vertical. But if the top of the wall were smooth a ball bearing would roll right off. Would you then say that gravity were wrong?
There are a lot of discussions here about infinity and how it applies to mathematics. You can say that an undefined slope approaches vertical but you can not say that it is absolutely vertical. And in this case it cannot be proven that this angle is 90°. (or -90°)

Math Is Fun Forum

#1 2005-12-10 18:40:07

stuck on angle between two lines

#2 2005-12-10 19:21:37

Re: stuck on angle between two lines

#3 2005-12-11 14:34:31

Re: stuck on angle between two lines

#4 2005-12-11 14:50:56

Re: stuck on angle between two lines

#5 2005-12-11 14:53:47

Re: stuck on angle between two lines

#6 2005-12-11 15:23:14

Re: stuck on angle between two lines

#7 2005-12-11 15:50:24

Re: stuck on angle between two lines

#8 2005-12-12 06:35:20

Re: stuck on angle between two lines

#9 2005-12-13 15:18:38

Re: stuck on angle between two lines

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