Basic graph - this ought to be simple

[Guest]

I think this must be very simple, but I just cannot see it.

This is a graph of [tex]y = x[/tex], and [tex]y = x + 50sin(x)[/tex]. The ‘50’ is just to scale the sine function to make the curve more obvious.



(If the picture doesn't appear, this is the URL: https://www.dropbox.com/s/gxccmsfrpa7j3j3/GraphQuestion.png

Now, relating the two lines together, delta y (in terms of x) is obviously just 50sin(x), and as you’d expect reaches a maximum at 90 degrees. But here is the question, in two parts:

1/ What is the expression for delta x (in terms of x)?

2/ Delta x peaks at 140 degrees (50 degrees past 90 degrees), and if I change the original formula to (say) [tex]y = x + 30sin(x)[/tex] then delta x peaks at 120 degrees (30 degrees past 90 degrees). So why is the scaling factor somehow being converted into the number of degrees past 90?

Any help with those questions (especially the first one) would be very much appreciated.

[SteveThackery]

Sorry, for some reason it logged me out while I was typing the question - it should have my name as the author.

Does anyone know why the image won't appear in-line?

Thanks,
Steve

[Guest]

1- in this example, you cannot express [tex]\Delta x[/tex] in terms of x

the idea of [tex]\Delta x[/tex] and [tex]\Delta y[/tex] is that when you express one of them with functions, you express the other with numbers

in this example [tex]\Delta x \approx 100 - 58 = 42[/tex]


2-your scaling is wrong. you cannot use degrees. you have to use radians

this means at [tex]180[/tex]

[tex]y_{1} = x = 180[/tex]

[tex]y_{2} = 180 + 50 \sin(180 \ {radians}) \approx 139.94[/tex]


-Jambo

[SteveThackery]

Thank you, Jambo, for your reply.

Some little clarifications:

in this example, you cannot express \displaystyle \Delta xΔx in terms of x

But surely I can? That horizontal distance is determined solely by the current value of x and the multiplier in front of sin(x). How about if I rephrased it and said,

[tex]y = z.sin(x), where z = 50[/tex]

Then I could ask how to express delta x in terms of x and z. Would that not be legitimate?

I accept I might be misusing the term "delta x", but it seems that if I know the two formulae, it must be possible to work out the horizontal distance between the two lines at any value of x on the [tex]y = x[/tex] line. I'm doing it in Excel using the XLOOKUP function, but I would have thought there must be a formula for it.

2-your scaling is wrong. you cannot use degrees. you have to use radians

Ah, interesting, but I don't understand. What is wrong with using degrees? All I've done is added the first half of a sine wave to the y values of the original [tex]y = x[/tex] graph. As I look at it, the graph will be exactly the same shape if the x axis goes from 0 to pi radians, rather than 0 to 180 degrees. I realise I've probably misunderstood what you were saying!

[Guest]

you are right, the horizontal distance is determined solely by the current value of x, but there is no mathematical expression that will let you solve for x

for example, if [tex]y_1 = \sin x_1[/tex] and [tex]y_2 = x_2[/tex]

yes here you can solve for [tex]x[/tex], and say

[tex]x_1 = \sin^{-1} y_1[/tex]
[tex]x_2 = y_2[/tex]

then

[tex]\Delta x = y_2 - sin^{-1} y_1[/tex]

this means there is a mathematical expression that let you solve for x

but with [tex]y = x + \sin x[/tex], there isn't


And for the degrees scaling

you cannot scale with degrees because degrees are not numbers

Try this, and i hope you get the idea

[tex]180° = \pi[/tex]

[tex]k_1 = 180° + \sin 180°[/tex]
[tex]k_2 = \pi + \sin \pi[/tex]

is [tex]k_1 = k_2?[/tex]


-Jambo

[SteveThackery]

Thank you so much, Jambo. I haven't had time to study your reply yet, but at first glance it seems to answer my question perfectly.

I will read it and study it carefully and reply when I have understood it properly! Meanwhile, thank you again - very much appreciated.

Steve