Sketch graph of sin(x^2)

Hi fellow forumers,

Would like to seek help in understanding the sketching of sin(x^2)=y

I've searched high and low on the internet to no avail, and I've left school for a gap year before applying to Cambridge for Economics so I've been attempting to brush up on mathemathics now.

What would be of great interest to me is the derivation of the exact shape of the graph (not the actual graph itself as it is easily done by any GC) and explanation of it's features and how they are derived (the decreasing period as x increases).

Thank you so very much!

Reply 1

Ateo

9

First of all, you know that the curve is between -1 and 1, (it took me around 10 minutes to realize something so simple in my interviews when I was asked a similar question). I will try to explain now how the period decreases.

In physics you may have encountered the equation:

sin(\omega t)

. also remember that

\omega = \frac{2\pi}{T}

where T is the period and omega is angular velocity.

Anyways think of the above as

sin(\frac{1}{T}x)

and your original equation as

sin(x \times x)

.

From there you can see that the period is inversely proportional to x and hence will decrease as x increases.

Here are some other ones to try;

y= sin(\frac{1}{x}), y=e^x sin(x), y=sin(\frac{1}{sin(x)})

Reply 2

dente

OP

2

Original post by Ateo

First of all, you know that the curve is between -1 and 1, (it took me around 10 minutes to realize something so simple in my interviews when I was asked a similar question). I will try to explain now how the period decreases.

In physics you may have encountered the equation:

sin(\omega t)

. also remember that

\omega = \frac{2\pi}{T}

where T is the period and omega is angular velocity.

Anyways think of the above as

sin(\frac{1}{T}x)

and your original equation as

sin(x \times x)

.

From there you can see that the period is inversely proportional to x and hence will decrease as x increases.

Here are some other ones to try;

y= sin(\frac{1}{x}), y=e^x sin(x), y=sin(\frac{1}{sin(x)})

Hey,

Thanks for your help, that information was great! I think I get it now.

For your additional questions, I get

1) y=sin(1/T*x) - Hence, period T is proportional to X and hence period increases as x increases, also since 1/x tends to infinity as x tends to 0. Hence we have a vertical asymptote at x=0

2) I'm not too sure how to continue with this, but I'm pretty certain the first step involves putting the e^x into the brackets of the trigo function.

3) y=sin(1/T*sin(x)) I'm pretty sure T is proportional to the value of sin(x), but how this translates onto the graph is beyond me. I'm thinking the period oscilliates back and forth (as T is proportional to a value between and including 1 and -1), but I'm unable to figure out the period of this repeating function.

Thanks for your help once again! Would you be willing to share more tips, privately or otherwise, about the Camb econs interview? I think any info I can get would be a great help right now.

(edited 10 years ago)

Reply 3

Ateo

9

Original post by dente

...

My analogy may not have been the best. What I was trying to do was to split up

x^2

into two variables so that I could match one of them to the period. I could have been clearer.

I wouldn't use what I mentioned as a general rule of thumb to determine the behavior of trigonometric curves. It is much more important to consider the behavior of the function. Determine what happens as x tends to infinities and 0 from both sides, what points it crosses the axes at, is any part of the function periodic and quite importantly, the rate at which the variable will change. These are the main questions to ask.

Sometimes it may not be easy to answer those questions and I was trying to illustrate how to answer one about the periodicity/rate of change of the function. I admit that my example is not ideal.

1. If we are to go with the previous analogy;

sin(\omega t)

(t is like your normal variable x in a sine function, in physics this equation is taken with omega being a constant, what I was trying to do was make it variable), in general I guess we want to make x=t and put everything else as omega. Consider

sin(\frac{1}{x})=sin(\frac{1}{x^2} x)

so we have x=t and

\frac{1}{x^2} = \frac{2 \pi}{T} \Rightarrow T \propto x^2

.

From this we can see that as x increases, the period will increase quite rapidly (think of y=x^2 graph) - this is where the rate part comes in. Thus the period is wild at the start and then calms a bit down and eventually becomes infinite. You were correct about the asymptote. I asked a student to explain a solution to this question to me, it was much more concise then what I'm writing :biggrin:

. He said the way to approach this is to think of the sine curve from 0 to 1 and from 1 to infinity and flip switch the entire curve in those regions around so that whatever was from 1 to infinity is now between 0 and 1 etc.

2. Think of what the curve is BOUND by. What if you were to consider the graph of xsin(x) - this is a simpler case of an almost identical problem.

3. For this one I would completely forget about my analogy. Focus on determining the answers to the questions I have mentioned. When does sin(x)=0? What happens at this value? Think about the periodicity of the sine function. What is the function bound by. (This is an Oxford maths interview question as is 1)

To summarize I would like to emphasize on how much more important it is to determine the behavior of the function. I have no idea how universal that analogy is (and it's not that great).

I'm really bad at explaining things that are more elaborate than A'level but I hope that helps. I have some Oxford interview questions scribbled in a notebook somewhere. I think you could get best advice from Cambridge applicants themselves.

EDIT: Just a small note I wanted to add. Do not use that equation I suggested for questions involving anything other than a polynomial or some simple reciprocal like 1/x .

(edited 10 years ago)

Reply 4

james22

16

Say you want the points where it crosses the x axis, then you need to solve sin(x^2)=0. This means that x^2=n*pi for n is a (+ or -) integer. The roots are therefore x=pi*sqrt(n) for integers n. You can find the turning points in a similar way.

Reply 5

Ateo

9

Original post by james22

Say you want the points where it crosses the x axis, then you need to solve sin(x^2)=0. This means that x^2=n*pi for n is a (+ or -) integer. The roots are therefore x=pi*sqrt(n) for integers n. You can find the turning points in a similar way.

Definitely worth checking where the curve will cross the axes, but is there much point in finding turning points if the domain is not constrained? You usually get infinite turning points for these types of curves.

Reply 6

james22

16

Original post by Ateo

Definitely worth checking where the curve will cross the axes, but is there much point in finding turning points if the domain is not constrained? You usually get infinite turning points for these types of curves.

You can find the general turning point, no hard than finding a specific one.

You do the opposite

Original post by dente

Hi fellow forumers,

Would like to seek help in understanding the sketching of sin(x^2)=y

I've searched high and low on the internet to no avail, and I've left school for a gap year before applying to Cambridge for Economics so I've been attempting to brush up on mathemathics now.

What would be of great interest to me is the derivation of the exact shape of the graph (not the actual graph itself as it is easily done by any GC) and explanation of it's features and how they are derived (the decreasing period as x increases).

Thank you so very much!

Your squaring every x value but as the transformation is in the brackets you do the opposite. So the periodicity increases between 0-2pi

Sketch graph of sin(x^2)

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