graphing sinx/x

[Boots]

How would u draw the graph of sinx/x without using a graphics calc?
Like what key points would you need to graph it?
thanks

[appianway]

Think about when it's positive and negative, and think about the limit of the equation when x-> 0

[Boots]

Think about when it's positive and negative, and think about the limit of the equation when x-> 0

yeh but that isnt enough to sketch the graph

[luffy]

How would u draw the graph of sinx/x without using a graphics calc?
Like what key points would you need to graph it?
thanks

Without a calculator, you would have to use 'product of ordinates.'

i.e. Draw the graphs of y = sin(x) and y = 1/x on the same set of axes. Make sure your graphs are roughly on the correct scale to each other.

Then, pick the key co-ordinates and multiply their respective y-values:

You will pretty much notice that y-> 0 as x-> infinity (due to y= 1/x graph), hence, towards the right of the graph, the y-values will also be close to 0, alternating between approaching it from the negative and positive sides (as y = sin(x) shifts between positive and negative frequently).
As x -> negative infinity, y -> 0 as well. Hence, a very similar graph will occur on the left hand side.
Note that any x-intercepts on either of the two graphs will also result in intercepts on the resultant graph.
The middle of the graph is your main concern, and you might be slightly off the true graph in this, but it is a limitation of 'product of ordinates.'

Hope I helped. Let me know if you need a better explanation.

[paulsterio]

How would u draw the graph of sinx/x without using a graphics calc?
Like what key points would you need to graph it?
thanks

Without a calculator, you would have to use 'product of ordinates.'

i.e. Draw the graphs of y = sin(x) and y = 1/x on the same set of axes. Make sure your graphs are roughly on the correct scale to each other.

Then, pick the key co-ordinates and multiply their respective y-values:

You will pretty much notice that y-> 0 as x-> infinity (due to y= 1/x graph), hence, towards the right of the graph, the y-values will also be close to 0, alternating between approaching it from the negative and positive sides (as y = sin(x) shifts between positive and negative frequently).
As x -> negative infinity, y -> 0 as well. Hence, a very similar graph will occur on the left hand side.
Note that any x-intercepts on either of the two graphs will also result in intercepts on the resultant graph.
The middle of the graph is your main concern, and you might be slightly off the true graph in this, but it is a limitation of 'product of ordinates.'

Hope I helped. Let me know if you need a better explanation.

Luffy, good explanation, but you seem to forget that 1/x * sin(x) is basically a sin(x) graph with amplitude modulation
the amplitude will be 1/(x) so essentially, for the positive part first, you start off with an infinite amplitude at x = 0 and the amplitude gets smaller and smaller, so basically a normal (very tall) sin(x) graph that gets smaller and smaller, the period remains constant

similar for the negative part (:

[paulsterio]

if i didnt make that clear enough, here's a step by step guide as to how you would draw it

1) dot in the 1/x for x>0 (positive hyperbola)
2) dot in its reflection in the x-axis ie. -1/x for x>0
3) now fill in between the two graphs with a sin curve

4) repeat for the other side (:

[luffy]

How would u draw the graph of sinx/x without using a graphics calc?
Like what key points would you need to graph it?
thanks

Without a calculator, you would have to use 'product of ordinates.'

i.e. Draw the graphs of y = sin(x) and y = 1/x on the same set of axes. Make sure your graphs are roughly on the correct scale to each other.

Then, pick the key co-ordinates and multiply their respective y-values:

You will pretty much notice that y-> 0 as x-> infinity (due to y= 1/x graph), hence, towards the right of the graph, the y-values will also be close to 0, alternating between approaching it from the negative and positive sides (as y = sin(x) shifts between positive and negative frequently).
As x -> negative infinity, y -> 0 as well. Hence, a very similar graph will occur on the left hand side.
Note that any x-intercepts on either of the two graphs will also result in intercepts on the resultant graph.
The middle of the graph is your main concern, and you might be slightly off the true graph in this, but it is a limitation of 'product of ordinates.'

Hope I helped. Let me know if you need a better explanation.

Luffy, good explanation, but you seem to forget that 1/x * sin(x) is basically a sin(x) graph with amplitude modulation
the amplitude will be 1/(x) so essentially, for the positive part first, you start off with an infinite amplitude at x = 0 and the amplitude gets smaller and smaller, so basically a normal (very tall) sin(x) graph that gets smaller and smaller, the period remains constant

similar for the negative part (:

I think he would rather have an explanation of how to do these questions + similar typed questions without a calculator, than to know how this graph looks. If he wanted to know that, he could easily just check with a graphics calculator. That is why I didn't mention the amplitude modulation.

[paulsterio]

I think he would rather have an explanation of how to do these questions + similar typed questions without a calculator, than to know how this graph looks. If he wanted to know that, he could easily just check with a graphics calculator. That is why I didn't mention the amplitude modulation.

true...but amplitude modulation is an easy way of sketching any sin/cos multiplied by another function

hmm, i wonder if theres a more "easy" way of thinking about product of ordinates...

[taiga]

draw y=sin(x) and y=x and divide the y values of the first equation by the one of the second, and fill them in.

[Boots]

@paulsterio thanks thats a very smart explanation.....thanks to everyone else to.....helped heaps