Harder Graphing...sinx cosx (1 Viewer)

[wrxsti]

is there like a specific way to graph lik (sin^3)x + (cos^3)x = y or lik sin(2x+5) cause the way i graph them is lik i draw up a table and plot values in and see what happens. does anybody know an easier way to do graph them.?

 

[ianc]

is there like a specific way to graph lik (sin^3)x + (cos^3)x = y or lik sin(2x+5) cause the way i graph them is lik i draw up a table and plot values in and see what happens. does anybody know an easier way to do graph them.?

there is no right or wrong way to graph i guess.....just as long as you get the shape right and show stuff like stationary points, intercepts and asymptotes.


However I see using a table of values as a bit of a last resort, and it can be pretty time consuming.

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for f(x) = sin³x + cos³x, I would first find the stationary points:
f'(x) = 3.sin²x.cosx - 3.cos²x.sinx
solve for f'(x)=0
3.sinx.cosx.(sinx-cosx) = 0
So sinx.cosx=0 and sinx = cosx
So sinx = 0, cosx = 0, tanx = 1

solve these 3 equations for the x values, then find the y values of the stationary points - not too difficult

Then plot these points on the graph

Then work out whether they are min/max points by quickly plugging in some values on your calculator

After that you should pretty much be able to sketch it, and it should look something like this:

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For f(x)=sin(2x+5): this should pretty much look like a sin(x) graph, but with a different period.

just simply solve these:
2x+5 = 0, ie at this value the sin graph is 0
2x+5 = pi/2, the sin graph is 1
2x+5 = pi, the sin graph is 0
2x+5 = 3pi/2, the sin graph is -1
2x+5 = 2pi the sin graph is 0

I hope you understood what i just did there......if not feel free to ask and I will explain it further.

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But as you can see, with graph sketching you pretty much just have to have your wits about you.

For a complicated graph like the first one, after solving stationary points, x and y intercepts and asymptotes, you should have a pretty good idea of what the graph should look like



Hope this was of some help!

 

[wrxsti]

thanxx very very helpful
sorry but um one more thing, u kno testing for minimum and max pt, shud i get the second derivative, or shud i jus plug in a point be4 and after?

 

[pLuvia]

Depends which method is easier as in exams time is of essence. Unless they ask you for the second derivative don't bother, just test either points of the stat point

 

[elseany]

wow that first one looks kinda hard, are you sure you saw it in a 3u book?

i would of thought that the best way around this question (and what they would've wanted you to do) is to use addition of ordinates, but im not sure if you learn that in 3u? :S

 

[ianc]

yeah only use the second derivative if it's straightforward - which in this case it's not.

in that question, the first derivative was 3sin²x.cosx - 3cos²x.sinx....so while not impossible, finding the second derivative will get a little bit messy and time consuming because you will have multiple applications of the chain rule. so yeah, either substituting values into the derivative to work out the gradient either side of the stationary point, or substituting values into the original function to work out the actual values either side of the stationary point will probably be much faster.

addition of ordinates is not in the extension 1 syllabus. but in this question i think the easiest thing to do was to find the stationary points, because it boiled down to a very nice quadratic which was quite easy to solve. (sinx=0,cosx=0,tanx=1)

so for an extension 1 question, i would imagine that it is considered a harder one but well within the capabilities of most students if they just gave it some thought.

one of the biggest challenges with maths is not being daunted by questions that look really hard, because by applying what you know you should get somewhere.....

 

[jb_nc]

what are you using to graph those graphs? matlab?

 

[ianc]

maple....just cos that's what we've been using at uni

but i think theyre all pretty much the same

 

[jemsta]

id rather mathematica then maple tbh (i have used both)