Math Is Fun Forum

hi bobbym

Wow! that's made my day!  I got something before you did.  I'm so happy.

:):):)

Bob

hi Identity,

I've got a problem with this too.  Am I interpreting your diagram properly?  See below.

My software (Sketchpad) chooses vertex letters as I measure angles so I've had to letter as shown.

I'm taking your A as my ACB  and your B as AEB.

I've chosen x to be 12 in the first diagram and 14 in the second.

This gives firstly A + B = 65 but secondly A + B = 89.

I can show that B = A - 2x if that's any help.

Bob

Hi bobbym

So what if he chooses an irrational number?

Bob

hi MathsIsFun,

Sorry I have not come back on any of your earlier posts.  I've been off-line in France.  Arrived in Paris today and a wi fi I can piggy back on.

All above mods very acceptable.  Thanks for considering my suggestion.  Let's review after a month or so.

Bob

hi Alive,

I'm in France at the moment and using the hotel wifi.  Don't know when I'll be back on so apologies if there's a long wait before the next post.

I'll start with a basic cosine curve.  Try it just to iron out any bugs with using Excel for this.  It's not the best software for trig graphs but I thought there would be a good chance you'd have it which means we have got a starting point.  Anyhow, you only want it so you can understand how the graphs behave; and it'll be good for that.

Excel assumes your angles are in Radians so I've kept with that.  If you want a degrees version you'll have to convert the degrees by x pi / 180 before applying the COS function.

Pi is stored in Excel as if it was a function.  The code to get it is PI()

To make my graph I let Excel work out the value of pi/6 (30 degrees) and then used formulas to calculate subsequent angles in equal steps.

To 'insert' the graph choose a scatter graph.  That might seem a bit odd but the reason is that Excel makes unwelcome assumptions if you ask for a 'line graph' so you don't get sensible x and y coordinates.

To add a line between points use 'Add a Trendline'.  Again, Excel hasn't got what we'd like as the types of line don't cover trig functions.  I found the best fit is to choose 'polynomial' with the 'order' set to 6 (max possible on my version of Excel)

The picture below shows the values I used and also the formulas so you should be able to reproduce these for a basic cosine curve.

I've just checked what the post looks like and see I haven't included the column and row in the shot to save space.  To help you locate which cells .... the zero rads  value is in cell D3.

I'll sign off with that; let me know how it goes.  I'll start work offline for a more advanced general version.

Bob

hi xsw001

Looks good to me but , I confess, it's years since I did this sort of thing.  Was it you who posted a request to find functions with given domains and ranges?  I put one idea back but haven't thought about the other two since.

Bob

Hi bobbym,

I've just completed it and put it on the suggestions page.

How to get people to read it?

You ask the toughest puzzles don't you!

I'd have big letters and a pretty graphic on the front page and maybe more graphics to support each point.

Plus, we'd have a reasonable excuse to respond "Please read the front page advice before continuing!."

You could make a macro to do that automatically at the touch of a button!:)

Bob

hi bobbym,

I'm working on a post for the suggestions slot which is intended to encourage posters to give more information when asking for help.  It's a long post and I'm only half way through constructing it.  Please look out for it and let me know what you think.

Best wishes,

Bob

hi Sephiroth Valentine

Help to attempt this:

Firstly make a good diagram so you can see what you have.  (see picture below)

I've had to choose where to put 'u' and 't'.  It's VERY unlikely I've put them in the right places so don't take the diagram as an answer!

You can box in any triangle like I've shown.  Getting its area is then a matter of working out the area of the box and subtracting the areas of the shaded (yellow) triangles.  They're all right angled so it shouldn't be too bad to calculate them.  You'll have to use these ideas to make an equation for 'k' and then solve it.  I haven't tried so I don't know how easy that will be.

second part. 

Calculate the equations for the lines st  and the parallel line that goes through t and u. (dotted on my diagram).

Use simultaneous equations to find where they cross."

That should do it.

Please post again if you need more details and, finally, to let us know you've done it.

Bob

Hi xsw001

3) How about y = (2/pi)arctan (x) ................  arctan is inverse function of tan taking 'principle value' -pi/2 < y < +pi/2

Let me know if what you think of this.

Still working on 1) and 2)

Bob

Hi Alive,

This is what I think you've asked:

The depth of water d metres in a  river, at a time t hours after 12 noon is given by d(t) = 5cos(πt / 12)
[π is pi]

a) sketch the graph of d(t) against t, showing one cycle on the axes, clearly marking in suitable areas

b) state the period and amplitude of depth of water

c) State the highest level of water and the time of day when this occurs

d) State the lowest level of water and time of day when this occurs

e) by solving an appropriate trigonometric equation, find time of day when level of water in river is equal to 2.5 metres

bobbym's graph in post 12 is the right shape.  The Cosine graph starts with  t = 0, cos(o) = 1, so d = 5x1=5

If the formula has a pi in it then one cycle is over when you get to cos(2pi) [360 degrees but your problem is in radians]

So t x pi / 12 = 2 pi   =>   t / 12 = 2 => t = 24.

So the complete cycle should have d back at 5 when t = 24 not 30.  Tides are not exactly repeated after 24 hours but near enough for a mathematical model.  (see graph below)

b)  So the period is 24 hours and the water level goes from +5 to -5 => amplitude = 5

c)    That'll be 5 metres then at t = 0, so at noon.

d)  That'll be - 5 metres?  I don't know how a river manages this.  I wonder if that's why you had a stray 400 in your post.
But it'll happen at t = 12 so at midnight.

Should it have been 400 + 5cos(πt / 12)
in which case the maximum is 405 and the minimum is 395.

e) But now I cannot do this with the 400 in there so I'll go back to the original

5cos(πt / 12) = 2.5 => cos (nt/12) = 0.5

this would be 60 degrees which is pi/3 in radians => pi t /12 = pi /3  => t = 4 ... 4.00 pm.

I did the graph below using Excel.  I've left the formula for d in the shot so you can re-create it if you have the software.

Hope that helps.  ps.  What exam is this for?  it looks like an Edexcel AS or higher level GCSE ??

If the latter, I have past papers so I could look it up.

Bob

hi bobbym

Ok, I admit I went 'all round the houses' to do this, but I was trying to avoid algebra. 

If the reason for the original question was to help someone who has an aversion to algebra your

might put them off.

You could argue that 'blocks' and 'groups' are just xs and ys really but they don't look like them.

I once got a remedial (at maths; she was very skilled at avoiding work!) child to construct formulas for an investigation but using jelly babies*.  Instead of T = 3x + 17 she was quite happy to write T = 3J + 17 because she thought the Js were just jelly babies.  I only told her at the end that she'd been doing algebra!

* In the UK jelly = jello and a jelly baby is a small sweet made of jelly.  They're quite popular and the first thing kids like to do is 'bite the heads off'.  What that tells us about children I would rather not speculate.  But at the end of the lesson you can tidy up by eating the visual aids.  That's got to be worth it!

Bob