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I think there's a geometric solution too, but I've been busy today in the garden. I'll have a go now.

LATER EDIT:

AS TPG = 90, => TG is a diameter. Let C be the centre of the circle, and let FPH be the tangent to the circle at P (F on the 'T' side of the circle)

FPC = 90

x = FPT = CPG and y = TPC = GPH = PTC.

WE know that TP has gradient 1/t => tan(y) = 1/t and, as x + y = 90, => tan(x) = -t

Bob

hi

Like that solution. Then you can use

to simplfy this

Bob

hi Stefy,

It would help me if you were to post your method. Thanks,

Bob

hi Andrew1930

5% credited monthly. ? This can mean one of two things.

(1) UK. You work out the monthly rate that is equivalent to 5% annually.

so r = 1.004074123

(2) But in some countries I have discovered that r = 1 + 0.05/12 = 1.004166666

Clearly, this will lead to very different answers. I think (1) is correct but you need to check.

I'll avoid the problem by just using r.

month 0: 25000

month 1: 25000xr - W where W is the amount withdrawn.

month 2:

month 3:

month n:

where a is the first term of the GP (written in reverse order) = 1

So you can put n = 8x12 and set this to zero to solve for W.

EDIT: Just tried with my r (as above) and got 315.18 rounded to the nearest cent.

and using (2) I get 316.50 which agrees with bobbym's answer.

Bob

FURTHER EDIT: In post 4 of your other thread, Phrontister gives a formula which has the term i/q. This is equivalent to (2)

You say you are sitting the exam in the UK. What syllabus? I'll look it up.

OK. Like this:

AP: a, a+e, a+2e, a+3e

GP: a-2, b-4, c-3, d+2 becomes a-2, a+e-4, a+2e-3, a+3e+2

Now using the common ratio r

The first pair leads to an equation in a and e and the first and third similarly.

Make a the subject of each and equate them to get a quadratic in e. I got

11 leads to the solution you are seeking.

Then my difficulty begins because the second value of e leads to an AP of 2,4,6,8 but not a GP ... values of 0,0,3,10.

So what's going on there.

I made up a second problem: AP 27,31,35,39 and GP 27,36,48,64.

Using the same method I got a quadratic:

e=4 gives the AP and GP as expected but e = -5 gives AP 0,-5,-10,-15 and not a GP of 0,0,3,10 again.

So I picked a third example:

AP 8,13,18,23 and GP 8,12,18,27

My quadratic this time was

e=5 gives the expected AP and GP but e = 1 gives an AP of 0,1,2,3 but not a GP of 0,0,2,7.

I don't understand why this is happening.

Bob

hi Stefy,

Thanks. But I'm getting strange results from the second solution to the quadratic. And I've tried two more examples with the same happening again. I'm just trying to make sense of all my paperwork; then I'll post what I've found.

Bob

hi Stefy

Bob

Several corrections later, I have a solution. The quadratic factorised to give two values of e.

solution 1 gave the required AP and GP.

solution 2 did not. I haven't worked out why not yet.

So I have a single solution. Do you?

LATER EDIT: Looks like solution 2 results from a division by zero and so is not valid.

Bob

You are welcome.

Post again when you next get stuck.

Bob

hi cooljackiec

I'm hoping I've slipped up with my arithmetic because I'm getting a complex number for the common difference. The method should work though so give it a try:

AP

with e = the common difference.

GP

with r = common ratio.

So eliminate b, c and d from the GP equations and then divide the first by the second and first by the third to eliminate r and leave a pair containing just a and e.

I found that it was easiest to then eliminate 'a' to leave a quadratic in e.

But it only had complex solutions so I'll try again carefully.

Bob

hi Andrew1930

Welcome to the forum.

Yes, it's a great site!

I don't get quite that value (few cents out) which may just be due to rounding after each year (I didn't but maybe the questioner did).

It doesn't say there is any initial investment, so let's say there isn't.

At the end of year 1 the investment is 500.

At the end of year 2 that has grown by a factor of 1.06 then another 500 = 500x1.06 + 500

At the end of year 3 that has grown by a factor of 1.06 then another 500 =

Then x by 1.06 again and add another 500 =

And one more time to get to the end of year 5

The bracket is a geometric progression with first term a = 1 and common ratio r = 1.06

Bob

hi shyclaw

Welcome to the forum.

There's a load of pictures of prisms here:

http://www.mathsisfun.com/geometry/prisms.html

That should give you an idea of what one is.

To get the surface area you need the area of each end and the area of the rectangular sides

So for a triangular prism surface area = (area of one end) x 2 + area of the three rectangular sides.

At the bottom of that page you'll find some questions so you can get some practice.

For a cylinder:

http://www.mathsisfun.com/geometry/cylinder.html

you need the area of each circular end plus the area of the curved surface. If you imagine it 'unfolded' it is a rectangular shape with length equal to the length of the cylinder and width equal to the circumference of the circle.

Hope that helps.

If you are stuck with a question, post it here.

Bob

hi Beverrrn

Take a look at this page especially (but there's loads of backup material on the connected trig. pages.)

http://www.mathsisfun.com/algebra/trig- … ircle.html

Bob

hi mrpace,

Not really my area so I've been waiting for someone else to have a go at this. But no one has so I'll give you my thoughts. Maybe this will attract someone else to do it properly .

To solve second order differential equations like this you are suppose to look at the 'equivalent' quadratic ... in this case

To get real solutions, you'd want lambda to be negative, so let's say

Then a solution is

and another one is

I think, and this is where I'm getting 'shaky', this means the general solution is:

where A and B are constants. This certainly does fit the differential equation OK.

So then you put in the boundary conditions to find A and B. Maybe I made a mistake, but so far that has just given me nonsense.

I'll start again on a fresh sheet of paper and see if I can convert this into sense.

LATER EDIT:

here's a typical graph:

This obeys the first constraint but won't simultaneously obey the second.

Or is this two separate questions in one?

Bob

hi PatternMan

What is the expression?

Bob

Two vectors can be parallel too.

I would have defined coplanar vectors like this:

Two vectors that lie in the same plane.

Stefy's definition comes to the same as you can always translate the vector until it does lie in the plane.

Two non-parallel vectors in a plane will form a basis for all vectors that lie in that plane. If the vectors are **a** and **b** you can write any other vector in that plane as a linear combination thus:

In the question, any two vectors will always generate a plane unless they are parallel (eg. OA and OC in my explanation above)

You can choose the third so that it is not in that plane provided you avoid the two vectors that are parallel to the chosen ones.

Bob

hi Niharika,

Did you mean {-1,1} ?

I'll write vectors as rows (rather than columns) to save having to construct the right forms.

So a possible vector would be (1,-1,1)

So that gives you the 8 vectors.

Sketch a cube, centred on (0,0,0) with these 8 points as the corners.

And then I come 'unstuck' because every three that I select are coplanar.

Wait a mo. I'm thinking these are coordinates, not vectors.

Thinking ..................

Got it! I think!

If you choose any two, say OA = (1,1,1) and OB = (-1,1,1) then these form the plane that goes through (0,0,0) as well. The plane is OAB.

There are exactly 2 points chosen from the other 6, that are also in this plane. They are C = (-1,-1,-1) and D = (1,-1,-1)

So with that choice for the first two points there are 4 non coplanar choices for the third.

So proceed like this:

How many choices for the first vector?

How many choices for the second ?

How many non coplanar choices for the third?

You may end up counting the same solutions more than once so reduce by division.

Bob

ps. yes that worked

hi mathstudent2000

Yes, bobbym has correctly recognised these. thedarktiger posted these last March. Looks like you're getting the same exercises.

You'll find solutions if you do a search on tdt's topics.

Bob

hi pari_alf

Did you mean this?

I've got 4 rectangles and I've shaded just one (they overlap).

Bob

hi auyeungyat

When you start the fire in the middle why does it spread westward against the wind direction? This puzzle has come up before and I gave an alternative solution in post 6 of

http://www.mathisfunforum.com/viewtopic.php?id=20864

Bob

hi Justin Johns

Differential calculus and integral calculus are just two tools that mathematicians use.

What they have in common is they both consider what happens to the value of an expression as things tend to a limit.

In differentiation, when you want the gradient on a graph at a point A, you start by choosing a nearby point B and working out the gradient of AB. Then you let B get closer and closer to A and see what happens to the gradient. For 'well-behaved'* graphs the gradient gets closer and closer to a fixed value, so it seems reasonable to assume the gradient of the tangent at A is that value.

In integration, you construct an expression for a 'small amount' of something and then add up lots of those to get an approximation for what you're trying to find. One way is to under-estimate and hence get a lower bound and then repeat with over-estimates to get an upper bound. If you find that improving the approximations makes the lower bound higher and the upper bound lower then the answer you want gets sandwiched between ever closer amounts and so, again, you can get a value. Again, the function needs to be 'well-behaved'.*

One surprising, and very useful, result of the two processes is that with one small proviso differentiation and integration are opposite processes. So if you start with a function and integrate it, then differentiate the result, you get back to where you started. This is called **the fundamental theorem of calculus.** You can work out rules for differentiation and use them to work out many rules for integration.

*Some functions are so odd that they cannot be differentiated. And if you just make up a random function to integrate there's no guarantee that you'll be able to find a function that would differentiate to it, so integration may be difficult or even impossible.

Most calculus courses start with differentiation and then go on to integration. My teacher when I was doing this said that integration is the opposite of differentiation (as a definition) but that's not really true. It's a summation process that can be **shown** to be the opposite. That's another thing I can show you if you want, but it's probably best to do some examples first so you are familiar with how it all works.

Bob

hi Justin Johns

Welcome to the forum.

what really integration is other than saying it's just adding slices together to get a full piece.

Cannot do that because that is what it is. The integral sign is a fancy S to stand for sum.

In this case:

Try thinking about the graph y = 2x.

From 0 to 3, it is a diagonal line going from (0,0) to (3,6). The 'area under the line' is the area of a right angled triangle with a base of 3 and a height of 6, so the area is 0.5 x 3 x 6 = 9.

You will never get the right answer just by adding the y coordinates of the whole numbered x values.

In general, it doesn't have to be areas. Anything that can be written:

can be evaluated using integration. If you want, I'll show how to work out (for example) the volume of any pyramid (any base shape) using integration.

Bob

hi josiahb

Welcome to the forum.

I've only just looked at this problem. It's a long thread and Reuel seems to sort it out in post 32.

That dA should be du by the way.

Bob

Oh. You actually want me to work it out.

Ok, then.

Yes I get 10.26

Extra suggestion. They tell you AB so if you cannot work that out you are allowed to do part b and at least get the marks for that.

Bob