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hi iamaditya

In differential calculus, we are trying to develop a rule for how the gradient of a function varies with x. The usual method for working this out is this:

(1) Pick a point on the curve for the function, say ( x, f(x) )

(2) Move to a point close to ( x, f(x) ). The way this is written is ( x + Δx, f(x+Δx) ) meaning ("x plus a little bit more in the x direction", "value of the function at this new value of x")

note: Δx should be regarded as a single variable, not as two variables multiplied together. Thus Δy/Δx must not be simplified to y/x by cancelling! This should be considered as "a little bit in the y direction"/"a little bit in the x direction"

(3) The gradient of the chord joining these points is then considered, simplifying where possible.

(4) Then Δx is reduced towards zero, to see if the gradient found in step (3) tends towards a recognisable gradient function.

(5) This gradient function is usually written dy/dx. Again these two, dy and dx, should be treated as two single variables so the 'd's cannot be cancelled.

Some texts avoid this confusion by using x + h and f(x+h) where 'h' is a small amount in the x direction.

eg https://diversity.umn.edu/multicultural … nRules.pdf

Hope that helps,

Bob

hi Dark Tangent

I get this:

Q (2,7) rotates to Q' ((0,5) and then reflects to (2,5)

thickhead wrote:

So the coordinates of various points will be N(0,-4) ,P( 2,-4) and Q(2,0).

Is this correct? The rotation is 270 clockwise.

Bob

hi DarkTangent

First draw a diagram. Then make use of some tracing paper. In the UK in GCSE exams, tracing paper is allowed.

Bob

hi samuel.bradley.99

I've been away so I've only just seen this problem. I had a vague memory that the volume of a tetrahedron could be calculated using vector cross and dot products. A quick 'google' took me to this page:

https://math.stackexchange.com/question … ot-product

Your tetrahedron is a regular one by the way which fixes the directions of the vectors IA, IB and IC. So it looks like you can use this formula to get a result. But post back if you need a quick lesson in cross and dot products.

Bob

Can do, but I was thinking of doing 0.5 CB.CA

Bob

OK. Not quick enough to catch before you logged out. Never mind I have it sorted. If AB is the hypotenuse, and C is 90 degrees, there's a property of circles that AB is also a diameter. The theory is here:

http://www.mathisfunforum.com/viewtopic.php?id=17799

see post 6

So draw coordinate axes, mark A and B and find the midpoint, O. This is the centre of a circle that goes through A, B and C. Draw a line at right angle to AB, through O to cut the circle at C. From there it should be straight forward, but post back if you are still having difficulties.

Bob

hi Lolz

welcome to the forum.

I think I can help. Stay on -line while I make a diagram

Bob

hi DarkTangent

I find it best to get a really good diagram first. I use some software called Sketchpad but you can also use geogebra which is free.

In the first question you can write an equation for DA in terms of ED, and then CD in terms of ED. Add and re-arrange to get ED.

In the second you should be able to work out YOX. The angle at the edge, YZX, is half this using angle properties of a circle.

Bob

I had a look at other on-line sites for compound proportion. There are formulas but I wouldn't recommend trying to learn and use them. Compound proportion means a series of simple proportion steps, so you can always do a problem by considering each step separately. Here's a example to show what I mean:

example. A fence company employs people to dig the post holes. There are three variables:

How many people to dig the holes.

How much time they will take.

How many holes are needed.

If the number of holes is fixed, then more people means less time to dig, so that is inverse proportion.

If the number of people is fixed, then more holes will take more time, so that is direct proportion.

If the amount of time available if fixed, then more people means more holes, so that is direct proportion.

Suppose we known that 10 people can dig 18 holes in 5 hours.

How long will it take for 15 people, to dig 12 holes ?

Let's keep the number of holes fixed first. How long will 15 people take to dig 18 holes?

Holes fixed is inverse proportion. The number of people has gone up from 10 to 15, ie times 15/10, so the time will go down by times 10/15

New time = 5 x 10/15 = 10/3 hours.

Now we'll keep the number of people fixed. How long will 15 people take to dig 12 holes ?

People fixed is direct proportion. The number of holes has gone from 18 to 12 ie. times 12/18, so the time will go down by times 12/18

New time = 10/3 x 12/18 = 10/3 x 2/3 = 20/9 hours.

Hope that helps.

If you really want the formulas, 'google' compound proportion and you'll find lots of sites.

Bob

I expect MIF wanted to use 'ScienceIsFun' but that domain is already taken.

HQ is an abbreviation for Head Quarters.

Bob

hi Zeeshan 01

You were asking about ratio and fractions and now you have jumped to compound proportion.

May I assume you are ok now with ratio and fractions?

Before you jump to compound proportion, are you ok with simple proportion, both direct and inverse?

Easier if you have an example.

Bob

hi DarkTangent

Welcome to the forum.

Bob

It may be a cost thing. The forum is free to join but the web-space has to be funded somehow. There are discrete, relevant adverts that do this. I wouldn't want the forum to become overwhelmed with adverts.

And you have to allow MIF himself to have a life outside the forum. Don't forget all those excellent pages take time to develop.

Bob

hi NakulG

Yep! Got it now. That's an inspired substitution! Thanks once again. colon bracket with no minus gives a on this forum.

Bob

hi Alex23E

Welcome to the forum.

If you 'google' it you'll find several sites that have calculators for this. You just have to enter the data.

Bob

hi NakulG

Many thanks for that. I'll give it a try.

Bob

Sorry, Zeeshan 01, but what I'm seeing are not the same.

I'm seeing a divide symbol and a colon. A colon is commonly used to indicate a ratio. Ratio and division(fraction) are not the same, although they are closely and simply connected.

Bob

hi NakulG

I hadn't got anywhere with this. Please would you give a few hints/. Thanks

Bob

Analytical geometry is another term for coordinate geometry.

You probably know about that already, but here's a link in case you don't:

http://www.mathsisfun.com/data/cartesia … nates.html

It is also called 'cartesian' because it was invented by the mathematician René Descartes

https://en.wikipedia.org/wiki/Ren%C3%A9_Descartes

Bob

Zeeshan 01 wrote:

So what is 3÷1

This reads as 'three divided by one' Did you mean that, or did you intend to type 3:1 ?

Bob

hi Zeeshan 01

Are you still looking for these topics?

Calculus here:

http://www.mathsisfun.com/calculus/index.html

Geometry here:

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

Bob

Zeeshan 01 wrote:

denomin

ator is the whole

The whole has 4 squares.

Zeeshan 01 wrote:

numerator is part we have

If you want the blue fraction that is 3 squares so blue fraction is 3/4

If you want the yellow fraction that is 1 square so yellow is 1/4

Bob

I have a 'Euclidean' method.

I have extended BA to D, BQR to S and constructed CSD parallel to PA.

As BP = PC the triangle BCD is similar and twice the size of BPA.

Therefore if AQ = QP = x, then SC = 2x

In triangles AQR and CSR, AP is parallel to CS, AR and RC are part of the same line, and similarly QR and RS, so these triangles are similar. As CS = twice AQ then RS is twice RQ.

Let BQ = 3y = QS. R divides QS in the ratio 1:2 then QR = y and RS = 2y.

thus BQ = 3.QR

Bob

Ok Srikantan,

Here we go.

I'll use bold letters for the vectors.

First choose an origin. Any point will work but as the question is about the line BQR I'll choose B.

And for a 2-D problem you need two base vectors. Any will do providing they are not parallel so I choose **BA** = **a** and **BC** = **c**

The position of every other point can be written as some combination of an amount in the **a** direction and an amount in the **c** direction.

P is the midpoint of BC so **BP** = 0.5**c**

And **AP** = **AB** + **BP** = -**a** + 0.5**c**

=> **AQ** = 0.5(**AP**) = -0.5**a** + 0.25**c**

=> **BQ** = **BA** + **AQ** = **a** -0.5**a** + 0.25**c** = 0.5**a** +0.25**c**

**AC** = **AB** + **BC** = -**a** + **c**

Now we know R lies on AC but we don't know where, so suppose it is fraction f, along from A (normally I'd use a Greek letter lambda here but it gets too complicated to try and show that)

=> **AR** = f( -**a** + **c**) = -f**a** + f**c**

=> **BR** = **BA** +**AR** = **a** -f**a** + f**c** = (1-f)**a** + f**c** ......(1)

WE also know that R is some way along BQ, let's say g along where g is a number over 1.

=> **BR** = g (0.5**a** +0.25**c**) = 0.5g**a** + 0.25g**c** .......(2)

We now have two ways to describe **BR** so they must be equal. In vectors, each component must be equal so we have a pair of simultaneous equations:

**a** components: 1-f = 0.5g

**c** components: f = 0.25g

adding gives 1 = 0.75g so g = 4/3 and so f = 1/3

In (1) gives **BR** = 2/3. **a** + 1/3. **c** and in (2) gives **BR** = 2/3. **a** + 1/3. **c**

I've worked out both ways to check for mistakes. I did find one, so these are the corrected values.

As g = 4/3, we know BQ:BR = 1:4/3 = 3:4 => 3.QR = BQ as required.

Bob

Will do, but I need to know how far back to go.

Do you know what a vector is?

Have you used a vector approach to find triangle properties? eg. That the 3 medians go though a single point.

Bob