brightsky's Maths Thread

[brightsky]

Is there are quick way to find the points of intersection between two circles? (e.g. (x-15/2)^2 + (y-19/2)^2 = 25/2 and (x-5)^2 + (y-7)^2 = 25)

[Planck's constant]

The 'quick' way would be to solve the 2 equations simultaneously on CAS.
The analytical way would be to somehow make y the subject of both equations.
Equating those will leave you with a quadratic in x. (this may involve some squaring of both sides, which introduces domain issues)
Solve in the normal manner and you will get either 0,1,2 solutions depending on the number of points of intersection.
Using either of the original equations, you can now find the corresponding y's

[brightsky]

thanks, yeah only way i could think of was to solve the two equations simultaneously, although, as argonaut pointed out, this involves dealing with very messy equations. i guess essentials wants us to use the CAS.

btw, anyone know how to solve a system of equations using the ti-nspire. i am hopeless with calcs.

[Planck's constant]

thanks, yeah only way i could think of was to solve the two equations simultaneously, although, as argonaut pointed out, this involves dealing with very messy equations. i guess essentials wants us to use the CAS.

btw, anyone know how to solve a system of equations using the ti-nspire. i am hopeless with calcs.

Press Menu, press 3, press 7, thaaaaan press 1, place the equations in each of the boxes, press enter, and bazzam two answers with both variables solved.

(from memory) you can also try,

solve(equation1 and equation2,x,y)

'and' is a required keyword

[kamil9876]

One possible way: Expand out the squares, subtract the equations and you get a linear equation relating x and y. Express y in terms of x and plug back into one the equation and you have yourself a quadratic equation in x, which you know how to solve.

[brightsky]

One possible way: Expand out the squares, subtract the equations and you get a linear equation relating x and y. Express y in terms of x and plug back into one the equation and you have yourself a quadratic equation in x, which you know how to solve.

You can do that? Won't you get a redundant solution?

[Planck's constant]

One possible way: Expand out the squares, subtract the equations and you get a linear equation relating x and y. Express y in terms of x and plug back into one the equation and you have yourself a quadratic equation in x, which you know how to solve.

Yes thats the most straightforward analytical solution.


.

[brightsky]

just a general question: are all the questions in essential/mq designed to be non-calc?

[pi]

just a general question: are all the questions in essential/mq designed to be non-calc?

Well for Multiple Choice and Extended response are considered to be 'calc'.
The Short-Answers are considered to be non-calc'.

+1

Furthermore, in exercises, the calc qs usually have some sort a 'technology' symbol next to them (a pic of a calc or something)

[brightsky]

anybody know how to convert from 'radians' to 'degrees' on a ti-inspire?

[b^3]

Alternative shortcut that doesn't rely on calculator mode.

Enter what you want to convert = Ans
Put in the "conversion" symbol, this is a right block arrow, can be selected using the caltalouge, press [shift] + [catalouge] (the one with the book). In the fourth tab it is in the top row and 3rd from the right.

Now to convert to degrees type, DMS
To convert to radians type RAD.

Personally I found it quicker to manually convert by typing * 180/pi or pi/180, but its up to you with what you feel better doing.

[pi]

I use a scientific calc for this, way faster conversion.

[brightsky]

thanks guys!

also, any help would be appreciated for q6 d).

[kamil9876]

Let C be the centre of the circular arc. We know that the segment CS is perpendicular to AB. Thus if we let and be the radius of the circular arc then

(by looking at the right angle triangle CSB)

Now you know the angle BSP (you know from previous questions SBP and SPB and so just subtract) Hence you know that the angle CSP= angle BSP + pi/2. Now apply the cosine rule to the triangle CSP to get another equation and you should be able to solve from there

[brightsky]

can someone give me the intuition on linear dependence/independence?