tutoring thread

[JCB]

<div class='quotetop'>QUOTE (cpawfan)</div><div class='quotemain'></p>

Remember sin(0) = 0, sin(pi) = 0, sin(2pi) = 0 etc</p>

There are multiple solutions for x</p>

</div></p>

That's what I was thinking with the sin curve and such, but working with that equation, and solving for x algebraically, just gives you 0. Using the sin curve will give you infinite answers, but algebraically, you just get one answer, 0. Right? Or am I missing something completely obvious?</p>

I've got a few derivative questions I'll ask tomorrow. This thread could be useful.</p>

</p>

 

[pegs]

<div class='quotetop'>QUOTE (cpawfan)</div><div class='quotemain'></p>

Remember sin(0) = 0, sin(pi) = 0, sin(2pi) = 0 etc</p>

There are multiple solutions for x</p>

</div></p>

any within [pi/10, 3pi/10]?</p>

i can't figure any out within those parameters. I don't think there is a right answer there.</p>

</p>

 

[cpawfan]

<div class='quotetop'>QUOTE (Kidd)</div><div class='quotemain'></p>

Why I am having a hard time with multiplying, dividing, adding radicals? It's like 7th grade math isn't it? That's proof that math is no where close to being one of my strong points.</p>

</div></p>

Got an example?</p>

</p>

 

[cpawfan]

<div class='quotetop'>QUOTE (peg182)</div><div class='quotemain'></p>

<div class='quotetop'>QUOTE (cpawfan)</div><div class='quotemain'></p>

Remember sin(0) = 0, sin(pi) = 0, sin(2pi) = 0 etc</p>

There are multiple solutions for x</p>

</div></p>

any within [pi/10, 3pi/10]?</p>

i can't figure any out within those parameters. I don't think there is a right answer there.</p>

</div></p>

2pi/10 because sin( 5*2pi/10) = sin(pi) = 0</p>

</p>

</p>

</p>

 

[pegs]

<div class='quotetop'>QUOTE (cpawfan)</div><div class='quotemain'></p>

<div class='quotetop'>QUOTE (peg182)</div><div class='quotemain'></p>

<div class='quotetop'>QUOTE (cpawfan)</div><div class='quotemain'></p>

Remember sin(0) = 0, sin(pi) = 0, sin(2pi) = 0 etc</p>

There are multiple solutions for x</p>

</div></p>

any within [pi/10, 3pi/10]?</p>

i can't figure any out within those parameters. I don't think there is a right answer there.</p>

</div></p>

2pi/10 because sin( 5*2pi/10) = sin(pi) = 0</p>

</div></p>

</p>

SHIT!!!</p>

my calculator was in degrees. Wow, I was wondering why it wasn't coming out right. Duh.</p>

Don't know why it was degrees either...we never use degrees, always radians. Weird. Someone must've used it or something.</p>

Oh well, thanks Cpaw and JCB.</p>

 

[cpawfan]

Learn the unit circle, you don't need a calcuulator for this stuff</p>

 

[GMJ]

Damn cpaw beat me too it, and yes he's correct.</p>

I just got back from school, so I can answer some questions if need be</p>

 

[Kid Chocolate]

The unit circle? oh boy... </p>

 

[pegs]

Ha, I suck at remembering shit. God knows how I pass any tests.</p>

 

[lukewarmplay]

<div class='quotetop'>QUOTE (Kid Chocolate)</div><div class='quotemain'></p>

The unit circle? oh boy... </p>

</div></p>

You're a member.</p>

</p>

 

[JCB]

Can someone explain how to find the derivative of:</p>

f(x) = cos^4(sin(3x))</p>

I have a calc test tomorrow and this is the only type of problem I don't understand. Any help?</p>

 

[JCB]

<div class='quotetop'>QUOTE (JCB)</div><div class='quotemain'></p>

Can someone explain how to find the derivative of:</p>

f(x) = cos^4(sin(3x))</p>

I have a calc test tomorrow and this is the only type of problem I don't understand. Any help?</p>

</div></p>

Nevermind. Figured it out. I must've copied down the problem wrong in my notes because what I had written didn't make any sense. But I got it now.</p>

</p>

 

[bbwMax]

ok got this for maths.</p>

An (infinitely small) ball starting out in the middle of a 5 pointed star table (outer 5 points - 10m radius..... inner 5 points - 5m radius) has a starting angle of a random value from 0 to 360 degrees. The ball is now set loose and travels around the table.
On average, how many sides will have been hit once the ball has travelled 1000m ?</p>

heres the pic</p>

</p>

 

[GMJ]

What are meters? :-D</p>

 

[lukewarmplay]

<div class='quotetop'>QUOTE (Max)</div><div class='quotemain'></p>

ok got this for maths.</p>

An (infinitely small) ball starting out in the middle of a 5 pointed star table (outer 5 points - 10m radius..... inner 5 points - 5m radius) has a starting angle of a random value from 0 to 360 degrees. The ball is now set loose and travels around the table.
On average, how many sides will have been hit once the ball has travelled 1000m ?</p>

heres the pic</p>

</p>

</div></p>

OK, how the hell do you do this?</p>

</p>

 

[Денг Гордон]

Any help on...</p>

Let y= (x^2 + 3) / x. Find dy/dx (a.) by using the Quotient Rule, and (

by first dividing the terms in the numerator by the denominator then differentiating.</p>

The derivative is (x^2 -3) / (x^2)</p>

I got a, but I have no idea how to do b.</p>

 

[lukewarmplay]

<div class='quotetop'>QUOTE (BG7 Lavigne)</div><div class='quotemain'></p>

Any help on...</p>

Let y= (x^2 + 3) / x. Find dy/dx (a.) by using the Quotient Rule, and (

by first dividing the terms in the numerator by the denominator then differentiating.</p>

The derivative is (x^2 -3) / (x^2)</p>

I got a, but I have no idea how to do b.</p>

</div></p>

For b they want you to differentiate x^2/x + 3/x, or x + 3x^-1.</p>

Does that help?</p>

</p>

 

[GMJ]

<div class='quotetop'>QUOTE (lukewarmplay)</div><div class='quotemain'></p>

<div class='quotetop'>QUOTE (BG7 Lavigne)</div><div class='quotemain'></p>

Any help on...</p>

Let y= (x^2 + 3) / x. Find dy/dx (a.) by using the Quotient Rule, and (

by first dividing the terms in the numerator by the denominator then differentiating.</p>

The derivative is (x^2 -3) / (x^2)</p>

I got a, but I have no idea how to do b.</p>

</div></p>

</p>

For b they want you to differentiate x^2/x + 3/x, or x + 3x^-1.</p>

Does that help?</p>

</p>

</div></p>

I was thinking the same thing, but I think you should keep the first denominator in there. So its all of that over x, and then its just another quotient rule problem.</p>

 

[pegs]

divide each term in the numerator by the term in the denominator. so that the problem looks like:</p>

y=(x^2/x)+(3/x)</p>

...does that help?</p>

 

[pegs]

<div class='quotetop'>QUOTE (GMJigga)</div><div class='quotemain'></p>

<div class='quotetop'>QUOTE (lukewarmplay)</div><div class='quotemain'></p>

<div class='quotetop'>QUOTE (BG7 Lavigne)</div><div class='quotemain'></p>

Any help on...</p>

Let y= (x^2 + 3) / x. Find dy/dx (a.) by using the Quotient Rule, and (

by first dividing the terms in the numerator by the denominator then differentiating.</p>

The derivative is (x^2 -3) / (x^2)</p>

I got a, but I have no idea how to do b.</p>

</div></p>

</p>

For b they want you to differentiate x^2/x + 3/x, or x + 3x^-1.</p>

Does that help?</p>

</p>

</div></p>

I was thinking the same thing, but I think you should keep the first denominator in there. So its all of that over x, and then its just another quotient rule problem.</p>

</div></p>

</p>

I don't see the point in that...? It seems much more easy to solve using the power rule when you do it the way luke has it.</p>

 

[Денг Гордон]

OKay...</p>

so you get d/dx (x) + d/dx (3/x)</p>

And then for the derviatives you get 1 + (-3/x^2). Multipley the 1 by x^2/x^2....okay, got it.</p>

 

[Денг Гордон]

Have to find the deriv for</p>

y = x /(1 + cos x)</p>

Is there a trigonometric identity for that 1 + cos x ?</p>

 

[GMJ]

I don't know one off the top of my head, but wouldn't it just be (0 + -sinx) so -sinx?</p>

 

[JCB]

There's no trig identity for that, I don't think.</p>

</p>

 

[Денг Гордон]

Anyone can figure out the path for.</p>

</p>

Prove d/dx csc(x) = -csc(x)cot(x)</p>

Not sure how to go about that one. (It notes the derivitive of sin x, as being cos x, and cos x being -sin x).</p>

So first thing I'd do, is make csc x into 1 / sin x. Then use the quotient rule.</p>

So you get:</p>

sin x (0) - 1 (cos x)
-------------------
(sin x) ^2</p>

Which gets you</p>

-cos x
------
sin^2 x</p>

to</p>

-cot x (1/ sin x)</p>

to</p>

-csc(x)cot(x)</p>

Yay, got that one....so yeah...no help needed, figured it out, when trying to explain how much I had done on it.</p>

 

[Денг Гордон]

Hmm...so a harder one...</p>

Find the points on the curve y = tan x, -pi/2 < x < pi/2, where the tangent is parallel to the line y = 2x.</p>

WTF....</p>

 

[GMJ]

<div class='quotetop'>QUOTE (BG7 Lavigne)</div><div class='quotemain'></p>

Hmm...so a harder one...</p>

Find the points on the curve y = tan x, -pi/2 < x < pi/2, where the tangent is parallel to the line y = 2x.</p>

WTF....</p>

</div></p>

Find the derivative of tan x,</p>

set it equal to 2x</p>

solve for x</p>

only use the x-value that falls between -pi/2 and pi/2</p>

</p>

 

[Денг Гордон]

So you get</p>

sec^2(x) = 2x</p>

Or</p>

1/cos^2 (x) = 2x</p>

How do you solve for x on a cosine or secant?</p>

I used the solve function on my calculator, and the only values that came up for x when setting the secant squared of x = 2x, a ton of values came up, but no values in that domain came up....</p>

 

[bbwchingy0007]

Here's a few for you to try, from my practice British Mathematical Olympiad papers (non-calculator):</p>

1. Let n be an integer greater than 6. Prove that if n-1 and n+1 are both prime, then n^2(n^2+16) is divisible by 720. Is the converse true?</p>

2. Adrian teaches a class of six pairs of twins. He wishes to set up teams for a quiz, but wants to avoid putting any pair of twins into the same team. Subject to this condition:
i) In how many ways can he split them into two teams of six?
ii) In how many ways can he split them into three teams of four?</p>

3. In the cyclic quadrilateral ABCD, the diagonal AC bisects the angle DAB. The side AD is extended beyond D to a point E. Show that CE=CA if and only if DE=AB.</p>

4. The equilateral triangle ABC has sides of integer length N. The triangle is completely divided (by drawing lines parallel to the sides on the triangle) into equilateral triangular cells of side length 1.
A continuous route is chosen, starting inside the cell with vertex A and always crossing from one cell to another through an edge shared by the two cells. No cell is visited more than once. Find, with proof, the greatest number of cells which can be visited.</p>

5. Let G be a convex quadrilateral. Show that there is a point X in the plane of G with the property that every straight line through X divides G into two regions if and only if G is a parallelogram.</p>

6. Let T be a set of 2005 coplanar points with no three collinear. Show that, for any of the 2005 points, the number of triangles it lies strictly within, whose vertices are points in T, is even.</p>

That is one full paper, that you get 3 1/2 hours to do. We were told that we would be doing very well if we manage to answer 2.</p>

Two more from another paper:</p>

1. Find four prime numbers less than 100 which are factors of 3^32 - 2^32.</p>

3. The number 916238457 is an example of a nine-digit number which contains each of the digits 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not. How many such numbers are there?</p>

 

[Денг Гордон]

Anyone good with the chain rule.</p>

I solved the following (find dy/dx)</p>

y = sin (3x + 1)</p>

y= cos (square root (3)*x)</p>

y= 5 cot (2/x)</p>

y= (x + square root (x)) ^ -2</p>

y= sin^-5(x) - cos^3(x)</p>

but then I got to</p>

y= sin^3(x)tan(4x)</p>

y= 3 / (square root(2x +1))</p>

y= (1 + cos^2(7x))^ 3</p>

couldn't solve those three....</p>