Mathematics HL/SL/Studies Help

[Drake Glau]

The answers are :

a) (15^2)(2) = 225 (cm^2)

b)Area OAB = 15^2 Sin 2 = 102.3

Area = 225-102.3 = 122.7 (cm^2)

= 123 (3sf) <-- I don't know what this means!!

Then I have no idea what they did or I have my diagram wrong...and 3sf is 3 significant figures...

Fixed part b, just didn't do it their way I guess...

Still not sure on part a, am I finding the area of the triangle AOB where AO and BO are both 15cm radii? and the angle AOB=2 radians?

Edited April 19, 2011 by Drake Glau

[Ice Cream is really yummy]

Hmmm... I forgot to say that it's triangle AOB.... sorry

[genepeer]

You must know these formulae

Area of a sector = theta/2*r²... theta must be in radians.

Area of the triangle = 1/2*ab sin(theta)... where a=b=r in the diagram.

Therefore:

Segment = theta/2*r² - 1/2*r²sin(theta)

= (theta-sin(theta))/2*r²

another useful formula would be:

Length of an arc = theta*r (theta in radians)

[dessskris]

Write down, in an exact form, all the 8th roots of 1 in the form a+bi.

My working so far:

z⁸ = 1

z⁸ = cis(0)

z = cis(0), cis(π/4), cis(π/2), cis(3π/4), cis(π), cis(-π/4), cis(-π/2), cis(-3π/4)

Then express them in the form of a+bi. (I can do it myself later.)

What I worry about is the cis(π) part

I know it's the one on the left hand side, on the negative real axis, but what is the argument? Is it π or -π? Actually when I express cis(π) and cis(-π) in a+bi I will eventually get -1 from both, but what is actually the argument? Does it make any difference?

Thank you!

PS. Hey Gene-Peer, long time no see!

[annaanna]

In this question, a unit vecot represents a displacement of 1 metre.

Am miniature car moves in a straight line, starting at the point (2,0). after t seconds, its position, (x,y), is given by the vector equation:

(x,y)=(2,0)+t(0.7, 1)

a) how far from the point (0,0) is the car after 2 seconds?

b) find the speed of the car

c) obtain the equation of the car's path in the form ax+by=c.

Another miniature vehicle, a motorcycle, starts at the point (0, 2), and travels in a straight line with a constant speed. The equation of the path is:

y=0.6x +2 (x is greater than or equal to 0)

Evantually, the two miniature vehicles collide.

d) find the coordinates of the collision point.

e) if the motorcycle left point (0,2) at the same time as the car left the point (2,0), find the speed of the motorcycle.

[Drake Glau]

In this question, a unit vecot represents a displacement of 1 metre.

Am miniature car moves in a straight line, starting at the point (2,0). after t seconds, its position, (x,y), is given by the vector equation:

(x,y)=(2,0)+t(0.7, 1)

a) how far from the point (0,0) is the car after 2 seconds?

(2,0)+2(0.7,1)

(2,0)+(1.4,2)

(3.4,2) is the new coordinate of it's location after 2 seconds.

Distance formula is sqrt[(x₁-x₂)²+(y₁-y₂)²]

plug in your stuff and you get 4.31

b) find the speed of the car

Do distance formula again but use the original position instead of the origin that was specified in part a to get a distance from the start to the new position, divide it by 2 and you get your unit length per second. I got 1.22m/s

c) obtain the equation of the car's path in the form ax+by=c.

Make an equation use the two points (the new position point and the beginning displacement point)

(y₁-y₂)/(x₁-x₂)=your slope

Another miniature vehicle, a motorcycle, starts at the point (0, 2), and travels in a straight line with a constant speed. The equation of the path is:

y=0.6x +2 (x is greater than or equal to 0)

Evantually, the two miniature vehicles collide.

d) find the coordinates of the collision point.

Change your equation from part c to y=mx+b form and set it equal to 0.6x+2 and solve for x. Then plug the x back into 0.6x+2 to obtain your y value for the coordinate.

e) if the motorcycle left point (0,2) at the same time as the car left the point (2,0), find the speed of the motorcycle.

Use distance formula to find how far (2,0) is from the collision point and then divide by the car's speed you get on part b and this gives you how many seconds has passed.

Now find the distance between (0,2) and the collision point to find how far the motorcycle traveled and divide by the time you calculated just a second ago and that gives you a speed in unit length per second

Edited April 20, 2011 by Drake Glau

[genepeer]

What I worry about is the cis(π) part

I know it's the one on the left hand side, on the negative real axis, but what is the argument? Is it π or -π? Actually when I express cis(π) and cis(-π) in a+bi I will eventually get -1 from both, but what is actually the argument? Does it make any difference?

Thank you!

PS. Hey Gene-Peer, long time no see!

I know! Not on the internet as much as before, our ISP got cut off and the new internet sucks

As for your question, the argument doesn't make any difference. Angles are periodic, 5π/2 = π/2 = -3π/2 or in degrees -20° = 340° = 700° but we usually try to keep in the [0,2π[ or ]-π,π] ranges for simplicity's sake.

[dessskris]

Complex Numbers

if -cis(-4π/5)=cis(π/5),

wouldn't cis(π/5)+cis(-4π/5) be zero??

if yes, then what is this...?

can anybody explain that to me? thank you!!

because I need to simplify:

4+cis(-4π/5)+2cis(-3π/5)+cis(-2π/5)+2cis(-π/5)+2cis(π/5)+cis(2π/5)+2cis(3π/5)+cis(4π/5)

and the answer is 5 :/

thanks!

---

@Gene-Peer aww poor you I hope you're back because I need some help thanks for helping with the previous question!

[Hexa]

A factory makes calculators. Over a long period, 2 % of them are found to be faulty. A random sample of 100 calculators is tested.

Find the probability that three calculators are faulty.

Find the probability that more than one calculator is faulty.

Lost completely

[Abu]

A factory makes calculators. Over a long period, 2 % of them are found to be faulty. A random sample of 100 calculators is tested.

Find the probability that three calculators are faulty.

Find the probability that more than one calculator is faulty.

Lost completely

It's a binomial distribution with X ~ B(100,.02), where X is your random variable, 100 is your n and .02 is your p.

For the first part, you have to calculate P(X=3) and for the second, you have to calculate P(X>1). To calculate P(X=3), use the binomial formula in your formula booklet with x =3 and. For the second part, you will need to calculate (1 – P(X≤1)) which can be calculated by (1 – (P(X=0)+P(X=1)).

Once you see it is a binomial, it's pretty easy. A binomial distribution is any where the probability of success (p) is fixed, with n trials. Whenever they ask for more than something, you will have to calculate the probability of 1–probability of something because more than can go on to infinity.

[Emmi]

I take SL Math, and this is a question from a past paper my teacher assigned us for homework:

I'm terrible with vectors, so some help would be much appreciated! Thanks!

[Summer Glau]

I take SL Math, and this is a question from a past paper my teacher assigned us for homework:

I'm terrible with vectors, so some help would be much appreciated! Thanks!

Alright, so you have two vector equations. I'm just going to use s as the variable to replace lambda since there is no lambda sign on my keyboard Also I'm just going to use coordinate form (x,y) to represent vectors since I can't type vectors either...

If you multiply the s and t into the vector equations, you'll get:

r1 = (5,1) + (3s,-2s)

r2 = (-2,2) + (4t,t)

Remember that the r in a vector equation represents (x,y) for vectors (the vector in the x axis and the vector in the y axis)

So if you write it as:

(x,y) = (5,1) + (3s,-2s)

(x,y) = (-2,2) + (4t,t)

then you can form parametric equations for these vectors. So by adding these sets of vectors together, you'll get

x1 = 5+3s

y1 = 1-2s

x2 = -2+4t

y2 = 2+t

Now, since you want a point of intersection, these two lines have to meet at the same point. Therefore they will have the same x and y coordinates when they do meet. So, you can equate the x and y equations that you just formed.

So,

x1 = x2

5+3s = -2+4t

3s-4t = -2-5

3s-4t = -7

y1 = y2

1-2s = 2+t

-2s-t = 2-1

-2s-t = 1

Now you can use substitution or elimination to solve for s and t

Sooooo I will use substitution...

-t = 1+2s

t = -1-2s

3s-4(-1-2s) = -7

3s+4+8s = -7

11s = -11

s = -1

t = -1-2(-1)

t = -1+2

t = 1

Now you have the values of s and t, so just plug in s or t into your original equations (pick one set of equations, it doesn't matter which one you pick since the answers should end up being the same either way, since they intersect)

So:

x1 = 5+3s

x1 = 5+3(-1)

x1 = 5-3

x1 = 2

y1 = 1-2s

y1 = 1-2(-1)

y1 = 1+2

y1 = 3

So the point of intersection is (2,3). This is in vector form by the way.

[genepeer]

Complex Numbers

if -cis(-4π/5)=cis(π/5),

wouldn't cis(π/5)+cis(-4π/5) be zero??

if yes, then what is this...?

can anybody explain that to me? thank you!!

3e-14 - 4e-14i = 0.000000000000003 - 0.000000000000004i

The calculator simply isn't accurate enough to give you exactly zero due to approximations in intermediate steps.

[Emmi]

Thanks Summer Glau! It makes sense now.

Only 9 days until the exam though.

[Ice Cream is really yummy]

You must know these formulae

Area of a sector = theta/2*r²... theta must be in radians.

In the SL formulae booklet, it says:

Area of a sector: A = (theta)(r²)/2

Was your way just a different way of representing the same formula, or did you just get mixed up?

[Chronofox]

You must know these formulae

Area of a sector = theta/2*r²... theta must be in radians.

In the SL formulae booklet, it says:

Area of a sector: A = (theta)(r²)/2

Was your way just a different way of representing the same formula, or did you just get mixed up?

It's just a different way of representing the same formula. Order of operations, remember?

[Ice Cream is really yummy]

You must know these formulae

Area of a sector = theta/2*r²... theta must be in radians.

In the SL formulae booklet, it says:

Area of a sector: A = (theta)(r²)/2

Was your way just a different way of representing the same formula, or did you just get mixed up?

It's just a different way of representing the same formula. Order of operations, remember?

Since it's all division and multiplication, it dosen't matter in what order I do it it, right? Because I had actually used what Gene-Peer wrote and I made a ton of mistakes

[Drake Glau]

You must know these formulae

Area of a sector = theta/2*r²... theta must be in radians.

In the SL formulae booklet, it says:

Area of a sector: A = (theta)(r²)/2

Was your way just a different way of representing the same formula, or did you just get mixed up?

It's just a different way of representing the same formula. Order of operations, remember?

Since it's all division and multiplication, it dosen't matter in what order I do it it, right? Because I had actually used what Gene-Peer wrote and I made a ton of mistakes

Make sure you square the radius first but then yea, it doesn't matter.

(theta)/2*r^2 is the same as [(theta)/2]*(r^2/1) which is the same as (theta)(r^2)/2

[timtamboy63]

Question is:

Simplify:

2a•(a×b)

The answer is 0, i'm not too sure how

[Keel]

Question is:

Simplify:

2a•(a×b)

The answer is 0, i'm not too sure how