Mathematics HL/SL/Studies Help

[dessskris]

(Math HL - Complex numbers)

How do I find arg(z) if

z = 1 - i cot(π/8)

?

I know I need to express z in polar form first but how do I get there? I can only get

z = [sin(π/8) - i cos(π/8)] / sin(π/8)

How do I proceed?

[Slovakov]

If you want to get a polar form, you need to find an angle and modulus of z.

When you have a form z=a+bi and you put it on Argand diagram, you obtain a line, and the nagle you need is the angle between this line and Re axis. So, just as in case of linear function, b/a=tan(theta), and hence theta=arctan(b/a)

For the number z you gave: arg(z)= arctan(-cotπ/8)

Modulus is just (a^2 + b^2)^(1/2), so here it is (1 + (-cotπ/8)^2)^(1/2)=(cosec(π/8)^2)^(1/2)=cosecπ/8

So the polar form of z is z= cosecπ/8(cos(arctan(-cotπ/8))+isin(arctan(-cotπ/8))

Typing that in my TI is easier than doing it here I hope at least part of it is useful, anyway...

Edited April 16, 2011 by Slovakov

[dessskris]

OHHH right,

I kind of forgot that arg(z) is simply arctan(b/a)!

I thought I needed to express it in polar form first and then arg(z)=θ!

That's very detailed and helpful. Thank you very much!

[Markee]

I have a couple of questions.

-How would I go about in calculating the area of the triangle OPQ, where O is the origin with these? Tangent to the curve y=3x sq.root of 1+2x where the point (4,36) meets the x-axis at P and y-axis at Q.

-How can I find the coordinates of the two points on the curve y=x/1+x with the gradient being 1/9?

[Drake Glau]

I have a couple of questions.

-How would I go about in calculating the area of the triangle OPQ, where O is the origin with these? Tangent to the curve y=3x sq.root of 1+2x where the point (4,36) meets the x-axis at P and y-axis at Q.

-How can I find the coordinates of the two points on the curve y=x/1+x with the gradient being 1/9?

The first one, if I'm understanding this right...

You have O at (0,0) and P and Q are both on the derivative of 3xsqrt(1+2x) at point (4,36).

Find your derivative, plug in the x value of 4, this gives you the slope of the PQ segment of your triangle. With the slope you can form an equation for the tangent line. To get the y-intercept of that line (point Q of your triangle) plug in 0 into that tangent equation. To get the x-intercept (point P of your triangle) plug in 0 for Y and solve for X. This has now given you coordinates for all 3 lines segments of your triangle, do some distance formula, 0.5*b*h and you should be set.

2nd question:

Find derivative of your curve, set it equal to 1/9, solve for x, plug in the 2 values of x you get into the original equation and you get your two y coordinates to go with them

[Markee]

I have a couple of questions.

-How would I go about in calculating the area of the triangle OPQ, where O is the origin with these? Tangent to the curve y=3x sq.root of 1+2x where the point (4,36) meets the x-axis at P and y-axis at Q.

-How can I find the coordinates of the two points on the curve y=x/1+x with the gradient being 1/9?

The first one, if I'm understanding this right...

You have O at (0,0) and P and Q are both on the derivative of 3xsqrt(1+2x) at point (4,36).

Find your derivative, plug in the x value of 4, this gives you the slope of the PQ segment of your triangle. With the slope you can form an equation for the tangent line. To get the y-intercept of that line (point Q of your triangle) plug in 0 into that tangent equation. To get the x-intercept (point P of your triangle) plug in 0 for Y and solve for X. This has now given you coordinates for all 3 lines segments of your triangle, do some distance formula, 0.5*b*h and you should be set. --> About this one, finding the derivative, do I just simply plug in the x value? Then find the tangent to the curve? then plug in 0 to the tangent equation I got to Y so I can solve for X? Then, just find the area? - how did you get 0.5???

2nd question:

Find derivative of your curve, set it equal to 1/9, solve for x, plug in the 2 values of x you get into the original equation and you get your two y coordinates to go with them Ok, the only problem with this is, how do I derive my curve? differentiation?

Edited April 18, 2011 by MARK10
Making annotations clearer

[Drake Glau]

I have a couple of questions.

-How would I go about in calculating the area of the triangle OPQ, where O is the origin with these? Tangent to the curve y=3x sq.root of 1+2x where the point (4,36) meets the x-axis at P and y-axis at Q.

-How can I find the coordinates of the two points on the curve y=x/1+x with the gradient being 1/9?

The first one, if I'm understanding this right...

You have O at (0,0) and P and Q are both on the derivative of 3xsqrt(1+2x) at point (4,36).

Find your derivative, plug in the x value of 4, this gives you the slope of the PQ segment of your triangle. With the slope you can form an equation for the tangent line. To get the y-intercept of that line (point Q of your triangle) plug in 0 into that tangent equation. To get the x-intercept (point P of your triangle) plug in 0 for Y and solve for X. This has now given you coordinates for all 3 lines segments of your triangle, do some distance formula, 0.5*b*h and you should be set. --> About this one, finding the derivative, do I just simply plug in the x value? Then find the tangent to the curve? then plug in 0 to the tangent equation I got to Y so I can solve for X? Then, just find the area? - how did you get 0.5???

2nd question:

Find derivative of your curve, set it equal to 1/9, solve for x, plug in the 2 values of x you get into the original equation and you get your two y coordinates to go with them Ok, the only problem with this is, how do I derive my curve? differentiation?

the 0.5 is because the area of a triangle is one half times the base times the height...

Yes, find the derivative of your curve that it gave you, THEN plug in the x value of 4 and that will give you the slope of that tangent line which according to your question is the line segment PQ.

To get the x and y intercepts you plug in 0 for x and solve for y to get your y intercept and you plug in 0 for y and solve for x to get your x intercept.

For the 2nd problem, yes differentiate your curve using the quotient rule. Should get 1/(1+x)^2 if I did that right

Then you set that equal to 1/9 so:

1/(1+x)^2=1/9

9/(1+x)^2=1

9=(1+x)^2

square root is + or - so you end up with

-3=1+x

3=1+x

x=-4

x=2

Plug these into the original equation.

y=x/1+x

y=-4/1-4

y=4/3

and

y=2/1+2

y=2/3

(-4, 4/3)

(2, 2/3)

I think that's right

[Ice Cream is really yummy]

Man, I'm not getting Trig Does anyone have any suggestions... analogies... formulaes that they can share with me so that I can wrap my head around all of this?? Perhaps names of formulaes, or other stuff I can read?? I'm really struggling... math has never been easy for me. I'm great with everything else... and I LOVE chem I'm doing really well in it (predicted 7) buut.... Science and Math are forever linked together.. so if I don't get my math going... chem is slowly going to slip away from me....

PS, I AM getting myself a tutor pronto!! .... and I would LOVE to get any "help-get-trig-and-math-in-general-wrapped-around-my-math-resistant-brain" from any IBers too .... THANKS

[Drake Glau]

How deep into trig do you need? Math St? SL Math? HL Math?

There's the basic SOH CAH TOA for triangles...sin(x)=opposite/hypotenuse; cos(x)=adjacent/hypotenuse; tan(x)=opposite/adjacent

Also learn the unit circle, it helps. I don't really know what else to say unless you know what you need help with besides "trig" =/

[Ice Cream is really yummy]

How deep into trig do you need? Math St? SL Math? HL Math?

There's the basic SOH CAH TOA for triangles...sin(x)=opposite/hypotenuse; cos(x)=adjacent/hypotenuse; tan(x)=opposite/adjacent

Also learn the unit circle, it helps. I don't really know what else to say unless you know what you need help with besides "trig" =/

I'm in math SL I know the SOH CAH TOA and I'm feeling confident with that. Then we learnt the unit circle just last week... it kinda makes sense. I mean, it does.... but my teacher is a little stressed out right now and is going WAY too fast... we finished up like a whole unit in like, 5 days And I'm the type of person who needs to see the big picture and let everything sink in until I hear bells ringing in my brain Then I can make all the connections and I feel confident again

Edited April 19, 2011 by egr12resa

[Drake Glau]

Think of the unit circle like a normal graph almost but with radians for units. the cos(x) gives you the x coordinate and the sin(x) is like the y coordinate.

sin(90)=1 for example. At ninety degrees your point is at the top of circle and the sin means you're kind of looking at the y-coord so it's 1 whereas the cos(90)=0, the x-coord for the same point on the circle. Make sense?

Would help if you had specific examples of what you need help on Trig is big...

Edited April 19, 2011 by Drake Glau

[Summer Glau]

Ooo I got a question:

Find the coordinates of the point where the line with parametric equations x=1-t, y=3+t, and z=3-2t meets the YOZ plane.

And another one:

Find points on the line with parametric equations x=2-t, y=3+2t and z=1+t which are 5(sqrt)3 units from the point (1,0,-2)

This is a math SL question by the way. Thanks for any help

[Ice Cream is really yummy]

Thanks Drake Let's see... We're working with circles and triangles. And circular sectors, and arcs, and triangles overlapping circles, and we are supposed to find the shaded area (which would be the part of the triangle that is not overlapping with the circle). I'm not really good at explaining.... lots of my questions as to "refer to the picture below". But yeah, the questions look/sound somewhat like that.

How do I convert degrees into radians and vice versa? I've checked around... but I'm kinda confused now

[Drake Glau]

Use your unit circle for the conversions. every 45 degrees is 1pi/4, every 30 degrees is 1pi/6. 90 degrees is 1pi/2, 180 degrees is pi, 270 is 3pi/2, and 360 degrees is 2pi.

You can give the specifications of the picture if that's easier. Say a circle has radius r and the triangle ABD has AB=4 BD=6 or something, we can probably draw our own pictures and help you

Sam's conversion makes more sense...I just remember the circle =/

Edited April 19, 2011 by Drake Glau

[Summer Glau]

How do I convert degrees into radians and vice versa? I've checked around... but I'm kinda confused now

To convert radians to degrees, you multiply the radians by 180/pi. So if you have 2 pi radians and you want to find what that is in degrees, multiply 2 pi radians by 180/pi. So you should have:

= 2 pi x 180/pi

= 360pi/pi

The pi on the bottom and on the top will cancel, so you get

=360 degrees.

To convert from degrees to radians, the formula just works the other way around. So if you have 180 degrees and you want to find its radian equivalent, multiply by the reciprocal of the previous formula, which is pi/180. So you will have:

=180 x pi/180

=180pi/180

This time, there is 180 on top and on the bottom, so they will cancel. so then you will get

=pi radians

Hope this helps

[Ice Cream is really yummy]

Thanks you two!! It's starting to click a little....

HOw about... (I gotta translate this to english.. so hold up a sec)....

OH NEVERMIND!!! I'm just going to try to explain to you this diagram:

It's like a portion of a circle (like a quarter) and there's an arc as well as a chord. The radius is 15 cm and the angle (theta) is 2 radians.

a) calculate the area of the sector AOB <-- That's the triangle that the pizza-shaped-picture makes... the chord = opposite of O (theta)

b) calculate the shaded area of the circular segment <-- this is the area between the chord and the arc.

(I hope that you were able to make some sense out of that...)

[Drake Glau]

Yea, I understand the description, but 2 radians is 360 degrees...that's the whole circle O.o

2*180/pi=about 115 degrees...

Edited April 19, 2011 by Drake Glau

[Ice Cream is really yummy]

Isn't a whole circle 2(Pi)???

[Drake Glau]

Meh, just leave it in radians...

c²=b²+a²-2abcosC

a=15

b=15

C=2

chord length is about 25.24 (make sure your calculator is set to radians...)

Not sure why I did this, I feel like it's important for part b which I don't know how to do.

Area of sector=r²(theta)/2

15²(2)/2

15²=225cm^2

Wikipedia is awesome...

Circular segment area=(r²/2)(theta-sin(theta))

In your case that's (15²/2)(2-sin2) leave it in radians still

This gives you the area of that space between the chord and the arc. I misread wikipedia...

So yea, 122.7 or 123 to 3sf

Edited April 19, 2011 by Drake Glau

[Ice Cream is really yummy]

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!!