man0005's specialist question thread :)

[man0005]

Give parametric equations for the following:
1) (x-1)^2/9  + (y+3)^2 / 4 = 9

Determine the cartesian equation of the curve determined by the following parametric equation:
x = 1 - sec(2t)
y = 1 + tan (2t)
t : (pie/4 , 3pie/4)

[xZero]

firstly
you rearrange
x = 1-sec(2t) into x-1 = -sec(2t)
y = 1+tan(2t) into y-1 =t an(2t)

then you square them so (x-1)^2 = sec^2(2t) and (y-1)^2 = tan^2(2t)

then using the trig identity: 1 + tan^2(x) = sec^2(x),
1+ tan^2(2t) = sec^2(2t)
1+ (y-1)^2 = (x-1)^2
(x-1)^2 - (y-1)^2 = 1

now since pi/4< t < 3pi/4, we can see that pi/2< 2t < 3pi/2 (multiply it by 2 to match the 2t)
the range of 2t is the third quadrant so sec(2t) will always be negative or 0 so the domain of the equation is 1<x<infinite
the range of the equation should be R <- need verify

[m@tty]

1.

[man0005]

Does it matter which one is x and which one is y?

[kamil9876]

You mean, would x=9sint+1 and y=6cost -3 also work? yes it would

edit: However it is a good habit to associate with and with if possible. Here is an example where it is important:

parametrize  with the additional constraint that (half of an ellipse). State the domain of any parameter used.

[man0005]

oh do you get the wrong y-intercepts if you do it the other way?

[kamil9876]

the point is, you can do it two ways:

Solution 1:

and the domain of is

Solution 2:

and the domain of is

So you see, the domains for are different if you do it differently, but solution 1 is easier to visualize because of your experience with circular functions.

[man0005]

oh okay thanks heaps

im stuck on this question now :/

Show that every circle that touches the x axis and the y-axis has an equation of similar form to x^2 + y^2 - 2ax - 2ay + a^2 = 0

[xZero]

ok firstly the standard equation of a circle is x2 + y2 = a2 (where a = radius of the circle), for the circle to only touch the x axis and the y axis, you can move the circle by a units right and up to the first quadrant (you can move the circle to any other quadrant)

(x-a)2 + (y-a)2 = a2
x2 - 2ax +a2 + y2 - 2ay + y2 = a2
x2 + y2 - 2ax -2ay +a2 =0

you can then move the circle to other quadrant and repeat the steps above

[kamil9876]

your approach works for the first and third quadrant only.

The real question should be show that every circle that touches the x axis and the y axis has the form:

[xZero]

your approach works for the first and third quadrant only.

The real question should be show that every circle that touches the x axis and the y axis has the form:

how come?
1st quadrant x2 + y2 - 2ax - 2ay +a2 = 0
2nd quadrant x2 + y2 + 2ax - 2ay + a2 =0
3rd quadrant x2 + y2 + 2ax + 2ay + a2 = 0
4th quadrant x2 + y2 -2ax + 2ay + a2 = 0

[kamil9876]

Firstly let me restate what I wanted to say just to remove ambiguity, every such circle has either one of these forms:

(1)
(2)

For example if it is a circle or radius r centre at (r,r) then we have form (1) with a=r. If it is a circle of radius r centred at (-r,-r) then we have form (1) with a=-r. If it is a circle or radius r centred at (r,-r) then we have form (2) with a=r, while if it is at (-r,r) then it is form (2) with a=-r.

The problem with the original problem statement is that if we have the centre at what can we set to? we need the option of form (2) to allow for the coefficients of and to have different sign (as you have) because the way they have written it we need those coefficients to both be the same ()

[golden]

Express the equation in the modulus–argument form, 0 < theta < pi/2:

1 + tan i

The solutions state to use the modulus argument, which gets sec(theta). But it then multiplies sec(theta) by cis(theta). How did the cis(theta) come about?

Thanks.

[kamil9876]

Are you saying that you want to write in polar form over that domain?. Well you must write it in the form you have figured our correctly, now you must find the argument , but from a simple sketch and trig you see and so since your complex number is in the first quadrant and this gives so:

[man0005]

got a really good one for you kamil d:

(sin x + sin 3x + sin 5x)
____________________      =   tan 3x

(cos x + cos 3x + cos 5x)