VCE Specialist 3/4 Question Thread!

[lzxnl]

I never ever used that method. Probably would have saved me lots of time hahahaha.

[Sanguinne]

The graph of the circle with parametric equations x = 2 cos(2t) and y = 2sin(2t) is dilated by a factor 3 from the x-axis. For the image curve state:
a) the possible pair of parametric equations

Wouldn't we change the y to 3y making 3y=2sin(2t) or is this wrong as the answer says y=6sin(2t)

[brightsky]

You change y to y/3.

[Sanguinne]

You change y to y/3.

Why?

[Ancora_Imparo]

If you are dilating by a factor of 3 from the x-axis, every new y-value would be 3 times larger than it was previously. If we let the new y-values be y', then:


Rearranging to make y the subject gives:


Substituting this back in the parametric equation gives:



Changing y to 3y would cause a dilation by a factor of 1/3 from the x-axis.

[Cort]

I'm sorry, but would anyone give me a step by step, otherwise, an idea how to tackle this? I've been slowly doing the easier ones for the past 3-4 hours..needless to say it isn't helping too much. I can understand drawing circular functions for sin/cos/tan... but somehow I'm getting these all screwed up.

Furthermore, do you have recommendations for how I can improve on this? Do more questions, or get other textbooks with the easy san/cos/tan and then do its reciprocal?

Cheers,
Cort

[revcose]

Learn the basic shape, then apply transformations. You only really need to pay attention to assymptotes, special points, and shape (reflections). With the form they are in , you do dilations and reflections, and then translations.

Hopefully these graphs demonstrate how you change the graphs. The red is the basic form, the blue is after the dilations and reflections, and the green is the final graph.
https://www.desmos.com/calculator/aixybmqpup
https://www.desmos.com/calculator/mpctsbsahq

[lzxnl]

If you seriously get stuck, take the analogous sine function and work out how to take the reciprocal of that. So when sine x increases csc x decreases, when sin x=0 csc x is undefined, when sin x is at its max csc x is at its min etc.

[Alex.vce]

Hey guys, could someone walk me through this problem:


Thanks,
~ Alex

[IndefatigableLover]

Hey guys, could someone walk me through this problem:
(Image removed from quote.)

Thanks,
~ Alex

a)
However we want it in terms of 'a', 'b', and 'c' which has been given already. Now vector AG is the same as vector BC which gives us an answer in terms of a,b and c.




Now we can substitute in the vectors for a,b and c.

Spoiler

Therefore answer = a+c-b

b). (Vector BG is the same as CD when looking at the diagram).

Spoiler

a+c-b-b = a+c-2b

[Alex.vce]

I think I see where I went wrong: Instead of treating the origin algebraically, I assumed it needed a definite position so I got lost.

Thanks,
~Alex

Update:
So for part b I worked out:

[Sanguinne]

Question 6 of 6E in cambridge book

For f:[0,20] ---> R, f(x) =

c) Describe the rate of change of gradient for the function.
I assumed that the rate of change of the gradient was decreasing for [20/3, 20] as this is when the second derivative is below the x-axis. However the answer states [0,20], so where have I gone wrong?

Thanks.

[Ancora_Imparo]

The function is . The gradient of the function is given by . The rate of change of gradient for the function is thus .





The graph of is a straight line with a negative gradient (). Thus, it is decreasing over its entire domain, ie: (0, 20). What you have calculated is when the rate of change of the gradient is negative, which, as you said, is when the second derivative is below the x-axis.

[Nato]

Just looking over the conjugate factor theorem, I am having trouble understanding why when the coefficients of P(z) are real, complex roots occurring in conjugate pairs. Pretty much, why do the roots occurring in conjugate pairs?

[lzxnl]

I'll give you a few ways of looking at it.
By the fundamental theorem of algebra (one version of it at least), any real polynomial can be factorised as a product of real linear terms (like z-1) and irreducible quadratic terms (quadratic expressions whose discriminants are negative, like z^2+1). Now, any complex roots must therefore arise from factorising these irreducible quadratic terms. As the roots of an irreducible real quadratic are complex conjugates, and as all of the other roots of this polynomial are real (due to the linear terms), the only complex roots occur in conjugate pairs if the original polynomial has only real coefficients.

Or, by the factor theorem, if P(z=a+bi)=0, then z-a-bi is a linear factor of the real polynomial. Only by multiplying z-a-bi by z-a+bi can you possibly get a polynomial that has real coefficients (you can try this if you want; multiply z-a-bi by z-c-di and you'll find that for this polynomial to be real for all z, a=c and b=-d). Thus, another complex root has to be a-bi. Et cetera.