VCE Specialist 3/4 Question Thread!

[Mattjbr2]

When graphing the parametric equations: x=tan(t) and y=sec(t) for all real values of t into CAS, why do the asymptotes show up as solid lines and why is the bottom half of the hyperbola shown, when the range is supposed to be [1, infinity)? Attached is a photo of my CAS. Only the top portion of the graph should be shown, not the asymptotes nor the bottom part. Why is this happening?

Sec(x) is only [1,infinity) if the domain is [0,pi/2] (for example). If the domain is all real values, then the range is (-infinity,-1]U[1,infinity)

[VanillaRice]

When graphing the parametric equations: x=tan(t) and y=sec(t) for all real values of t into CAS, why do the asymptotes show up as solid lines and why is the bottom half of the hyperbola shown, when the range is supposed to be [1, infinity)? Attached is a photo of my CAS. Only the top portion of the graph should be shown, not the asymptotes nor the bottom part. Why is this happening?

Not sure why the asymptotes are drawn in. Might be some convention that the CAS uses for parametric functions?

The range of your function is the same as the range of y=sec(t) i.e. the range is all the possible values of sec(t). The range of sec(t) is both [1, inf) and (-inf,-1].

Hope this helps

[Mattjbr2]

Not sure why the asymptotes are drawn in. Might be some convention that the CAS uses for parametric functions?

The range of your function is the same as the range of y=sec(t) i.e. the range is all the possible values of sec(t). The range of sec(t) is both [1, inf) and (-inf,-1].

Hope this helps

I realized my error when I graphed sec(x) haha

[noregret]

Hello, how do I solve integration of 8artanx/1+x^2? I tried let u=artanx and du/dx=1/1+x^2, but I could not get the answer the book says, which is 4(artanx)^2+k

[RuiAce]

Hello, how do I solve integration of 8artanx/1+x^2? I tried let u=artanx and du/dx=1/1+x^2, but I could not get the answer the book says, which is 4(artanx)^2+k

[recess]

Hi just a quick question. How come for the POI, the y value is the same as the x value eventhough y=4x so shouldn't the y value be infact, 4 times the x value? Or am I just getting this wrong?

[Sine]

Hi just a quick question. How come for the POI, the y value is the same as the x value eventhough y=4x so shouldn't the y value be infact, 4 times the x value? Or am I just getting this wrong?

you are correct the graph doesn't match the maths shown

[dr_jl47]

Hi just a quick question. How come for the POI, the y value is the same as the x value eventhough y=4x so shouldn't the y value be infact, 4 times the x value? Or am I just getting this wrong?

is equivalent to the graph shows and

[noregret]

Hello, how do I solve this question.

Find the exact area between the y axis and each of the following curves

y=x^3 where x is a set of [1,2]?

[VanillaRice]

Hello, how do I solve this question.

Find the exact area between the y axis and each of the following curves

y=x^3 where x is a set of [1,2]?

HINT: Draw out the relevant graph section. Shade in the relevant area. You can see that the area lies in between the curve and the y-axis. However, your function is in terms of x. So, you must rearrange your function (in terms of y) to make x the subject. Now, you can integrate as normal (against the y-axis.

Note: another way to view the area is as an area between two curves (integrating against the x-axis). However, for this method you'll need two seperate integrals.

Hope this helps, post if you get stuck

[noregret]

HINT: Draw out the relevant graph section. Shade in the relevant area. You can see that the area lies in between the curve and the y-axis. However, your function is in terms of x. So, you must rearrange your function (in terms of y) to make x the subject. Now, you can integrate as normal (against the y
Note: another way to view the area is as an area between two curves (integrating against the x-axis). However, for this method you'll need two seperate integrals.

Thanks for the help and you always subtract from the higher y value or the higher x value?

Hope this helps, post if you get stuck

[VanillaRice]

Thanks for the help and you always subtract from the higher y value or the higher x value?

In terms of integrating? Not always. It will depend on whether your area is "above" or "below" the axis. If it's below the axis (i.e. on the negative side), you can swap the terminals around to negate the negative number (recall areas of curves using integration from Methods).

[TheAspiringDoc]

Is this correct:

Arccos, arctan etc should be used in the exam instead of cos^-1, tan^-1 etc, because 'arc-' implies that it is one to one but the ^-1 doesn't?

I don't really get how this makes it one to one anyway:
Arccos = sin^-1: [-1,1]-->R, sin^-1x=y

Couldn't that have many sin^-1 's stacked on top of each other, and thus not pass the vertical line test, and thus not be a function?

Thanks

[VanillaRice]

Is this correct:

Arccos, arctan etc should be used in the exam instead of cos^-1, tan^-1 etc, because 'arc-' implies that it is one to one but the ^-1 doesn't?

I don't really get how this makes it one to one anyway:
Arccos = sin^-1: [-1,1]-->R, sin^-1x=y

Couldn't that have many sin^-1 's stacked on top of each other, and thus not pass the vertical line test, and thus not be a function?

Thanks

By definition, the inverse trig functions have restricted ranges (allowing us to call them functions!). Many textbooks/mathematicians often use capital letters (Arcsin, Sin^-1) to define the principle domain/range that you see on your formula sheet. However, many don't use these capital letters (including your formula sheet and many calculators). In VCE, I would say that it is safe to assume that any inverse trig function is in the domain/range as defined in your formula sheet.

Hope this helps

[TheAspiringDoc]

RTP: cosec(x) + cot(x) = cot(x/2) where sin(x) =/= 0.

So far all I have done is:

Spoiler

But now I don't even know if that's of any use...

Any thoughts?