VCE Methods Question Thread!

[BeadieG]

could someone please explain this question to me?

find the antiderivative of (3x + 1) / x using the absolute value function in your answer.

im confused as to what you do with the 3x + 1 when using the antiderivative formula thing. thanks.

[JIN1N]

I think it might be a typo given it looks like A is suggesting you take the area of the rectangle and subtract the integral within the same bounds. Note that integrating with respect to x or y doesn't actually matter, since they are just using the change of variables x = y. Pretty sure it should be \(27 - \int_{-8}^1 \sqrt{-y+1} \ dy\).

Hope this helps

Thanks for the reply!

[fun_jirachi]

Note that \(\frac{3x+1}{x} = 3+\frac{1}{x}\). This can be seen by splitting up the fraction and simplifying. Hopefully this clears things up

Hence, \(\int \frac{3x+1}{x} \ dx = \int 3+\frac{1}{x} \ dx = 3x + \ln |x| + C\).

[BeadieG]

ohhh, duh
thankyou!

[Ruchir]

Hey can anyone help in this question
Find area common to y=x-2 and x=y^2
I can’t really understand how to properly approach this question
I found the coordinates they are equal at to be (1,1) and (4,2)
I used this integral
But I did not get the right answer
I think the reason is because I did not use.  -sqroot(x)
But I don’t know how to use or approach it

[fun_jirachi]

I have a few questions in return for you that will help you with this question:

- Does the question say area or the integral? Integrals use signed area, 'area' does not imply that
- If in doubt, graph the functions. Is there common area outside the bounds you've used?
- Is there an easier way to find this? Hint: change of variables

Hope this helps

[Ruchir]

I have a few questions in return for you that will help you with this question:

- Does the question say area or the integral? Integrals use signed area, 'area' does not imply that
- If in doubt, graph the functions. Is there common area outside the bounds you've used?
- Is there an easier way to find this? Hint: change of variables

Hope this helps

This is the exact wording of the question
Find the exact area of region bounded by the graph with the equation....
When I roughly graphed it I saw zero as well, but I wasn’t sure how to use so I stuck with what I found.
Also I tried using y as the variable to integrate but I wasn’t sure about the lower and upper limits for it

[fun_jirachi]

I wasn't actually asking for the question; but that does help - you should now have a clearer understanding of what you should be doing. Be careful with integration, you want the unsigned area, not the signed area as integration would give you.

You don't have to use y to integrate. Would the area common to \(y=x-2, x=y^2\) be the same as the area common to \(x=y-2, y=x^2\)? Is this an easier area to find? It's often easier to reframe the question rather than tackle it head on.

I'm glad you've addressed the zero; that was the reasoning behind my last hint, which I've more explicitly restated

[Ruchir]

Thanks a a lot!!!! 😃
That helped me get the right answer.
It was really interesting that the inverse functions had the same area in common.
Also on a curious note what other method could you use to solve this question.

[fun_jirachi]

It was really interesting that the inverse functions had the same area in common.

Really, this is just a result of the chain rule - if you're curious and willing to spend more time on exploring why, definitely go ahead and do it

Also on a curious note what other method could you use to solve this question.

You could also have just integrated with respect to the y-variable. The area bounded by the two curves lies between y=2 and y=-1; similar logic applies to the way you found it; just treat the y-axis as the x-axis and vice versa like you would have for more 'conventional integration' (at least, the way you're taught initially with respect to x). 

If you really wanted to punish yourself, you could find the area between \(\sqrt{x}, y = 0, x = 4\), the area between \(-\sqrt{x}, y = 0, x = 1\), then add and subtract the areas of the relevant triangles.

Hope this helps

[arnavg2207]

What the question really tells you is that \(P(X \leq 1.8 ) = 0.9\) and \(P(X \leq 1.35) = 0.15\). We can standardise this to obtain \(P\left(Z \leq \frac{1.8-\mu}{\sigma}\right)  = 0.9\) and \(P\left(Z \leq \frac{1.35-\mu}{\sigma}\right)  = 0.15\). Using your table of values or whatever software you have available to you, we can deduce that \(\frac{1.8-\mu}{\sigma} = z_1\) and \(\frac{1.35-\mu}{\sigma} = z_2\), where \(z_1, z_2\) are the z-scores associated with 0.9 and 0.15 respectively on the standard distribution.

Because we now have two simultaneous equations and two unknowns, you can solve this using your favoured method for simultaneous equations. The resultant values for \(\mu\) and \(\sigma\) will give you the answer

You're definitely right in pointing out that the next question uses similar ideas; give it a shot and let us know if you get stuck!

Hope this helps

Thank you so much! I finally got up to doing the second question and can do it now

[pans]

Hello,
This is a 1 mark Q but Im having trouble understanding how s=1/r. I tried to find the inverse but my CAS gave me a weird answer. I wanted to approach this Q by finding the inverse and seeing how the relationship changes between s and r but that didnt work out : (

[james.358]

Hey Pans!

I remember doing this exact question last year! (MAV 2016 last question IIRC?)

While doing the question I gave up trying to solve it mathematically after a couple min, and just guessed that s = 1/r from the previous question. (which had r=1/4 and s=4 to give 3 intersections). I'm assuming thats what they want you to do, since after the exam I solved the question properly playing around with Desmos sliders and using calculus, and there were like 5 cases, and s = 1/r was only one of the cases.

In the future if you just use an "educated guess" for a lot of these separator 1 markers you'll be surprised how many of them you can get correct.

Hope this helps!
James

[schoolstudent115]

Hey Pans!

I remember doing this exact question last year! (MAV 2016 last question IIRC?)

While doing the question I gave up trying to solve it mathematically after a couple min, and just guessed that s = 1/r from the previous question. (which had r=1/4 and s=4 to give 3 intersections). I'm assuming thats what they want you to do, since after the exam I solved the question properly playing around with Desmos sliders and using calculus, and there were like 5 cases, and s = 1/r was only one of the cases.

In the future if you just use an "educated guess" for a lot of these separator 1 markers you'll be surprised how many of them you can get correct.

Hope this helps!
James

The way I solved this is that there is a point of inflection at (1/r, s). If the two equations intersect, this is identical to having one of the equations interesting with the line y=x. Hence, due to the concavity of the cubic, if the line y=x intersects with the POI, then you must have 3 intersections. So this means there is an intersection at the point (1/r, 1/r), meaning the y coordinate (s) is 1/r.

I should also have noted, r<0 (needs to be if you notice discriminant is itself -r^3). This makes the y=x line an ‘equivalent’ intersection.

[amyzzwq]

 Is there an error on the answer sheet here? Shouldn't the last translation be pi/3 units in the negative direction of the x-axis?