VCE Specialist 3/4 Question Thread!

[j.wang]

Sorry for the late response :/

I end up getting the latter answer

my syntax input is

Thanks so much Sine, I was staring at the cas for what felt like hours

[Gogo14]

Help with question 4.Answer is 8pi. When you rotate x^2=4-y, arent you rotating both branches? So the volume will overlap, so shouldnt you half the volume?

[Syndicate]

Help with question 4.Answer is 8pi. When you rotate x^2=4-y, arent you rotating both branches? So the volume will overlap, so shouldnt you half the volume?

Yes, there is an overlap, but does this really affect the amount of volume you will get? (No, as you are basically finding the same volume over the same region)

[Joseph41]

Hey there! Welcome to the VCE Specialist Maths Question Thread.

If you have a Spesh question you'd like answered, you'll first need to make an ATAR Notes account - it should take about four seconds. Then, simply scroll down and ask your questions in the "Quick Reply" box. I've included a picture below to show you what it looks like.

Alternatively, you have in this thread almost 600 pages of Specialist questions and answers! All the best!

[vasuk]

Could someone please explain why it would be wrong to see this question as a hemisphere, minus the volume of the solid of revolution formed between the curve y = x^2 and the x axis between zero to one?

[RuiAce]

Could someone please explain why it would be wrong to see this question as a hemisphere, minus the volume of the solid of revolution formed between the curve y = x^2 and the x axis between zero to one?

A parabola does not look like a circle. When a circle is rotated, the solid of revolution will be a sphere. When a parabola is rotated, the solid of revolution will be a paraboloid

[vasuk]

A parabola does not look like a circle. When a circle is rotated, the solid of revolution will be a sphere. When a parabola is rotated, the solid of revolution will be a paraboloid

thanks a lot!!

[KLRah]

For the function f(x)= 2cos(ax)+1, where x is between [0,4]. Find the largest value of a to three decimal places for which their is only one point on the graph at which the gradient is equal to the average gradient of the function over the specified domain. Any help would be appreciated

[Willba99]

hey guys.
so long story short, i bit off more than i could chew with the take-home part of my specialist SAC and i have to solve

192600pi+60000=1254pi(x)+12500sin(pi(x)/100)-40000sin(pi(x)/200)

any help solving this would be greatly appreciated. Basically I'm looking for solutions of something in the form of a=bx+csin(dx)-esin(fx)

[peterpiper]

Hey, I need some help 

I was wondering where you might get the z 'value' for the confidence intervals. Like for the approximate 95% confidence interval, the z value is 1.96. I was wondering how you might get the z value for a 90% interval or an 85% interval etc.

Apparently we use the inverse normal function on our CAS, but I can't seem to recall it correctly...

[keltingmeith]

hey guys.
so long story short, i bit off more than i could chew with the take-home part of my specialist SAC and i have to solve

192600pi+60000=1254pi(x)+12500sin(pi(x)/100)-40000sin(pi(x)/200)

any help solving this would be greatly appreciated. Basically I'm looking for solutions of something in the form of a=bx+csin(dx)-esin(fx)

You already have something of that form.

Hey, I need some help 

I was wondering where you might get the z 'value' for the confidence intervals. Like for the approximate 95% confidence interval, the z value is 1.96. I was wondering how you might get the z value for a 90% interval or an 85% interval etc.

Apparently we use the inverse normal function on our CAS, but I can't seem to recall it correctly...

Yup, you do. What you've gotta do is remember WHERE the z-value comes from - and from that, what even the hell a confidence interval is in the first place.

The easiest thought is that it's the interval in which you are "x% confident the true mean is". More strictly speaking, it's the interval over which there's a x% probability that the mean exists. Using notation (where Z is the standard normal distribution), for the 0.95 case this becomes:



Using our knowledge of normal distributions, we can rewrite this like so:



And now we'll use the fact that the sum of all probabilities is 1:



And now, the symmetry of the normal distribution:



Now, we simplify:



Which you can use the inverse normal calculator for! Try it out, you should get 1.96. To get this to work for some other percentage, just follow the same steps, let me know if you have any issues.

EDIT: TeX works differently to how I remember. D: Give me a bit to fix it up...
EDIT 2: Fixed!

[peterpiper]

You already have something of that form.
Yup, you do. What you've gotta do is remember WHERE the z-value comes from - and from that, what even the hell a confidence interval is in the first place.

The easiest thought is that it's the interval in which you are "x% confident the true mean is". More strictly speaking, it's the interval over which there's a x% probability that the mean exists. Using notation (where Z is the standard normal distribution), for the 0.95 case this becomes:



Using our knowledge of normal distributions, we can rewrite this like so:



And now we'll use the fact that the sum of all probabilities is 1:



And now, the symmetry of the normal distribution:



Now, we simplify:



Which you can use the inverse normal calculator for! Try it out, you should get 1.96. To get this to work for some other percentage, just follow the same steps, let me know if you have any issues.

EDIT: TeX works differently to how I remember. D: Give me a bit to fix it up...
EDIT 2: Fixed!

Thank you! That helped a lot!

[peanut]

When finding confidence intervals, is it ok just to use the z-value up to two decimal places? I don't remember the question, but there was a question where using the calculated z-value (using inverse normal calculator) and 1.96 gave different 95% confidence intervals (to two decimal places)

[peterpiper]

I have another question: I was wondering how they got the "n-1" part of the formula to get the sample variance. Or simply just what the formula means haha.


I thought it'd just be 'divide it by N' because mean X is equal to X/n... a bit confused with this tbh lmao.

Thanks in advance!

[Gogo14]

How do u do this question using volumes of revolution?