Specialist 1/2 Question Thread!

[Gogo14]

You can solve this question using the work-energy principle (I am not sure if it is part of the specialist curriculum).

Don't what that is lol.
But how do u do it by resolving vectors?

[anotherworld2b]

Could i please get help qith these 2 questions?

[Syndicate]

Could i please get help qith these 2 questions?

Have been really busy since the past week..  sorry about the late reply.

Q15)







Therefore it have been proven that x^5-x is divisible by 5, since x^5-x = 5n.

Q14 can be proved using a similar approach.

[anotherworld2b]

Thank you for the help.
I was also windering how would yoj solve this question?

Have been really busy since the past week..  sorry about the late reply.

Q15)







Therefore it have been proven that x^5-x is divisible by 5, since x^5-x = 5n.

Q14 can be proved using a similar approach.

[camrenis]

Thank you for the help.
I was also windering how would yoj solve this question?

Start with the formulas
\begin{align}
A&=2\pi rh+2\pi r^2\\
V&=\pi r^2 h
\end{align}
Substitute the volume in to \(V\) and solve for \(h\) in terms of \(r\).  Substitute this relationship into \(A\), and then you now should have \(A\) in terms of \(r\) only.  Derive \(A\) with respect to \(r\) and set to 0 to find the minimum value of \(r\).  The substitute this value of \(r\) into \(V\) with the volume to find \(h\).

Spoiler

\(h\) in terms of \(r\) is
\begin{align}
h=\frac{535}{\pi r^2}
\end{align}
Sub into \(A\) to get
\begin{align}
A&=2\pi r\frac{535}{\pi r^2}+2\pi r^2\\
&=\frac{1070}{r}+2\pi r^2
\end{align}
Derive with respect to \(r\)
\begin{align}
\frac{dA}{dr}&=-\frac{1070}{r^2}+4\pi r
\end{align}
Set to 0 to find where the minimum of \(A\) is (plotting \(A\) reveals that it is a local minimum and not a local maximum when \(\frac{dA}{dr}=0\)).
\begin{align}
0&=-\frac{1070}{r^2}+4\pi r\\
\implies r&=\left(\frac{535}{2\pi}\right)^\frac{1}{3}\\
&\approx 4.4
\end{align}
Sub into the volume equation to find \(h\):
\begin{align}
535 &= \pi \left(\frac{535}{2\pi}\right)^\frac{2}{3} \times h\\
\implies h&\approx 8.8
\end{align}

[anotherworld2b]

Thank you very much for your help
I was wondering would you do the same for this question?

Start with the formulas
\begin{align}
A&=2\pi rh+2\pi r^2\\
V&=\pi r^2 h
\end{align}
Substitute the volume in to \(V\) and solve for \(h\) in terms of \(r\).  Substitute this relationship into \(A\), and then you now should have \(A\) in terms of \(r\) only.  Derive \(A\) with respect to \(r\) and set to 0 to find the minimum value of \(r\).  The substitute this value of \(r\) into \(V\) with the volume to find \(h\).

Spoiler

\(h\) in terms of \(r\) is
\begin{align}
h=\frac{535}{\pi r^2}
\end{align}
Sub into \(A\) to get
\begin{align}
A&=2\pi r\frac{535}{\pi r^2}+2\pi r^2\\
&=\frac{1070}{r}+2\pi r^2
\end{align}
Derive with respect to \(r\)
\begin{align}
\frac{dA}{dr}&=-\frac{1070}{r^2}+4\pi r
\end{align}
Set to 0 to find where the minimum of \(A\) is (plotting \(A\) reveals that it is a local minimum and not a local maximum when \(\frac{dA}{dr}=0\)).
\begin{align}
0&=-\frac{1070}{r^2}+4\pi r\\
\implies r&=\left(\frac{535}{2\pi}\right)^\frac{1}{3}\\
&\approx 4.4
\end{align}
Sub into the volume equation to find \(h\):
\begin{align}
535 &= \pi \left(\frac{535}{2\pi}\right)^\frac{2}{3} \times h\\
\implies h&\approx 8.8
\end{align}

[Gogo14]

Does sin(pi/2+pi/3) give cos(pi/3) or -cos(pi/3)

[Syndicate]

Does sin(pi/2+pi/3) give cos(pi/3) or -cos(pi/3)

[Gogo14]

(Tech free)Determine the locus of points such that if P=(x,y) ,A=(-2,0) and B=(2,0) the PA+PB=4
Help! My exam is in 3 days!

[jamonwindeyer]

(Tech free)Determine the locus of points such that if P=(x,y) ,A=(-2,0) and B=(2,0) the PA+PB=4
Help! My exam is in 3 days!

Disclaimer: No idea if this working out is VCE friendly. Just thought I'd lend a hand.

Hey Gogo! We'll take the points and substitute them into expressions for the distances PA and PB to start:




Now let's play with that relationship a bit so we can bring in the expressions without the square roots:



Notice we got a whole heap of cancellation into that last line. That's what we wanted to see. We can continue:



So this locus lies on the x-axis; however, it isn't the whole x-axis. We need to check something else to find the missing restriction, remember that for our condition to make sense, PA must be less than or equal to 4, and so must PB. Let's check that condition:




These two regions overlap only for \(-2\le x\le2\), so that's where our locus lies. Putting it together, this locus is just the line segment joining A and B.

Note that you could do this last bit just with some clever thinking. Clearly we can't venture beyond A or B with our locus, because that just doesn't make sense. As soon as we do, one of our distances is immediately, clearly, too large. You could probably not do any algebra for this question if you are good at picturing loci, but I thought I'd show everything sorry if this isn't VCE friendly, but I thought I'd lend a hand anyway!

[Gogo14]

Disclaimer: No idea if this working out is VCE friendly. Just thought I'd lend a hand.

Hey Gogo! We'll take the points and substitute them into expressions for the distances PA and PB to start:




Now let's play with that relationship a bit so we can bring in the expressions without the square roots:



Notice we got a whole heap of cancellation into that last line. That's what we wanted to see. We can continue:



So this locus lies on the x-axis; however, it isn't the whole x-axis. We need to check something else to find the missing restriction, remember that for our condition to make sense, PA must be less than or equal to 4, and so must PB. Let's check that condition:




These two regions overlap only for \(-2\le x\le2\), so that's where our locus lies. Putting it together, this locus is just the line segment joining A and B.

Note that you could do this last bit just with some clever thinking. Clearly we can't venture beyond A or B with our locus, because that just doesn't make sense. As soon as we do, one of our distances is immediately, clearly, too large. You could probably not do any algebra for this question if you are good at picturing loci, but I thought I'd show everything sorry if this isn't VCE friendly, but I thought I'd lend a hand anyway!

Thanks!

[samuelbeattie76]

I have attached the question and its answer. What I do not understand is the 195 different sums possible. Isn't there 18 different possible sums from 1 to 18? Can some please explain this question I do not understand the answer given in the solutions manual.

Thanks a lot, any help is appreciated.

[Shadowxo]

I have attached the question and its answer. What I do not understand is the 195 different sums possible. Isn't there 18 different possible sums from 1 to 18? Can some please explain this question I do not understand the answer given in the solutions manual.

Thanks a lot, any help is appreciated.

I think you got a bit mixed up with the player and the number they were assigned
If each of the 35 players gets assigned a different number between 1 and 99, there  could be a player with the number 1 and another player with the number 2, so the sum of their numbers would be 3 (the lowest sum possible). There could also be a player with the number 99 and another player with the number 98, so the sum of their numbers would be 197 (the highest sum possible). So since the players could get any number between 1 and 99, the different sums could be anywhere between 3 (the lowest) and 197 (the highest), resulting in a possible 195 different sums
Does this help?

[codebreaker1_91]

Hi!

A girl at the bottom of a 100m high cliff throws a tennis ball vertically upwards. At the same instant, a boy at the very top of the  cliff drops a golf ball so that it hits the tennis ball while both balls are still in the air. The acceleration of both balls can be taken as 10.0 m/s^2 downwards.

a. If balls collide when tennis ball is at top of path, what is position of the tennis ball when it strikes the golf ball?
b. With what speed is the tennis ball thrown for this to occur?
c. What is the speed of the golf ball when it strikes the tennis ball?
d. For how long has each ball been in motion when they collide?

Thankyou!

[pha0015]

Dumb question alert!

I'm not too sure if I recall this question, but it's been baffling me:
If z=-4i what is the modulus-argument of Iz^3I

Wouldn't the modulus be 64 and the argument be 0?
However the teacher said that the argument was pi/2

Please help this poor soul