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#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4125 on: July 02, 2019, 11:25:25 am »
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The acceleration of a particle at a position x on the number line is given bya=-25x.
Initially the particle is at the origin and has a velocity of 10m/s.
Find the velocity after pi/15 seconds.

This was Q9 of 2016 Syd Girls Trial MC if anyone's interested.

#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4126 on: July 02, 2019, 11:36:14 am »
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An iceblock with temperature -14 degrees Celsius is left out in the sun. The air temp is a constant 25 degrees Celsius and after 40 seconds the temperature of the iceblock has reached -5 degrees Celsius. Find a) its temperature after 5 mins b) when the iceblock will start to melt (ie when its temp will reach 0 degrees Celsius).

The eqn I got to was T = 25-39e^(-0.007..t) -- specifically k=1/40 ln(29/39). For a, I had 8.9..., but the answer is 7.2 For b, I had 60.03..., but the answer is 68.

#### julz_roha

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##### Re: 3U Maths Question Thread
« Reply #4127 on: July 02, 2019, 12:40:00 pm »
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...

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4128 on: July 02, 2019, 04:27:34 pm »
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well I did this question, but i feel like this is very wrong, and when i put it in maths calculators.. they dont use the same u sub too. pls help.

question and image containing ans i did attached. edit: for some reason the file with my working out is not being able to be uploaded............

qn is  u= e^3x, integral(1/3,0) e^3x/e^6x+1 dx.

$\text{Let} \ u=e^{3x} \\ \text{Then} \ du=3e^{3x} \ dx \\ \text{Then for} \ x=\frac{1}{3}, u=e \ \text{and for} \ x=0, u=1 \\ \int_0^{\frac{1}{3}} \frac{e^{3x}}{e^{6x}+1} \ dx = \frac{1}{3} \int_1^e \frac{du}{1+u^2} \\ =\frac{1}{3} \left[\tan^{-1} u\right]_1^e \\ = \frac{1}{3} \left(\tan^{-1} (e) - \frac{\pi}{4}\right)$

hi i need help with this question......i was able to the first part, but for the second part, there seems to be no rule to tackle this question.........so how do i got about doing this question and what is the working out for this question...........

The key here is that every derivative gives us another integral and vice versa ie.

$\text{If} \ \frac{d}{dx} x\tan^{-1}x = \tan^{-1}x + \frac{x}{1+x^2} \\ \text{Then} \ \int \tan^{-1}x + \frac{x}{1+x^2} \ dx = x\tan^{-1}x + C \\ \text{Knowing this,} \ \int \tan^{-1}x \ dx= \int \tan^{-1}x + \frac{x}{1+x^2} - \frac{x}{1+x^2} \ dx \\ = x\tan^{-1}x - \frac{1}{2} \ln |x^2+1| + C \\ \text{(By the reverse chain rule)}$

Hope this helps

The acceleration of a particle at a position x on the number line is given bya=-25x.
Initially the particle is at the origin and has a velocity of 10m/s.
Find the velocity after pi/15 seconds.

This was Q9 of 2016 Syd Girls Trial MC if anyone's interested.

$\frac{d\left(\frac{1}{2}v^2\right)}{dx} = -25x \\ v^2 = -25x^2 + 2C \\ v^2 = 100-25x^2 \ \text{Since when} \ x=0, v=10 \\ v^2 = 5^2(2^2-x^2) \\ \text{This is now in the form} \ v^2=n^2(a^2-x^2) \\ \text{Given} \ n=5, a=2 \ \text{and that when} \ t=0, x=0 \\ x=2\sin (5t) \ \text{(Since the sine function starts at zero, subbing in values for n and a)}$

You can differentiate from here to find dx/dt, then sub in

An iceblock with temperature -14 degrees Celsius is left out in the sun. The air temp is a constant 25 degrees Celsius and after 40 seconds the temperature of the iceblock has reached -5 degrees Celsius. Find a) its temperature after 5 mins b) when the iceblock will start to melt (ie when its temp will reach 0 degrees Celsius).

The eqn I got to was T = 25-39e^(-0.007..t) -- specifically k=1/40 ln(29/39). For a, I had 8.9..., but the answer is 7.2 For b, I had 60.03..., but the answer is 68.

Your constant k I think is slightly off, check it again! I've got k=1/40 ln(13/10). Your k is negative, remember that subbing that value k will yield a positive exponential and thus your terminal temperature will balloon towards negative infinity. It's a good way of checking if your k value makes sense.

The answer you've got there for a) I think is wrong; try again with the new constant that you get. I'm getting 19.55 degrees
With the new constant you get you should also get the answer for b) correct.

Hope this helps
« Last Edit: July 02, 2019, 04:58:08 pm by fun_jirachi »
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#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4129 on: July 02, 2019, 05:48:56 pm »
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$\text{Let} \ u=e^{3x} \\ \text{Then} \ du=3e^{3x} \ dx \\ \text{Then for} \ x=\frac{1}{3}, u=e \ \text{and for} \ x=0, u=1 \\ \int_0^{\frac{1}{3}} \frac{e^{3x}}{e^{6x}+1} \ dx = \frac{1}{3} \int_1^e \frac{du}{1+u^2} \\ =\frac{1}{3} \left[\tan^{-1} u\right]_1^e \\ = \frac{1}{3} \left(\tan^{-1} (e) - \frac{\pi}{4}\right)$

The key here is that every derivative gives us another integral and vice versa ie.

$\text{If} \ \frac{d}{dx} x\tan^{-1}x = \tan^{-1}x + \frac{x}{1+x^2} \\ \text{Then} \ \int \tan^{-1}x + \frac{x}{1+x^2} \ dx = x\tan^{-1}x + C \\ \text{Knowing this,} \ \int \tan^{-1}x \ dx= \int \tan^{-1}x + \frac{x}{1+x^2} - \frac{x}{1+x^2} \ dx \\ = x\tan^{-1}x - \frac{1}{2} \ln |x^2+1| + C \\ \text{(By the reverse chain rule)}$

Hope this helps

$\frac{d\left(\frac{1}{2}v^2\right)}{dx} = -25x \\ v^2 = -25x^2 + 2C \\ v^2 = 100-25x^2 \ \text{Since when} \ x=0, v=10 \\ v^2 = 5^2(2^2-x^2) \\ \text{This is now in the form} \ v^2=n^2(a^2-x^2) \\ \text{Given} \ n=5, a=2 \ \text{and that when} \ t=0, x=0 \\ x=2\sin (5t) \ \text{(Since the sine function starts at zero, subbing in values for n and a)}$

You can differentiate from here to find dx/dt, then sub in

Your constant k I think is slightly off, check it again! I've got k=1/40 ln(13/10). Your k is negative, remember that subbing that value k will yield a positive exponential and thus your terminal temperature will balloon towards negative infinity. It's a good way of checking if your k value makes sense.

The answer you've got there for a) I think is wrong; try again with the new constant that you get. I'm getting 19.55 degrees
With the new constant you get you should also get the answer for b) correct.

Hope this helps

Thanks a million!! For the first question I asked, is that how you tell whether you use x=acos(nt+alpha) or x=asin(Nt+alpha) - by looking at where it starts?

Also:

The velocity of a particle is given by: v=1/(4cos2x) and initially the particle is at the origin. Find the total time of motion.

I got to t=2sin2x, not sure why the answer is 2 though? Something about the amplitude?

#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4130 on: July 02, 2019, 06:07:42 pm »
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Attached question: I got the v and a, but for x I have x=sqr48 /3 cos (3t+pi/6), when the answers say it's 3t-pi/6. I have no idea where I've made the mistake, help please!

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4131 on: July 02, 2019, 06:09:14 pm »
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Thanks a million!! For the first question I asked, is that how you tell whether you use x=acos(nt+alpha) or x=asin(Nt+alpha) - by looking at where it starts?

Also:

The velocity of a particle is given by: v=1/(4cos2x) and initially the particle is at the origin. Find the total time of motion.

I got to t=2sin2x, not sure why the answer is 2 though? Something about the amplitude?

Yep, in general:
- If a particle in SHM starts at the origin, use the sine function
- If a particle in SHM starts at a peak (centre of motion + amplitude) use the positive cosine function
- If a particle in SHM starts at a trough (centre of motion - amplitude) use the negative cosine function

Alternatively, you could use the cosine function for the first case, and the sine function for the last two cases, then shift it using f(x+h)+k for constants h and k, but it's a lot easier this way

I assume you did this?
$\frac{dx}{dt}=\frac{1}{4\cos 2x} \\ dt = 4\cos 2x dx \\ \int_0^t dt = 4 \int_0^x \cos 2x \ dx \\ t = 2\sin 2x$

You're right here, it's got something to do with the amplitude: 2sin2x has range -2<=f(x)<=2, and since time cannot be negative, it has a maximum at T=2, the total time of motion. You can also observe this by finding x in terms of t.

Attached question: I got the v and a, but for x I have x=sqr48 /3 cos (3t+pi/6), when the answers say it's 3t-pi/6. I have no idea where I've made the mistake, help please!

No attachment (yet)

Hope this helps
« Last Edit: July 02, 2019, 06:11:39 pm by fun_jirachi »
Failing everything, but I'm still Flareon up.

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#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4132 on: July 02, 2019, 07:40:02 pm »
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Oops, sorry. Attached is the previous question and another one!

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4133 on: July 02, 2019, 08:18:15 pm »
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$\frac{d\left(\frac{1}{2}v^2\right)}{dx} = -9x \\ \int_{2\sqrt 3}^v d\left(\frac{1}{2}v^2\right) = \int_2^x -9x \ dx \\ \frac{1}{2} \left[v^2\right]_{2\sqrt 3}^v = -\frac{1}{2}\left[9x^2\right]_2^x \\ v^2 - 12 = -(9x^2-36) \\ v^2=48-9x^2 \\ v^2=3^2\left(\left(\frac{4\sqrt 3}{3}\right)^2 - x^2\right) \\ n=3, a=\frac{4\sqrt 3}{3} \\ x=\frac{4\sqrt 3}{3}\cos (3t + \alpha) \\ \text{When} \ t=0, x=2 \\ 2=\frac{4\sqrt 3}{3}\cos (\alpha) \\ \cos \alpha = \frac{\sqrt 3}{2} \\ \alpha = \pm \frac{\pi}{6} \\ \text{Now} \ v=\frac{dx}{dt}=-4\sqrt 3 \sin (3t + \alpha) \\ \text{When} \ t=0, v=2\sqrt 3 \\ \text{Hence} \ \sin \alpha = -\frac{1}{2} \ \text{and hence} \ \alpha \ \text{must be a fourth quadrant angle since} \ \sin \alpha<0 \ \text{and} \ \cos \alpha>0 \\ \alpha = -\frac{\pi}{6} \\ x=\frac{4\sqrt 3}{3}\cos \left(3t - \frac{\pi}{6}\right)$

For the next one, try using the fact that ABF and DEF are similar, as well as the fact that EFD and EBC are similar. The key here is that BF and EF are common to both sets of similar triangles in the ratio 2:1. Try it again and see how you go! Answer should be B

Hope this helps
« Last Edit: July 02, 2019, 08:23:06 pm by fun_jirachi »
Failing everything, but I'm still Flareon up.

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#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4134 on: July 02, 2019, 11:39:23 pm »
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I still don't get how to work out the similar triangles one

Two more questions coming through, (attached)

Q1. In the solutions, after obtaining 2 different angles, we are to differentiate and then choose the one where x>0. I don't understand how to decide to differentiate to check for the right solution. How do you get to that step? What thinking is involved?

Q2. Why can't we integrate (3)^2 - (e^2y)^2 ?

Thanks so much for answering my questions especially when you have trials study going on!!

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4135 on: July 03, 2019, 05:42:21 pm »
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I still don't get how to work out the similar triangles one

Two more questions coming through, (attached)

Q1. In the solutions, after obtaining 2 different angles, we are to differentiate and then choose the one where x>0. I don't understand how to decide to differentiate to check for the right solution. How do you get to that step? What thinking is involved?

Q2. Why can't we integrate (3)^2 - (e^2y)^2 ?

Thanks so much for answering my questions especially when you have trials study going on!!

For the first question I assume you got this?
$\text{Since the plane must just clear the hill} \ \triangle = 0 \\ (m-8)^2-4\times 12 = 0 \\ m^2-16m+16=0 \\ m=\frac{16\pm \sqrt{16^2-4\times 16}}{2} = 8\pm \sqrt{48}=8\pm 4\sqrt 3$

I don't think you really need to differentiate here. Each value of m, when substituted back into the equation from part i) should yield a different parabola both of which are incidentally perfect squares. One value of m has one positive root and the other has one negative root, and it's pretty obvious that the root should be positive since the plane touches the hill at 2<x<6. Then you choose the correct value of m, then use m=tan theta.

Not sure what you mean by your second question (my brain is fizzing out right now, I might get back to this later) but this is what I get for the volume:
$\text{Remember that} \ V=\pi \int_a^b x^2 \ dy \\ \text{In this case the volume is} \ 3^2 \pi \times \ln 3 - \pi \int_0^{\ln 3} e^{2y} \ dy$

For the similar triangles one, AB is parallel to CE and AD is parallel to BC. In triangles EFD and EBC you have two corresponding angles in parallel lines, so they're equiangular, and thus similar. In triangles ABF and EDF, you have two alternate angles in parallel lines, so they're equiangular and thus similar. Since CD:DE is 2:1, BF:FE must also be 2:1 since similar triangles have corresponding sides in the same proportion as other corresponding sides. Since BF:FE = 2:1, AF:FD = 2:1 as well for the same reason. Then AF is 2/3 of AD, which is equal to BC. ie. AF:BC = 2:3

Hope this helps
« Last Edit: July 03, 2019, 11:31:24 pm by fun_jirachi »
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#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4136 on: July 03, 2019, 08:25:17 pm »
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Attached Q1; I don't understand why the velocity is negative in part a)
Q2; not sure how to work out c) - the solutions give an exact answer 3sqr5 whereas I can only get a decimal!

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4137 on: July 04, 2019, 12:39:59 pm »
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For Q1, the velocity shouldn't be just negative. When a particle is in SHM and is x distance away from the centre of motion on either side, the velocity has equal magnitude but the opposite direction for each side. I'm not too sure; must be a typo or something.

For Q2,
$\text{The amplitude} \ a = \sqrt{a^2+b^2} = \sqrt{(2)^2 + (-1)^2} = \sqrt 5 \\ \text{Hence, from} \ v^2=n^2(a^2-x^2), \ v^2 = 9(5-x^2) \\ \text{Since the maximum velocity occurs when} \ x=0, \ v_{\text{max}} = 3\sqrt 5$
« Last Edit: July 04, 2019, 12:41:52 pm by fun_jirachi »
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#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4138 on: July 05, 2019, 02:09:53 pm »
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THANKS A MILLION!

attached are some more questions!
q1: i got a negative discriminant so that's a bit weird. the answer is 63 deg 6 mins or 50 deg 52 mins
q2: i got around 8.6 seconds (7.22-0.58+2), but the answer's 2.7 seconds
q3: b) sorry, i think this one's kinda long. I found the cartesian equation for both and equated to get x=16.57 but the answer is x=11.5, y=8.2 and no, they pass that at different times.

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4139 on: July 05, 2019, 06:48:20 pm »
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Q1:
$\text{The equations of motion (not going to go over derivation unless you ask again) that you should get are} \\ x=12t\cos \theta \ \text{and} \ y=12t\sin \theta - \frac{gt^2}{2} \\ \text{Hence} \ y=x\tan \theta - \frac{g}{2} \frac{x^2}{144\cos^2 \theta} \\ \text{Subbing in the point (9, 4) since the ball passes through that point} \\ 4=9\tan \theta - 5 \times \frac{81}{144\cos^2 \theta} \\ \text{We now have a quadratic in} \ \tan \theta , \ \text{so have a go at finishing that off.}$

Q2:
$\text{Here, the only equation of motion you need should be} \ y=Vt\sin \theta - \frac{gt^2}{2} \\ \text{Subbing in values for} \ \theta \ \text{and} \ g \\ y=\frac{Vt}{\sqrt{2}}-4.9t^2$
From here, it's only a matter of subbing in V=45 and V=50, and solving the quadratic since when they land y=0. Comparing the times should yield you the time difference and then add that on to two seconds since the second object is launched two seconds after the first.

Q3:
$\text{For the first ball} \\ x=Vt\cos \theta = 15t\cos 60 = \frac{15}{2}t \\ \text{For the second ball} \\ x=30-Vt\cos \theta = 20t\cos 45 = 30-10\sqrt{2}t$
Since the particles have the same x-value, y-value and time upon collision, you can solve for t by equating the x-values. Subbing back in for t will get you the x-value for the collision, which I get to be 10.396. But at that point in time, subbing into the vertical displacement equation they have different y-values ie. they must not collide. You can just use the displacement equations to form a quadratic in x and y and solve to find a crossing point.
« Last Edit: July 05, 2019, 07:03:19 pm by fun_jirachi »
Failing everything, but I'm still Flareon up.

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