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Author Topic: 3U Maths Question Thread  (Read 1230273 times)  Share 

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RuiAce

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Re: 3U Maths Question Thread
« Reply #3495 on: June 21, 2018, 06:28:43 am »
0
Hello,
Could I get some help with this question:
Find the volume of the solid formed by rotating the region enclosed by y=e^(5x)+3, y=0, x=0, x=0.3 about the y-axis.
This cannot be done using 3U methods as there is no way to compute integrals of the form \( \int (\ln x)^2\,dx\). What is the source of this question?

vikasarkalgud

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Re: 3U Maths Question Thread
« Reply #3496 on: June 21, 2018, 05:48:42 pm »
0
Hey, just started projectile motion, needed some help plz
A stone is thrown horizontally with a velocity of 25m/s from the top of a cliff and hits the ground after 3 seconds. How high is the cliff and how far from its base does the stone hit the ground.

so, i got to the stage where i got y = 30t - 5t^2 - 45 (from vertical component of acceleration = -10)
from this, i got y = -45
of course, the answer is +45, given it's a height, but can i say that because the magnitude of y is 45, the height must be 45, thus ignoring the negative? or is there something wrong with my method? Got how far all g.

Thankyou for any help

clovvy

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Re: 3U Maths Question Thread
« Reply #3497 on: June 21, 2018, 06:00:53 pm »
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Hey, just started projectile motion, needed some help plz
A stone is thrown horizontally with a velocity of 25m/s from the top of a cliff and hits the ground after 3 seconds. How high is the cliff and how far from its base does the stone hit the ground.

so, i got to the stage where i got y = 30t - 5t^2 - 45 (from vertical component of acceleration = -10)
from this, i got y = -45
of course, the answer is +45, given it's a height, but can i say that because the magnitude of y is 45, the height must be 45, thus ignoring the negative? or is there something wrong with my method? Got how far all g.

Thankyou for any help

I might be wrong but as far as my understanding goes, this is how I did it
« Last Edit: June 21, 2018, 06:12:19 pm by clovvy »
2018 HSC: 4U maths, 3U maths, Standard English, Chemistry, Physics

clovvy

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Re: 3U Maths Question Thread
« Reply #3498 on: June 23, 2018, 08:01:53 pm »
0
Projectile Question HSC 1987
A pebble is projected from the top of a vertical cliff with velocity 20m/s at an angle of elevation of 30 degrees. The cliff is 40m high and overlooks a lake.
(a) Take the Origin O to be point at the base of the cliff immediately below the point of projection. Derive expressions for the horizontal component x(t) and vertical component y(t) of the pebble's displacement from O after t seconds (ignore AR)
(b) Calculate the time which ellapses before the pebble hits the lake and distance of the point of impact from the foot of the cliff.[Assume g=10m/s/s]

2018 HSC: 4U maths, 3U maths, Standard English, Chemistry, Physics

RuiAce

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Re: 3U Maths Question Thread
« Reply #3499 on: June 23, 2018, 08:07:07 pm »
+2
Projectile Question HSC 1987
A pebble is projected from the top of a vertical cliff with velocity 20m/s at an angle of elevation of 30 degrees. The cliff is 40m high and overlooks a lake.
(a) Take the Origin O to be point at the base of the cliff immediately below the point of projection. Derive expressions for the horizontal component x(t) and vertical component y(t) of the pebble's displacement from O after t seconds (ignore AR)
(b) Calculate the time which ellapses before the pebble hits the lake and distance of the point of impact from the foot of the cliff.[Assume g=10m/s/s]


a) is definitely fine.

All that part b) is asking for you to do is find the time of flight and the range of flight. Recall that our usual method of doing this is to set \( y(t) = 0\) for the time of flight, and then plug whatever value of \(t\) we get in to \(x(t) \) for the range. Note that you will need to evaluate \( \cos 30^\circ = \frac{\sqrt3}{2} \), \( \sin 30^\circ = \frac12\) and sub in \(g = 10\). You may or may not require the quadratic formula since you're projecting off a cliff.

Never.Give.Up

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Re: 3U Maths Question Thread
« Reply #3500 on: June 24, 2018, 10:33:41 pm »
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Hello could I please have some help with the attached question?
I would really appreciate it! ;D

Opengangs

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Re: 3U Maths Question Thread
« Reply #3501 on: June 24, 2018, 11:21:06 pm »
+3
Hello could I please have some help with the attached question?
I would really appreciate it! ;D
Hey, what they really want you to do is to note that \(\frac{d}{dx}\left(\cos (\pi x)\right) = -\pi \sin (\pi x)\) because in doing that, it's really obvious what substitution we can make. So, whenever \( k \neq -1 \), we can make the substitution \( u = \cos (\pi x) \).

Case 1: \(k \neq -1 \)


Case 2: \( k = -1 \)


Whenever we have these product of trig functions, it's extremely useful to have some intuition behind these subtle integration techniques because it makes our computation a whole lot easier. And this is one such example. It's extremely reassuring to note these relationships especially when it comes to exams, so it's in your best interest to familiarise yourself to these types before trials and the HSC.
« Last Edit: June 24, 2018, 11:22:49 pm by Opengangs »

Never.Give.Up

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Re: 3U Maths Question Thread
« Reply #3502 on: June 25, 2018, 04:04:57 pm »
0
Thankyou,
also how do I find the area between
y= 0.2 sin(pi.x) and y= 0.6 (cos pi.x)^2

both the big and little parts on the graph attached
POI are (0.322,0.1695), (0.6782,0.1695), (2.322,0.1695) and so on.... ;D
thanks heaps (I'm meant to have it done by tomorrow but I just dont really get it :-[)

thankuuu
« Last Edit: June 25, 2018, 04:09:41 pm by Never.Give.Up »

Never.Give.Up

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Re: 3U Maths Question Thread
« Reply #3503 on: June 25, 2018, 04:21:37 pm »
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Hey, what they really want you to do is to note that \(\frac{d}{dx}\left(\cos (\pi x)\right) = -\pi \sin (\pi x)\) because in doing that, it's really obvious what substitution we can make. So, whenever \( k \neq -1 \), we can make the substitution \( u = \cos (\pi x) \).

sorry im dumb... but could u pls explain how this works? i get the derivative of cos = -sin, but where do I get the pi out the front from? and then how can i make that substitution (im confused)

Case 1: \(k \neq -1 \)

and then how does -1/pi end up out the front??

sorry for all the questions

Opengangs

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Re: 3U Maths Question Thread
« Reply #3504 on: June 25, 2018, 04:59:16 pm »
+4
sorry im dumb... but could u pls explain how this works? i get the derivative of cos = -sin, but where do I get the pi out the front from? and then how can i make that substitution (im confused)
Sure, no problem!

So, basically it's an implicit chain rule. You can explicitly see this by making a subtle substitution here: \( u = \pi x \).


As long as you understand where the chain rule comes from, these types of functions should then be quite simple to differentiate. That is: \( \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\). In this case, our "outer" function, \( f(x) \), is \( \cos x \) and our "inner" function, \( g(x) \), is \( \pi x \).

and then how does -1/pi end up out the front??

sorry for all the questions

This one requires a bit more clever thinking! Notice that \( \frac{d}{dx}(\cos x) = -\sin x \). So if we had something like this: \( \int -\cos x \sin x\,dx \), then what we can do is a little substitution. That is, \( u = \cos x \), because when we differentiate, we find the following:
\[ u = \cos x \Rightarrow \frac{du}{dx} = -\sin x \\ \therefore du = -\sin x\,dx \]

Comparing our integral, we see that:
\[ \begin{align*} \int -\cos x \sin x\,dx &= \int \underbrace{\cos x}_{u} \cdot \underbrace{(-\sin x\,dx)}_{du} \\ &= \int u\,du \end{align*}\]

The same works with our integral of choice. Except it's a little bit more work. When we use the substitution \( u = \cos(\pi x) \), we find that:
\[ \frac{du}{dx} = -\pi \sin(\pi x) \Rightarrow du = -\pi \sin(\pi x)\,dx \].

Comparing our integral, we see that:
\[ \begin{align*} \int \cos^k(\pi x) \cdot (\sin(\pi x)\,dx &= \int \underbrace{\cos^k(\pi x)}_{u^k} \cdot (\sin(\pi x)\,dx)\end{align*} \]

In this case, our "\( du \)" is not exactly the same as the one we have in our integral. So, in this case, we need to manipulate our "\( du \)". Our integral is missing a \( -\pi \).

\[ \begin{align*} \int \cos^k(\pi x) \cdot (\sin(\pi x)\,dx &= \int \frac{-\pi \cdot \cos^k(\pi x) \cdot (\sin(\pi x)}{-\pi}\,dx \\ &= \int \frac{\cos^k(\pi x) \cdot (-\pi \sin(\pi x))}{-\pi}\,dx \\ &= -\frac{1}{\pi} \int \underbrace{\cos^k(\pi x)}_{u^k} \cdot \underbrace{(-\pi \sin(\pi x))\,dx}_{du} \\ &= -\frac{1}{\pi}\int u^k\,du \end{align*}\]

Never.Give.Up

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Re: 3U Maths Question Thread
« Reply #3505 on: June 25, 2018, 08:18:54 pm »
0
Sure, no problem!

So, basically it's an implicit chain rule. You can explicitly see this by making a subtle substitution here: \( u = \pi x \).


As long as you understand where the chain rule comes from, these types of functions should then be quite simple to differentiate. That is: \( \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\). In this case, our "outer" function, \( f(x) \), is \( \cos x \) and our "inner" function, \( g(x) \), is \( \pi x \).

This one requires a bit more clever thinking! Notice that \( \frac{d}{dx}(\cos x) = -\sin x \). So if we had something like this: \( \int -\cos x \sin x\,dx \), then what we can do is a little substitution. That is, \( u = \cos x \), because when we differentiate, we find the following:
\[ u = \cos x \Rightarrow \frac{du}{dx} = -\sin x \\ \therefore du = -\sin x\,dx \]

Comparing our integral, we see that:
\[ \begin{align*} \int -\cos x \sin x\,dx &= \int \underbrace{\cos x}_{u} \cdot \underbrace{(-\sin x\,dx)}_{du} \\ &= \int u\,du \end{align*}\]

The same works with our integral of choice. Except it's a little bit more work. When we use the substitution \( u = \cos(\pi x) \), we find that:
\[ \frac{du}{dx} = -\pi \sin(\pi x) \Rightarrow du = -\pi \sin(\pi x)\,dx \].

Comparing our integral, we see that:
\[ \begin{align*} \int \cos^k(\pi x) \cdot (\sin(\pi x)\,dx &= \int \underbrace{\cos^k(\pi x)}_{u^k} \cdot (\sin(\pi x)\,dx)\end{align*} \]

In this case, our "\( du \)" is not exactly the same as the one we have in our integral. So, in this case, we need to manipulate our "\( du \)". Our integral is missing a \( -\pi \).

\[ \begin{align*} \int \cos^k(\pi x) \cdot (\sin(\pi x)\,dx &= \int \frac{-\pi \cdot \cos^k(\pi x) \cdot (\sin(\pi x)}{-\pi}\,dx \\ &= \int \frac{\cos^k(\pi x) \cdot (-\pi \sin(\pi x))}{-\pi}\,dx \\ &= -\frac{1}{\pi} \int \underbrace{\cos^k(\pi x)}_{u^k} \cdot \underbrace{(-\pi \sin(\pi x))\,dx}_{du} \\ &= -\frac{1}{\pi}\int u^k\,du \end{align*}\]
okay thanks for that awesome explanation (i wasnt thinking before at all so please excuse the basic q) the only thing im still confused about is why you put it over pi in the first place?
 also, do you have any tips for the other question i put on here?  ;D ;D
ur a legend!!
thankyou :D

RuiAce

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Re: 3U Maths Question Thread
« Reply #3506 on: June 25, 2018, 08:20:34 pm »
0
okay thanks for that awesome explanation (i wasnt thinking before at all so please excuse the basic q) the only thing im still confused about is why you put it over pi in the first place?
 also, do you have any tips for the other question i put on here?  ;D ;D
ur a legend!!
thankyou :D
He used the substitution \( u = \cos \pi x\) the second time around. This gives \( du = -\pi \sin \pi x\), so since we have the sine expression but not the constant out in front, we need to consider \( \frac{du}{\pi} = -\sin \pi x\).

itssona

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Re: 3U Maths Question Thread
« Reply #3507 on: June 29, 2018, 11:14:12 am »
0
not really getting the right answer
HSC 2018 : Maths 3U, Maths 4U, English Advanced, Biology, Physics, Chemistry

Opengangs

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Re: 3U Maths Question Thread
« Reply #3508 on: June 29, 2018, 12:02:20 pm »
+2
not really getting the right answer
Hey there.

With every related rates problem, it's a good idea to give yourself a nice diagram that will aid your critical thinking about where to go next.
Note that the volume of the cone is given as:
\[ V = \frac{\pi}{3}r^2h \]



We know that the water flows from the apex at a constant rate: this tells us that the volume is changing at a constant rate. In our case, we know that it is decreasing:
\[ \frac{dV}{dt} = -0.2 \]

We then want to find the height's rate of change, \( \frac{dh}{dt} \), when the height is 4m.

However, we run into a problem. Our volume is in terms of two variables \( r \) and \( h \). Because we want the rate of the height, we will need to find some relationship between \( r \) and \( h \). Consider taking some random slice of \( h \) and now we see that similar triangles occur.



So now:
\[ \frac{r}{h} = \frac{2}{5} \Rightarrow r = \frac{2}{5}h \]

So, our volume of the cone becomes:
\[ \begin{align*} V &= \frac{\pi}{3} \left(\frac{2}{5}h\right)^2 h \\ &= \frac{\pi}{3}\left(\frac{2}{5}\right)^2 h^3\end{align*} \]
\[ \begin{align*} \frac{dV}{dh} &= \frac{4\pi h^2}{25}\end{align*}\]

\[ \text{Note that the chain rule tells us that: } \frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} \]
\[ \begin{align*}-0.2 &= \frac{4\pi h^2}{25} \cdot \frac{dh}{dt} \\ \frac{dh}{dt} &= \frac{-0.2}{\frac{4\pi h^2}{25}} \\ &= -\frac{5}{4\pi h^2} \\ &= -\frac{5}{64\pi} \text{, when } h \text{ is 4.}\end{align*}\]

\[ \text{So this tells us that the water level is dropping at a rate of } \frac{5}{64\pi}. \]

The key here really is to draw a diagram - it makes all of the working so much easier to understand :)

Mate2425

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Re: 3U Maths Question Thread
« Reply #3509 on: June 30, 2018, 02:21:50 pm »
0
Hey team, could i please have some help with this inverse function Question.
Q. If the domain is restricted to a monotonic increasing curve, find the inverse function, and state the domain and range of the inverse function.
The original function is y = 1 divided by x^2. Answer is y = -1 divided by root x, domain X>0 range Y<0.

I don't understand how they got their negative in front of the inverse function as well what is the best way to find the domain and range (is graphically a good method)?

Many Thanks!