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#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #17940 on: June 04, 2019, 11:42:00 pm »
+1
How do transformations change the values of circular functions? E.g. I know that for the base function sin(x), when you dilate it for 2 units from the x-axis and dilate it for 1/3 units from the y-axis, you have the transformed function y=2sin(3x), but in terms of the values of x and y, how are they different? What do I change? How do I change it?

I'm not sure I quite understand your question, so could you elaborate on what you mean by how transformations "change the values of circular functions"? And, when you say "how do I change it", what's it?

Edit: Are you asking about the overall effects of different transformations on the graphs of circular functions? Or, are you asking about how any given point say $(x,y)$ that is on the graph moves after a sequence of transformations are applied?
« Last Edit: June 04, 2019, 11:52:11 pm by AlphaZero »
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#### TyranT

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##### Re: VCE Methods Question Thread!
« Reply #17941 on: June 05, 2019, 07:35:24 am »
0
Hello,
Could i please get help with question 12b? I know the answer to question 12a but not 12b. The solutions say 19mm but with no working out; i want to understand the concept.
Thank you

#### MB_

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##### Re: VCE Methods Question Thread!
« Reply #17942 on: June 05, 2019, 09:47:30 am »
+1
Hello,
Could i please get help with question 12b? I know the answer to question 12a but not 12b. The solutions say 19mm but with no working out; i want to understand the concept.
Thank you
For part a, you get $\sqrt[3]{7} \text{ cm}$. Part b is just asking you to evaluate that answer and round it so $\sqrt[3]{7} \approx 1.9 \text{ cm} = 19 \text{ mm}$.
« Last Edit: June 05, 2019, 09:49:22 am by MB_ »
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#### EllingtonFeint

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##### Re: VCE Methods Question Thread!
« Reply #17943 on: June 05, 2019, 01:05:29 pm »
0
Hey,
I feel super self conscious posting on here but I need help, so...
There are a few questions which I am stuck with so I might be spamming this thread tonight. Sorry in advance! (these are practise qs for my upcoming SAC that I don't understand)
For question 17, I need to CTS right? Only thing is, I'm not sure how to do that... Could somebody please walk me through the working out please?

For the 2nd and 3rd attachments, I'm not really sure how to find the inverse of a log (bi) or find the rule for a composite function

Hope someone can help
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#### JR_StudyEd

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##### Re: VCE Methods Question Thread!
« Reply #17944 on: June 05, 2019, 01:16:23 pm »
0
I'm not sure I quite understand your question, so could you elaborate on what you mean by how transformations "change the values of circular functions"? And, when you say "how do I change it", what's it?

Edit: Are you asking about the overall effects of different transformations on the graphs of circular functions? Or, are you asking about how any given point say $(x,y)$ that is on the graph moves after a sequence of transformations are applied?

Sorry for not being clear. My question is about how any given point on the graph moves after a sequence of transformations are applied (specifically circular functions).

#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #17945 on: June 05, 2019, 01:43:31 pm »
+5
Hey,
I feel super self conscious posting on here but I need help, so...
There are a few questions which I am stuck with so I might be spamming this thread tonight. Sorry in advance! (these are practise qs for my upcoming SAC that I don't understand)
For question 17, I need to CTS right? Only thing is, I'm not sure how to do that... Could somebody please walk me through the working out please?

For the 2nd and 3rd attachments, I'm not really sure how to find the inverse of a log (bi) or find the rule for a composite function

Hope someone can help

Hey there, feel free to post as many questions as you like. As long as you've had a go at them first, we're here to help. Well done for reaching out for help

Question 17

For our convenience, let's define $f(x)=x^2-2bx+1$. Now, we wish to find the coordinates of the turning point. There are a few ways to go about this, and you're right, one way is to complete the square to put the rule of $f$ in 'turning point form': $f(x)=x^2-2bx+1=(x-b)^2-b^2+1\ \ \ \quad\text{(CAS or otherwise)}$ Hence, the coordinates of the turning point of $f$ is $\left(b,\ 1-b^2\right).$ Then, we would like an expression for the distance, $d$, from the origin to the turning point: $d(b)=\sqrt{b^2+(1-b^2)^2}.$ Now, the question has become a simple optimisation problem. Can you find the value(s) of $b$ for which $d$ is minimum?

Question 5

Going to preface this by saying that this is actually quite a tough question. Function theory is not well understood by the majority of Methods students, but no need to worry, let's try to figure this one out.

Part a.i
This question just asks us to find the rule of $h$:
$h(x)=f\big(g(x)\big)=\log_e(x^2+1).$

Part a.ii
The domain of $h$ is just the domain of the inner function $g$:
$\text{domain}(h)=\mathbb{R}.$ The range is a little trickier, but here's a nice diagram that should help with understanding:$\mathbb{R}\overset{g}{\longrightarrow}[1,\ \infty)\overset{f}{\longrightarrow}??$ What is the image of $f$ when its domain has been restricted to $[1,\,\infty)$?

Part a.iii
Evaluate the LHS and make some algebraic manipulations to reach the form of the RHS.$\text{Hint:}\ \ \log(a)+\log(a)=\log(a^2),\quad a>0.$

Part a.iv
Finding the coordinates of the turning point should be straight forward (don't forget the chain rule), but what about stating its nature? $\text{Hint:}\ \ \text{what is the range of }h?$

Part b.i
First, let  $y=k^{-1}(x)$. Then, we have $x=\log_e(y^2+1).$ Try to rearrange this expression for $y^2$. You should obtain $y^2=e^x-1.$ The question now becomes:  which root do we take - the positive or negative root? $\text{Hint:}\ \ \text{what does the domain of }k \text{ tell you?}$

Part b.ii
Finding the range of $k^{-1}$ should be easy - it's just the domain of $k$:  $(-\infty,\,0]$.  What about the domain though? $\text{Hint:}\ \ \text{what do you answers to part a.ii and part a.iv tell you?}$
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#### colline

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##### Re: VCE Methods Question Thread!
« Reply #17946 on: June 05, 2019, 01:49:38 pm »
+11
Hey,
I feel super self conscious posting on here but I need help, so...
There are a few questions which I am stuck with so I might be spamming this thread tonight. Sorry in advance! (these are practise qs for my upcoming SAC that I don't understand)
For question 17, I need to CTS right? Only thing is, I'm not sure how to do that... Could somebody please walk me through the working out please?

For the 2nd and 3rd attachments, I'm not really sure how to find the inverse of a log (bi) or find the rule for a composite function

Hope someone can help

Hey! For the composite function, remember you want to replace the 'x' of the outer function with the entirety of the inner function. For example in this case, h(x) = f(g(x)), so you just replace the 'x' in the original f(x) function with g(x). Here's my working out:

For Q5b, remember that for inverse you just swap the x and the y and that the inverse of log_e(x) is e^x. In this case, the working out would be:

For the first question, as it is multiple choice I assume you'd be allowed to use a calculator. I don't know about the classpad if that's what you use, but on the TI-Nspire, to complete the square you just go menu - 3 (algebra) - 5 (complete the square). Type in the function and remember to add ',x' at the end, and the CAS will give you the answer.
« Last Edit: June 05, 2019, 08:04:10 pm by colline »
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#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #17947 on: June 05, 2019, 02:12:39 pm »
+12
How do transformations change the values of circular functions? E.g. I know that for the base function sin(x), when you dilate it for 2 units from the x-axis and dilate it for 1/3 units from the y-axis, you have the transformed function y=2sin(3x), but in terms of the values of x and y, how are they different? What do I change? How do I change it?
Sorry for not being clear. My question is about how any given point on the graph moves after a sequence of transformations are applied (specifically circular functions).

The neat thing about simple transformations (dilations, reflections, translations) is that their actions are the same on every relation.

A dilation by factor $a$ from the $x$-axis will map any point $(x,\,y)$ on a graph to the point $(x,\,ay)$.  That is, the $y$-coordinate of every point on a graph will be multiplied by $a$.  For example, if $a=2$, every $y$-coordinate is doubled. Shown below are the graphs of $y=\sin(x)$ and $y=2\sin(x)$ for $x\in[0,\,2\pi]$.

Now, take a dilation by factor $b$ from the $y$-axis. This will map any point $(x,\,y)$ on a graph to the point $(bx,\,y)$.  That is, the $x$-coordinate of every point on a graph will be multiplied by $b$.  Given the graph of  $y=\sin(x)$, where $x\in[0,\,2\pi]$, could you draw the image relation after a dilation by factor $2$ from the $y$-axis?

In a nutshell

(1)  Dilation by factor $a$ from the $x$-axis
$(x',\,y')=(x,\,ay)\quad \text{and}\quad y=f(x)\overset{(1)}{\longrightarrow}y=a\,f(x)$

(2)  Dilation by factor $b$ from the $y$-axis
$(x',\,y')=(bx,\,y)\quad \text{and}\quad y=f(x)\overset{(2)}{\longrightarrow}y=f\left(\frac{x}{b}\right)$

(3)  Reflection in the $x$-axis
$(x',\,y')=(x,\,-y)\quad \text{and}\quad y=f(x)\overset{(3)}{\longrightarrow}y=-f(x)$

(4)  Reflection in the $y$-axis
$(x',\,y')=(-x,\,y)\quad \text{and}\quad y=f(x)\overset{(4)}{\longrightarrow}y=f(-x)$

(5)  Translation of $h$ units in the positive $x$-direction
$(x',\,y')=(x+h,\,y)\quad \text{and}\quad y=f(x)\overset{(5)}{\longrightarrow}y=f(x-h)$

(6)  Translation of $k$ units in the positive $y$-direction
$(x',\,y')=(x,\,y+k)\quad \text{and}\quad y=f(x)\overset{(6)}{\longrightarrow}y=f(x)+k$

(7)  Reflection in the line $y=x$
$(x',\,y')=(y,\,x)\quad \text{and}\quad y=f(x)\overset{(7)}{\longrightarrow}x=f(y)$
« Last Edit: June 05, 2019, 02:14:17 pm by AlphaZero »
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#### strugglebug

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##### Re: VCE Methods Question Thread!
« Reply #17948 on: June 05, 2019, 09:00:03 pm »
0
Hello all, I am doing a practice modeling task and struggling very badly I really need some help on all these questions, would really appreciate it!

(These are all separate questions)

1. h(x)= 4[3-cosh(x/50)] where cosh(x)= e^x+e^-x / 2
Side note: cosh(x) is all over 2, I'm not sure how to type proper equations on here
a. Express h(x) as an exponential function fully expanded.
b. Obtain the derivative d/dx h(x) and express in exponential functions in factorized form.
c. Sketch the graph of y=h(x) and state the x and y-intercepts.
d. What is the average value of h(x) between x-intercepts to 2 decimal places?
e. Determine the value of x when d/dx h(x)=0.
f. Determine the 1st positive value of x to 4 decimal places such that h(x)=g(x), where g(x)= -3x^3/140 + 9x^2/70 - 199x/140 + 33/2
g. Calculate the area of y=g(x) and y=h(x) to 4 decimal places.

2. g(x)=ax^3 + bx^2 + cx + d, where x ≥ 0
a. If g(0)=33/2, g(2)=14, g(4)=23/2 and g(7)=11/2, determine simultaneous equations and solve a, b, c and d algebraically, then show that g(x)= -3x^3/140 + 9x^2/70 - 199x/140 + 33/2 where x ≥ 0

3. g(x)= -9x^2/140 + 9x/35 = 199/140
a. Use the bisection method to derive the first positive root of g(x) between a=8 and a=9
b. Draw the graph of y { g(-x), -10⩽x⩽0 and g(x), 0⩽x⩽10

4. f(x)=ax^2 + bx + c where the co-ordinates of the graph are (-2,15), (0,16), (2,15), (-2,12), (0,12) and (2,12)
a. Find values of a, b and c using simultaneous equations and hence show that f(x)= -1x^2/4 + 16
b. What is the gradient of f(x) at x=0, 1, 2?

I know it is a lot but would really appreciate any help!

#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #17949 on: June 05, 2019, 11:46:06 pm »
+3
Hello all, I am doing a practice modeling task and struggling very badly I really need some help on all these questions, would really appreciate it!

...

I know it is a lot but would really appreciate any help!

Question 1

Part a
This part should be pretty straight forward. Substitute the definition of $\cosh()$ into $h(x)$. Ie. $h(x)=4\left(3-\frac{e^{x/50}+e^{-x/50}}{2}\right)=12-2e^{x/50}-2e^{-x/50}.$

Part b
Write $\dfrac{d}{dx}\big(h(x)\big)$ into your CAS.

Part c
Use your CAS to graph $h$. Find the coordinates of the axial intercepts and sketch.

Part d
The average value of a function $f$ on the interval $[a,\,b]$ is $\frac{1}{b-a}\int_a^b f(x)\,\text{d}x.$ You would have found the $x$-axis intercepts in part c. Plug all the information into the formula.

Part e
Write into your CAS $\texttt{solve}\left(\frac{d}{dx}\big(h(x)\big)=0,x\right)$

Part f
Write into your CAS $\texttt{solve}\Big(h(x)=g(x),x\Big)$ and choose the smallest positive value of $x$.

Part g
This question is broken as it doesn't actually describe a region in the plane. Have you perhaps misquoted the question?

Question 2

Part a
The question actually gives you the four equations. $\begin{cases}g(0)=33/2\\ g(2)=14\\ g(4)=23/2\\ g(7)=11/2 \end{cases}\implies \begin{cases} 0a+0b+0c+d=33/2\\ 8a+4b+2c+d=14\\ 64a+16b+4c+d=23/2\\ 343a+49b+7c+d=11/2\end{cases}$ Start with the first equation which gives $d$, then work through the algebra to solve for the remaining 3 variables. Use your CAS to guide you along the way.

I don't really have the time at the moment to go through the rest of the questions. If someone else hasn't discussed the remaining ones, I'll post tomorrow morning (I'm in serious need of a long sleep lol).
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#### JR_StudyEd

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##### Re: VCE Methods Question Thread!
« Reply #17950 on: June 06, 2019, 04:32:23 pm »
0
Hi, I'm struggling to relate the concept of the unit circle with the sin, cos, tan graphs. Like, I'm alright with exact values, but I can't really recall stuff outside  π/6, π/4, π/3, π/2. My teacher told me to 'memorise' them like you do your times tables, how does that work? Memorising, as you would imagine, is quite a pain and not how I like to recall things.

#### Sine

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##### Re: VCE Methods Question Thread!
« Reply #17951 on: June 06, 2019, 05:27:24 pm »
+1
Hi, I'm struggling to relate the concept of the unit circle with the sin, cos, tan graphs. Like, I'm alright with exact values, but I can't really recall stuff outside  π/6, π/4, π/3, π/2. My teacher told me to 'memorise' them like you do your times tables, how does that work? Memorising, as you would imagine, is quite a pain and not how I like to recall things.
Definitely, learn the principals behind the concept - that is the easiest way to "memorise"  without memorising something
The idea of the unit circle is that any point on that circle as a certain x value and a certain y value.
If you consider a radius from the centre of the circle (0,0) to a point (x,y) the distance is 1 since it is a unit circle. Then from that, you can produce a right triangle with one side parallel to the y-axis and the other side length parallel to the x-axis.
If you now think of the angle (theta) at the origin (from x-axis to the radius of 1 from the origin to (x,y)) you can determine that the two side lengths of this triangle are sin(theta) = y, cos(theta) = x.
This is clear if you use normal trig rules for that triangle.
That is the basic mechanism behind what values of sin and cos are.

The graphs of sin/cos just map out the changes as you move from the right most point on the unit circle (angle = 0) anticlockwise to increase the angle size.

Consider, the sin graph. we found sin(theta) = y.
Think about how the "height" of the triangle changes as you move anticlockwise around the unit circle.

Hopefully, this helps, sorry for the lack of diagrams.

You will use these values so much you will know them off by heart.
An easy way to learn them (or check) is to follow this trend

$sin(0) = \frac{\sqrt{0}}{2} = 0$
$sin(π/6) = \frac{\sqrt{1}}{2} = \frac{1}{2}$
$sin(π/4) = \frac{\sqrt{2}}{2}$
$sin(π/3) = \frac{\sqrt{3}}{2}$
$sin(π/2) = \frac{\sqrt{4}}{2} = \frac{2}{2} = 1$

And you can basically reverse it for cos, then for tan it is just sin divided by cos

#### JR_StudyEd

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##### Re: VCE Methods Question Thread!
« Reply #17952 on: June 06, 2019, 05:49:33 pm »
0
I'm also confused about what x-axis scale I should use when sketching trig graphs.

#### Geoo

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##### Re: VCE Methods Question Thread!
« Reply #17953 on: June 06, 2019, 08:58:04 pm »
0
Hello!
So I am doing a question on probability with Karnaugh maps, and was sable to solve everything around the table but the 4 squares in the middle.
Heres the question:
80 students attended a scout's camp where surfing was offered in the morning and bushwalking in the afternoon. Every student attended at least one activity. 44 students went surfing and 60 students went bushwalking.

So I inputted my table S for surfing and BW for bush walking, and I have answered 5 boxes. Since there was no number in the first 4 squares (If I did it right), I don't know how to input those numbers.
I also changed the whole numbers to decimals (percentage) since that's all I saw in examples.

I have no Idea what I am doing, and I don't have a teacher!
Help, pretty urgent

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#### Sine

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##### Re: VCE Methods Question Thread!
« Reply #17954 on: June 06, 2019, 09:45:49 pm »
+2
Hello!
So I am doing a question on probability with Karnaugh maps, and was sable to solve everything around the table but the 4 squares in the middle.
Heres the question:
80 students attended a scout's camp where surfing was offered in the morning and bushwalking in the afternoon. Every student attended at least one activity. 44 students went surfing and 60 students went bushwalking.

So I inputted my table S for surfing and BW for bush walking, and I have answered 5 boxes. Since there was no number in the first 4 squares (If I did it right), I don't know how to input those numbers.
I also changed the whole numbers to decimals (percentage) since that's all I saw in examples.

(Image removed from quote.)

I have no Idea what I am doing, and I don't have a teacher!
Help, pretty urgent

hmm I notice that the total number of students is not 100 so it wouldn't be accurate to conert 40 students to 0.40

S                S'

BW              24                36             60

BW'             20                 0              20

44              36               80

The key is the number of S' and BW' is 0.