Author Topic: VCE Methods Question Thread! (Read 5971301 times) Tweet Share

Mr. Study · « **Reply #690 on:** June 07, 2012, 07:12:03 pm »

Fine!

Jean has conducted an experiment controlling the amount of water given to the coffee plants using a drip system. The yield of the coffee beans in kg for a single coffee plant was found to follow the equation
$y=-0.072L^3+0.4068L-0.0216$

Where L is the number of litres per day given to the plant and Y is the yield in kg of coffee.

j.Sketch the graph of yield vs water over a appropriate domain.
Please use your calculator for this one.

k. Find the derivative of Y and hence find the maximum yield and the volume of water per day need to achieve this yield.
The derivative will be dY/dL = $-0.216x^2+0.4068$

Let dY/dL = 0 and solve for x. Which would be x=-1.37235 or x=1.37235

l. What is the minimum yield of coffee over the given domain. Explain your reasoning in one-two sentences.

Just sub both answers in and choose the lowest one.

CommanderElahi · « **Reply #691 on:** June 07, 2012, 07:33:41 pm »

Cheers buddy, greatly appreciated, you are a genius.

MelonBar · « **Reply #692 on:** June 19, 2012, 01:26:19 am »

Can someone please tell me what I did wrong? Essentials 11g q2. Pdf file of worked solutions stop at 11e.

An aeroplane is flying horizontally at a constant height of 1000m. At a certain instant the angle of elevation is 30 degrees and decreasing and the speed of the aeroplane is 480km/hr.

How fast is theta decreasing at this instant?

d(theta)/dt = dx/dt * d(theta)/dx

tan theta = 1km/x
x = (tan theta)^-1

dx/dtheta = -1/sin^(2)theta
therefore dtheta on dx = -sin^(2)theta (flip the fraction)

dx/dt = 480 as given in q

Using chain rule, and as 'at this instant' implies when theta = 30 degrees,
d(theta)/dt = 480*- 1/4 = -120 radians

Special At Specialist · « **Reply #693 on:** June 19, 2012, 05:38:30 pm »

I don't have the Essential's textbook. Would you please tell me what x represents? A diagram would be helpful, so that I don't misinterpret the question.
And the symbol for theta is: θ (feel free to copy and paste it in the future).

Kanon · « **Reply #694 on:** June 19, 2012, 11:04:04 pm »

Hey guys, was just revising Unit 3, in an equation such as
y = a^(-kx) sin(x)
what does the value of k represent? Would you still describe this as a dilation?

studynotes · « **Reply #695 on:** June 21, 2012, 01:05:24 pm »

Just wondering if you can solve circular functions without changing the domain thats given

brightsky · « **Reply #696 on:** June 21, 2012, 04:01:36 pm »

Quote from: Kanon on June 19, 2012, 11:04:04 pm

Hey guys, was just revising Unit 3, in an equation such as
y = a^(-kx) sin(x)
what does the value of k represent? Would you still describe this as a dilation?

doesn't represent anything as far as transformations are concerned.

Kanon · « **Reply #697 on:** June 21, 2012, 10:18:30 pm »

Quote from: brightsky on June 21, 2012, 04:01:36 pm

Quote from: Kanon on June 19, 2012, 11:04:04 pm
Hey guys, was just revising Unit 3, in an equation such as
y = a^(-kx) sin(x)
what does the value of k represent? Would you still describe this as a dilation?

doesn't represent anything as far as transformations are concerned.

Thank you!

Could someone help me solve

$cos(\pi x /2 ) = 1$

Jenny_2108 · « **Reply #698 on:** June 21, 2012, 10:26:35 pm »

whats the domain given?

HenryP · « **Reply #699 on:** June 22, 2012, 05:52:55 pm »

Without a domain I'll just assume they would want a general solution.
We have Cos(pi*x/2)=1
That means pi*x/2 = 2npi where n are integers.
Solving for x gives x=4n
Hopefully there aren't any mistakes in my work and I hope this helps!

paulsterio · « **Reply #700 on:** June 23, 2012, 02:04:37 am »

Quote from: Kanon on June 19, 2012, 11:04:04 pm

Hey guys, was just revising Unit 3, in an equation such as
y = a^(-kx) sin(x)
what does the value of k represent? Would you still describe this as a dilation?

It actually a dilation parallel to the x-axis, away from the y-axis

e^1 · « **Reply #701 on:** June 26, 2012, 06:10:16 pm »

Hey everyone, I am having trouble with this question (got it from the teacher). I'll use latex so that it doesn't look messy for the bits I do know how to go through. For parts d and e I am not sure about. If anyone can help that'll be great.

a) Finding the cost is equal to XA+ AZ. Hence:

Find the length of XA.
$10^2 + p^2 = (XA)^2$

$\sqrt{100 + p^2} = XA$

Cost of the cable of AX $= \$10000w(\sqrt{100 + p^2})$

Cost of the cable of AZ $= \$10000(20 - p)$

$\therefore C * \$10000 = \$10000(20 - p) + \$10000w(\sqrt{100 + p^2})$
$C = 20 - p + w\sqrt{100 + p^2}$

b) $\frac{dC}{dp} = -1 + \frac{pw}{\sqrt{100 + p^2}}$
To find minimum is when dC/dp = 0.

$0 = -1 + \frac{pw}{\sqrt{100 + p^2}}$

$\sqrt{100 + p^2} = pw$

$100 = p^2w^2 - p^2$

$100 = p^2(w^2 - 1)$

$\therefore p^2 = \frac{100}{w^2 - 1}$

c) Knowing $p^2 = \frac{100}{w^2 - 1}$

Transposing this equation for variable w gives:

$w^2 = \frac{100}{p^2} + 1$

$w = \sqrt{\frac{100}{p^2} + 1}$

$When &&w > \sqrt{2}$

$\sqrt{2} < \sqrt{\frac{100}{p^2} + 1}$

$2 < \frac{100}{p^2} + 1$

$p^2 < 100$

$p < 10$

When $w > \sqrt{2}$ , $p < 10$ , and so shows that the cable will pass partly or all of the resort, in this case.

#1procrastinator · « **Reply #702 on:** June 27, 2012, 03:35:30 pm »

Solve this for x

$\frac{1}{(x+1)} - \frac{4}{x-2} \ge 0$

After a bit of algebra, I ended up with

$-3(x+2)(x+1)(x-2) \ge 0$

So the solution I got was

$(-\infty, 2] \cup [1, 2]$

But my calculator gives both the intervals but says OR, so I'm assuming it means either only either one or the other , whereas I think my answer means and...assistance please!

pi · « **Reply #703 on:** June 27, 2012, 04:19:20 pm »

$\cup$ means "or" whilst $\cap$ means "and"

b^3 · « **Reply #704 on:** June 27, 2012, 04:23:12 pm »

Quote from: #1procrastinator on June 27, 2012, 03:35:30 pm

Solve this for x

$\frac{1}{(x+1)} - \frac{4}{x-2} \ge 0$

After a bit of algebra, I ended up with

$-3(x+2)(x+1)(x-2) \ge 0$

So the solution I got was

$(-\infty, 2] \cup [1, 2]$

But my calculator gives both the intervals but says OR, so I'm assuming it means either only either one or the other , whereas I think my answer means and...assistance please!

Quote from: VegemitePi on June 27, 2012, 04:19:20 pm

$\cup$ means "or" whilst $\cap$ means "and"

Also I might add I'm not sure if you've solved it correctly. When solving inequalities that aren't linear, you have to draw it out or do each case and remember when you solve it if you multiply by a negative number, then you have to flip the inequality.

So to start off
$\begin{alignedat}{1}\frac{1}{x+1} & -\frac{4}{x-2}\geq0 \\ \frac{1}{x+1} & \geq\frac{4}{x-2} \end{alignedat}$
Now when we multiply the $x+1$ and the $x-2$ up, they can be a negative or a postive number, so we will have four cases.

Case 1: Both Positive
$\begin{alignedat}{1}x+1>0\:\mathrm{and\:} & x-2>0 \\ x>-1\:\mathrm{and\:} & x>2 \end{alignedat}$

So we take this for $x>2$
$\begin{alignedat}{1}x-2 & \geq4x+4 \\ -6 & \geq3x \\ x & \leq-2 \end{alignedat}$
Now we know we can't have $x\leq2$ and $x>1$ at the same time, so we get nothing out of that.

Case 2: First Postive, Second Negative
$\begin{alignedat}{1}x+1>0\:\mathrm{and\:} & x-2<0 \\ x>-1\:\mathrm{and\:} & x<2 \end{alignedat}$
i.e. $-1<x<2$ , now we solve, flipping the sign as we are multiplying by a negative.
$\begin{alignedat}{1}x-2 & \leq4x+4 \\ -6 & \leq3x \\ x & \geq-2 \end{alignedat}$
So that means $-1<x<2$ the graph is greater than or equal to 0.

Case 3: First Negative, Second Positive
$\begin{alignedat}{1}x+1<0\:\mathrm{and\:} & x-2>0 \\ x<-1\:\mathrm{and\:} & x>2 \end{alignedat}$
Now this can't happen so we get nothing out of this.

Case 4: Both Negative
$\begin{alignedat}{1}x+1<0\:\mathrm{and\:} & x-2<0 \\ x<-1\:\mathrm{and\:} & x<2 \end{alignedat}$
So we will solve for when $x<-1$
BUT since we are multiplying by two negatives, we don't need to flip the sign.
$\begin{alignedat}{1}x-2 & \geq4x+4 \\ -6 & \geq3x \\ x & \leq-2 \end{alignedat}$
Now the intersection of the two is $x\leq-2$

So now we have $\frac{1}{(x+1)} - \frac{4}{x-2} \ge 0$ for $x\leq-2$ or $-1<x<2$ $\left(x\epsilon(-\infty,-2]\cup(-1,2)\right)$

And this is confirmed by the graph.

Hope that helps.

EDIT: F***ed up a negative in final answer, fixed.

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