VCE Methods Question Thread!

[LiquidPaperz]

i get the theory and how do to the question, but why do we need to differentiate it first and find the values of x (before subbing in for y) in part a?

[LiquidPaperz]

and if someone can step my through 13a and also the general gist of what these type of equations are asking.

Thanks 

[HermioneGranger29]

Wouldnt finding the equation of the tangent line =do the derivative ? Im curious

[LiquidPaperz]

yeah. i know how to do the question but i dont know what im doing, im simply rote learning the method , i really want to know what im doing in each step

[HermioneGranger29]

The points which are parallel to x axis, are the turning points of the function. First derivative gives us equation of tangent line. The tangent line (which I think is straight right?) will be parallel to the x axis when it is completely horizontal (y=c). So by finding coordinates of x of the turning points, then subbing it in to find y, we get the points of the turning points of the function and those are the points where when we find derivative, the tangent will be parallel to x axis. Im not sure if that helps or if u needed something else, but there is a very high chance I may be wrong (due to the fact that I have hardly had any calculus lessons before)

[Mieow]

Help for these two questions would be greatly appreciated 

When Yehudi and Carlos play racquetball the probability that Yehudi wins a point is 0.35.
(a) Find the probability that Yehudi wins fewer than four points from the first fifteen played.

Two brands of soap, Bubbles and Sweet Smell, are competing for market share. At the moment Bubbles
has 500 customers and Sweet Smell has 300 customers. However, Sweet Smell has started an aggressive
publicity campaign and the company hopes to attract 25% of Bubbles’ customers each month while
losing only 10% of their own. How many customers will Sweet Smell hope to have after two months
of their advertising campaign?

[keltingmeith]

The points which are parallel to x axis, are the turning points of the function. First derivative gives us equation of tangent line. The tangent line (which I think is straight right?) will be parallel to the x axis when it is completely horizontal (y=c). So by finding coordinates of x of the turning points, then subbing it in to find y, we get the points of the turning points of the function and those are the points where when we find derivative, the tangent will be parallel to x axis. Im not sure if that helps or if u needed something else, but there is a very high chance I may be wrong (due to the fact that I have hardly had any calculus lessons before)

Girl, you seem to know a lot about calculus for someone who isn't graduating for another 3 years.

Okay LP, for your question, consider the gradient in both of those cases. If our tangent line is parrallel to the x-axis, it has a gradient of 0. Since the derivative is the gradient of our tangent line, this means that dy/dx=0. So, we get f'(x)=3x^2-6x =0, x=0 or x=2 are the points at which the tangent line is parrallel to the x-axis. Similar for part b, but f'(x)=-3 (you should be able to tell me why)

13a will come tomorrow when I don't have to write up uni assignments.

[Frozone]

Could someone please help me understand question 5?

[kinslayer]

Could someone please help me understand question 5?

We know that



If Z is a standard normal RV, we also know that:



Using the transformation



we get the two simultaneous equations



to which the unique solution (correct to three decimal places) is

[plato]

yeah. i know how to do the question but i dont know what im doing, im simply rote learning the method , i really want to know what im doing in each step

Let me try.
The tangent at the point (1,7) is a straight line with equation y=mx+c

The gradient of this tangent is the same as the gradient of f(x)=5x²+3x-1 at the same point.

The gradient function for the curve = f'(x)=10x+3
This is the rule for the gradient at any point on the curve for f(x)

At (1,7), the gradient =

The equation of the tangent was y=mx+c
At (1,7) , this equation is now y=13x+c

To find c, substitute any point that this tangent goes through. How about the point (1,7)?
Substitue (1,7) into y=13x+c

From this, c=-6
The equation to the tangent is then

The normal at (1,7) is perpendicular to the tangent at that point.
The nomal is also a straight line with equation


If two lines are perpendicular, the rule m₁xm₂=-1 applies.
Here, m₁ is the gradient of the normaland
m₂ is the gradient of the tangent = 13
so

= the gradient of the normal.
The equation to the normal is now
Substitute a point this normal passes through. Again use (1,7)




The equation to the normal is
A better version is

[LiquidPaperz]

Thanks for that! Why do we sub 1 in to find the gradient, do we disregard the 7 when finding gradient? No more y2-y1...?

thanks

[Phenomenol]

Thanks for that! Why do we sub 1 in to find the gradient, do we disregard the 7 when finding gradient? No more y2-y1...?

thanks

Rise / run is the gradient only for a linear equation, or as a linear approximation. Remember, with any higher order polynomial you need to take the derivative, and evaluate the derivative at x = 1 (in this case) to find the instantaneous gradient at a point.

f'(x)=10x+3 is only in terms of the x ordinate, so y = 7 is not used to find the gradient of the tangent at a point, but is used to find the final tangent equation (or normal equation). You substitute in a full coordinate (1 , 7) to find c.

[LiquidPaperz]

Thanks for that Phenomenol!

I've got 1 quick questions and a normal questions

for Q12) for part d it asks if its increasing more rapidly, i know that its increasing but i tried to see if it was increasing more rapidly so i did when r=10 and r = 11, minused these two and got 6.2831.. (formula is dA/dt = 2pi r) and then i compared r= 30 and r = 31 and got 6.2831.. hence it is not increasing more rapidly but answer says it is?

For Q14, how do i do part a and b? how can you have a rectangular fish tank has a square base? 

https://drive.google.com/folderview?id=0B3uruRCX0QlQOGlHb2RhSHR1WHM&usp=sharing

thanks

[#J.Procrastinator]

Hi all,

I need help with this probability density function question, where f(x) = 6e^-6x, x= time in hours, and I want to find the average waiting time in minutes.

In the textbook example it states that for this random variable X, E(X)= 1/alpha and E(X^2)= 2/alpha^2 ... I'm slightly confused with this concept, so could anyone help explain this to me? Couldn't we just integrate and multiply the probability density function by x and solve?

Thank you!! 

[keltingmeith]

Hi all,

I need help with this probability density function question, where f(x) = 6e^-6x, x= time in hours, and I want to find the average waiting time in minutes.

In the textbook example it states that for this random variable X, E(X)= 1/alpha and E(X^2)= 2/alpha^2 ... I'm slightly confused with this concept, so could anyone help explain this to me? Couldn't we just integrate and multiply the probability density function by x and solve?

Thank you!! 

For the exponential distribution , it has mean and variance , which is what they've pulled out. These definitions can easily be proved via integration (as you've suggested doing), however note that exponential distributions aren't covered in methods, and so you do not need to memorise all of this. If asked about this stuff, I would simply do it as you would any other PDF.