VCE Specialist 3/4 Question Thread!

[mark_alec]

could someone help me with    f, g and h?  thanksssss

What's the question?

[Jaswinder]

sorry forgot to upload. did it on the first. help appreciated

[Alwin]

Hi guys, i just had a quick q about terminology:

what do we call the stationary point of x⁴ (or higher even power of x)?

for f(x) = x⁴, f'(0) = 0 clearly and when you graph it, looks like at local min at x = 0 so that's what i've always called it.
but, f''(0) = 0 which implies (in the VCE course) a point of inflection.

a "local minimum stationary point of inflection" just sounds wrong...

[lzxnl]

Actually, in the VCE course, a point of inflection only occurs when the first derivative changes sign. f''(x) = 0 is a necessary but insufficient condition. y=x^4 would have a global minimum at the origin. There, the first derivative is 4x^3 but does not change sign at the origin. The second derivative is zero, but so is the third derivative.
For the general nth derivative test, if the first n derivatives are all zero at a point, but the (n+1)th derivative is not, then if n+1 is even, the stationary point is a local extremum, but if n+1 is odd, then the stationary point is a point of inflection. If n+1 is even and the (n+1)th derivative is positive, then we have a local minimum and vice versa.

[Phy124]

In addition to what nliu1995 has said, this is a good explanation 

[Alwin]

Actually, in the VCE course, a point of inflection only occurs when the first derivative changes sign. f''(x) = 0 is a necessary but insufficient condition. y=x^4 would have a global minimum at the origin. There, the first derivative is 4x^3 but does not change sign at the origin. The second derivative is zero, but so is the third derivative.
For the general nth derivative test, if the first n derivatives are all zero at a point, but the (n+1)th derivative is not, then if n+1 is even, the stationary point is a local extremum, but if n+1 is odd, then the stationary point is a point of inflection. If n+1 is even and the (n+1)th derivative is positive, then we have a local minimum and vice versa.

okay, cool thanks nliu! (ofc it was you that replied , or b^3)
never knew about the (n+1)th derivative rule thanks, I used to always just look at the power.. x^(even) is a tp and x^(odd) is stationary poi which seemed so insufficient as a proof. . .

Add: Thanks 2/cos(c), ill check out the link now

[Aelru]

kk, looks like everyone's working hard as usual

"The gradient of the normal to a curve at any point (x,y) is three times the gradient of the line joining the same point to the origin."

Construct a differential equation, but do not attempt to solve it.

I nearly got it. Not sure where my logic's flawed though ><



1.As gradient of the line is three times to LH, .
2.Next is the part I got wrong. As "gradient of the normal x gradient of tangent = -1",

Yet the answer says
I assumed they just divided the 3 on RH to the LH, but not sure where the derivative bits disappeared to.

EDIT: unless i mistaken gradient of the normal , as differentiating the normal..

[Jeggz]

Aah this question was on our SAC this week!
Anyways, this is my logic -

We know the gradient of the curve at any point (x,y) is , hence the gradient of the normal would be
In regards to the gradient of the line, let's use the formula , where rise = and run = , which makes the gradient of the line .

Piecing together all the information we have we can conclude that
, re-arranging this gives the answer

[Aelru]

Aah this question was on our SAC this week!
Anyways, this is my logic -

We know the gradient of the curve at any point (x,y) is , hence the gradient of the normal would be
In regards to the gradient of the line, let's use the formula , where rise = and run = , which makes the gradient of the line .

Piecing together all the information we have we can conclude that
, re-arranging this gives the answer

wonderful explanation. Thanks Jeggz~

[bonappler]

Having trouble with this one.

[Conic]

Part a:
First we find the width of the rectangle:
(Check the diagram for the variables I used)













Now the area is simply , or .



Part b:





Using the initial conditions (t=0, x=4):





Part c:



Which is 42 hours and 40 minutes.

[scribble]

you can also find the surface area by writing the equation of the circle:

W^2+(h-2)^2=4
solve for w in terms of h

and then the width of the water is 2W

[Jaswinder]

could someone help me with    f, g and h?  thanksssss

sorry: forgot to upload

bump

[SocialRhubarb]

f.)

Spoiler

Eqn 1.

Eqn 2.

Alternatively, you could use the negative half of the ellipse.

Eqn 1 divided by Eqn 2 gives:

















Eliminate 5.58, since we generated an extra solution when we squared both sides, it doesn't actually work, which you can see from the graph.

Done with windows calculator, not sure if accurate.

g.)

Spoiler

i.)

ii.)

[saba.ay]

Could someone please explain how to do part d)? The midpoint, M, is ((3/4)*cos(θ) , (-3/2)*sin(θ)).

Please and thank you