VCE Methods Question Thread!

[HakunaMattata]

An entrance to a local suburban park has a series of posts connected with heavy chains

The chain between any two posts can be modelled by the curve defined by
h=0.295(e^x+e^−x),−0.6≤x≤0.6

where h metres is the height of the chain above the ground and x is the horizontal distance between the posts in metres. The x-axis represents the ground. The posts are positioned at x=−0.6 and x=0.6.

I've calculated the amount of sag in the chain (i.e. the difference in height between the highest points of the chain and the lowest point of the chain) = 0.1094 m

How do I calculate the angle the chain makes with the post positioned on the right hand side of the structure (i.e. at x=0.6) ?

Thanks in advance!

[wobblywobbly]

An entrance to a local suburban park has a series of posts connected with heavy chains

The chain between any two posts can be modelled by the curve defined by
h=0.295(e^x+e^−x),−0.6≤x≤0.6

where h metres is the height of the chain above the ground and x is the horizontal distance between the posts in metres. The x-axis represents the ground. The posts are positioned at x=−0.6 and x=0.6.

I've calculated the amount of sag in the chain (i.e. the difference in height between the highest points of the chain and the lowest point of the chain) = 0.1094 m

How do I calculate the angle the chain makes with the post positioned on the right hand side of the structure (i.e. at x=0.6) ?

Thanks in advance!

Recall that derivative = gradient at a point = tan(theta). Basically find the derivative at x=0.6, and solve tan(theta)=derivative for theta.

[@#035;3]

Quick Q, thanks in advance.
Show that (x+2a) is a factor of p(x)=x^3+(2*a-3)*x^2-2*(3*a-1)*x+4*a and hence solve the equation, p(x)=0

[keltingmeith]

Quick Q, thanks in advance.
Show that (x+2a) is a factor of p(x)=x^3+(2*a-3)*x^2-2*(3*a-1)*x+4*a and hence solve the equation, p(x)=0

You will find that the factor theorem will help you with the first part, and the second part can be done most obviously by polynomial division.

[cosine]

Quick Q, thanks in advance.
Show that (x+2a) is a factor of p(x)=x^3+(2*a-3)*x^2-2*(3*a-1)*x+4*a and hence solve the equation, p(x)=0

If (x+2a) is a factor of p(x), then this means that p(-2a) = 0. Prove this and you have shown that it is a factor.

p(-2a) = (-2a)^3 +(2a-3)(-2a)^2-2(3a-1)(-2a)+4a
p(-2a) = -8a^3 + (2a-3)(4a^2) + 12a^2 - 4a + 4a
p(-2a) = -8a^3 + 8a^3 - 12a^2 + 12a^2
p(-2a) = 0

As DickensFan101 said, you can use polynomial division to do the second part.

[upandgo]

sorry guys, got another question 

A piece of fencing 240m long will be used to enclose three sides of a rectangular field.
The fourth side has a brick wall.
Let L(m) be the length of the field. Let A (metre square) be the area of the field.

Express A as a function of L.

ive gotten 4 different answers and i have no idea if i'm on the right track! a huge thank you in advance 

[keltingmeith]

sorry guys, got another question 

A piece of fencing 240m long will be used to enclose three sides of a rectangular field.
The fourth side has a brick wall.
Let L(m) be the length of the field. Let A (metre square) be the area of the field.

Express A as a function of L.

ive gotten 4 different answers and i have no idea if i'm on the right track! a huge thank you in advance 

Is there a picture that goes with this question?

[qazser]

sorry guys, got another question 

A piece of fencing 240m long will be used to enclose three sides of a rectangular field.
The fourth side has a brick wall.
Let L(m) be the length of the field. Let A (metre square) be the area of the field.

Express A as a function of L.

ive gotten 4 different answers and i have no idea if i'm on the right track! a huge thank you in advance 

(240-L)/2=w

L(240-L)/2=A

[keltingmeith]

(240-L)/2=w

L(240-L)/2=A

Note that the structure looks something like this:
______________
|                          |
|                          |
|                          |
|_____________|

Where the green line is the brick wall that they've mentioned, and so will not include the fencing.

If you assume the length of the field is parrallel to the brick wall, you get:

240=L+2w
w=(240-L)/2

Which is what you got. HOWEVER, if you assume the length of the field is PERPENDICULAR to the brick wall, you get:

240=2L+w
w=240-2L

Which is VERY different to what you got.

Unfortunately, the vagueness of the question means we can't just jump in like this and assume we're doing things right.

[zsteve]

bump

CAS says yes yes it is

[Sanchita]

From the Cambridge Methods 3+4, Q5 in 1H:
'A rectangle ABCD is inscribed in a circle of radius a. Find an 'area-of-the-rectangle' function and state the domain'
Could anyone help me out with this? There is a diagram too; a rectangle perfectly within the boundary of a circle.

[Elizawei]

Hello!
First time posting on this board

How do I go about in doing this question:
"The graph of y=kx-2 intersects the graph of y=x^2+6x only once. Find possible values of k. (Tech free)"

Thanks in advance!!!! 

[natdogg]

How do I go about in doing this question:
"The graph of y=kx-2 intersects the graph of y=x^2+6x only once. Find possible values of k. (Tech free)"

Equate the two equations and then use the discriminant (if it intersects only once there's only one solution) to find an equation for k, which you can then solve for values of k

[Elizawei]

Thank you!!!! 

[YellowTongue]

How does 1 (attached) simplify to 2 (attached)? What am I missing here? Thanks