VCE Methods Question Thread!

[justsoundslikeaworn-outcliche]

 The equation of a hyperbola of a
building is (x)=E/(X+F)+G
It's height is=23, and meets ground 11.5m on either side of axis , curve also passes through (4,3)
   This info helps us find 3 equations...
Use height to find first equation.
   Use the points where it touches the ground on the right hand side to find the second
   Use pts (4,3) to find third
   Use i and ii to find G in terms of F only
   Show that F=1, G=2, and E=25
Need help with these :/ other post I had bumped too many times

[e^1]

a) Use height to find first equation.

Spoiler

When it's height is 23 cm, and we assume that the apex of the spire is at , then

b) Use the points where it touches the ground on the right hand side to find the second.

Spoiler

When (the ground), then . In this case I'm using the positive x-value because it only asks for the "right hand side".

c) Use pts (4,3) to find third

Spoiler

d) Use i and ii to find G in terms of F only

Spoiler

Take out e so that we only have to deal with variables f and g.

e) Show that F=1, G=2, and E=25

Spoiler

Knowing that and
NOTE: Including the third equation in this question is important so that the point (4,3) is of the hyperbola, since we did not use it in the last question. In other words, we only used two of the three system of equations.




Note that on your paper but this one is negative. If you plot the function it satisfies the three conditions -- not sure what's up with that.

Irrelevant

And oh yeah, you said you would love me forever. I love you too <3 haha

[justsoundslikeaworn-outcliche]

a) Use height to find first equation.

Spoiler

When it's height is 23 cm, and we assume that the apex of the spire is at , then

b) Use the points where it touches the ground on the right hand side to find the second.

Spoiler

When (the ground), then . In this case I'm using the positive x-value because it only asks for the "right hand side".

c) Use pts (4,3) to find third

Spoiler

d) Use i and ii to find G in terms of F only

Spoiler

Take out e so that we only have to deal with variables f and g.

e) Show that F=1, G=2, and E=25

Spoiler

Knowing that and
NOTE: Including the third equation in this question is important so that the point (4,3) is of the hyperbola, since we did not use it in the last question. In other words, we only used two of the three system of equations.




Note that on your paper but this one is negative. If you plot the function it satisfies the three conditions -- not sure what's up with that.

Irrelevant

And oh yeah, you said you would love me forever. I love you too <3 haha

Relevant: I LOVE YOU FOREVER haha thanks so much really cleared it up!

[Zealous]

Just wondering what answers others get for this question, not very difficult. I had this on a test for another subject, a little methods related.

Which one of the following sets of coordinates is a many-to-one relation?

     
     
     
     
     

[Professor Polonsky]

B and D, no?

[Zealous]

B and D, no?

Well I felt it was more than one, I actually thought B, D and E. But the test answers says it was E.
(it was like from a test builder or something.)

[academicbulimia]

Okay the area of a rectangle is A=50/(x+1)^2 -4
Find Max area and the dimensions of this rectangles door.
Neep help kind of urgent as well haha I know the max area is meant to be 24m
Oh and also length=25/x+1 -2 while width equals 2x

[Phy124]

Okay the area of a rectangle is A=50/(x+1)^2 -4
Find Max area and the dimensions of this rectangles door.
Neep help kind of urgent as well haha I know the max area is meant to be 24m
Oh and also length=25/x+1 -2 while width equals 2x

If the width is and the length is I don't see how the area can be ?

[Alwin]

Which one of the following sets of coordinates is a many-to-one relation?

          one-to-one relation since each x only one y
        many-to-one relation since x=-1, x=9 → y=6
                 one-to-many relation since one x=0 → y=4,5,6,7
          many-to-many relation since x=-2,2 → y=3 and x=-1,1→y=4
                 many-to-many relation since x=4,5,6,6 → y=1

Well I felt it was more than one, I actually thought B, D and E. But the test answers says it was E.

It appears as B, D, and E are all many-to-one functions. From the wording of the question, all 3 should be correct, but it's my guess that perhaps that mean many-to-one literally.. as in many x values to just the one y value (in this case, y=1 in E). If so, it is a poorly worded question

[academicbulimia]

If the width is and the length is I don't see how the area can be ?

It's meant to be  A=50x/x+1 -4

[Zealous]

....
It appears as B, D, and E are all many-to-one functions. From the wording of the question, all 3 should be correct, but it's my guess that perhaps that mean many-to-one literally.. as in many x values to just the one y value (in this case, y=1 in E). If so, it is a poorly worded question

Yep, that's exactly what I thought.
I asked one of the teachers and they said because, as you said, literally there are many x values that correspond with only 1 y value.
Meh, that was the only mark I lost on one of my tests. That's really annoying eh :0

It's meant to be  A=50x/x+1 -4

Ah your question still doesn't match up as 2/cos(c) mentioned.

Is there still a on the denominator now?

ummm..

Spoiler

  =  ?

[BubbleWrapMan]

There's no such thing as a 'many-to-one relation'...

[Planck's constant]

There's no such thing as a 'many-to-one relation'...

I can't see why not.
Functions are relations.

[aestheticatar]

Something that I'd like clarification on.
Let's say, I sketched a circle with radius 1 and centre (0,0).

Is it differentiable at x=1?
I.e. Are functions/relations differentiable at points where their tangent is a vertical line?

If so, how would I display my derivative function algebraically?
Thanks in advance

[Jeggz]

Something that I'd like clarification on.
Let's say, I sketched a circle with radius 1 and centre (0,0).

Is it differentiable at x=1?
I.e. Are functions/relations differentiable at points where their tangent is a vertical line?

If so, how would I display my derivative function algebraically?
Thanks in advance

How I see it, is that when you say "Is the function/relation differentiable at so and so point" you're asking whether or not the tangent at that point has a defined gradient. In this case the tangent is a vertical line at the point x=1 and hence the gradient is undefined right? Hence the circle at that particular point won't be differentiable either. I hope that's correct haha, and I'm hoping someone will correct me on this!