Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    12
    ReputationRep:
    Name:  image1.jpg
Views: 67
Size:  499.1 KB

    Can I please have some help with part (a) and hopefully it will naturally lead onto part (b). I'm pretty stuck with what they are asking. I've done a small amount of working/doodling near the bottom of the picture.

    Many thanks,

    Jamie
    Offline

    11
    ReputationRep:
    Here may be a clue. The first function is the set of spheres centre 0,0,0 and the second is a set of planes and in the case of the vector given, the sphere has radius sqrt(8) and the plane is the plane at 45 degrees where the line y=x extends into R3.

    The intersection is a circle also inclined at 45degrees where the plane intersects the sphere.

    All the points on this circle have the desired property that f1=8 and x-y =0


    Warning, I have not done much work recently on this kind of question but no one else replied
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by nerak99)
    Here may be a clue. The first function is the set of spheres centre 0,0,0 and the second is a set of planes and in the case of the vector given, the sphere has radius sqrt(8) and the plane is the plane at 45 degrees where the line y=x extends into R3.

    The intersection is a circle also inclined at 45degrees where the plane intersects the sphere.

    All the points on this circle have the desired property that f1=8 and x-y =0


    Warning, I have not done much work recently on this kind of question but no one else replied
    So this implies about the shape of the intersection of these functions, but what for finding the points that satisfy them?
    Offline

    11
    ReputationRep:
    I think that all that is wanted is a statement that the points on this inclined circle are the set of points where the vector function has the desired property of f1=8 and f2 =0.

    I think that part b is a matrix where the elements map the various derivatives you need for the gradient of a vector function.

    As I said, I rarely visit this stuff but I guess the normal gurus are not around.
    Offline

    19
    ReputationRep:
    (Original post by Jamie S)
    So this implies about the shape of the intersection of these functions, but what for finding the points that satisfy them?
    well well well ...


    Name:  H7.jpg
Views: 24
Size:  15.6 KB


    Shall I contribute?
    I think not ....
    • Thread Starter
    Offline

    12
    ReputationRep:
    I think there is more to part a?
    • Thread Starter
    Offline

    12
    ReputationRep:
    help
    Offline

    10
    ReputationRep:
    (Original post by Jamie S)
    help
    nerak99 Has provided the graphical situation.

    There may be more to part a) but:

    

2x^{2} + z^{2} = 8,

    is of a particular graph (a particular conic section (http://math2.org/math/algebra/conics.htm )).

    (Formula from the attachment in your original post).
    Offline

    13
    ReputationRep:
    (Original post by Jamie S)

    Can I please have some help with part (a) and hopefully it will naturally lead onto part (b).
    I think that others have already disposed of part (a). Part (b) appears to be independent of part (a) and just requires you to write down the matirix of partial derivatives.
    Offline

    11
    ReputationRep:
    In other words, all the points on that ' circle' (2x^2+z^2=8 is not a circle but an ellipse parallel to the y axis! which I can't picture from the functions! but certainly a conic) satisfy the requirements of part a.


    I think that this is the end of it. Following that, part b flows from the definition of a derivative of a factor valued function, which, so far as I can tell, is a matrix mapping the elements of the function and their partial derivatives.

    Edit: I think the function 2x^2+z^2 = 8 is a projection of the circle onto the z-x plane.
    Offline

    11
    ReputationRep:
    I think the set of points is a circle in 3D and the equation 2x^2+z^2=8 is a projection of the circle onto one plane. We are after a circle that lies on the plane defined by the line x=y projected in the z direction intersecting the sphere of radius \sqrt 8 centre (0, 0, 0).

    As I said above, my work on vector defined functions is pretty thin and possibly the curve needs to be defined as a vector defined function rather than the following. However, for what it is worth, to define an inclined circle the parametric form uses a unit vector perpendicular to the plane of the circle \bold n, a unit vector that is radial to the circle \bold u and the radius.

    The vectors are easy in that we can get our vectors in the xy plane by inspection. For \bold u we have\frac{1}{\sqrt 2}\begin{pmatrix} 1  \\1  \\0  \end{pmatrix}
    For \bold n we have\frac{1}{\sqrt 2}\begin{pmatrix} 1  \\-1  \\0  \end{pmatrix}

    The general form of a circle centre c [(0, 0, 0) in our case ] inclined in 3D space is \bold r = R\cos \theta \bold u + R \sin \theta \bold n + \bold c which in our case gives us

    \bold r = \sqrt 8 \cos \theta \bold u + \sqrt 8 \sin \theta \bold n as the curve which is the intersection of the two functions.

    Edit. I forgot to normalise the unit vectors
    • Thread Starter
    Offline

    12
    ReputationRep:
    Thanks, everyone! I was almost there then with my doodles. I think I was just expecting more depth to the question, thanks everyone.
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.