Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie

Maths HL question

  • 29-01-2014 10:23pm
    #1
    scallway
    Registered Users Posts: 20 ✭


    Hi,
    Does anyone know how to do question 9 on the 2014 SEC paper 2? With the tetrahedral pyramid in the cylinder? I'm totally stuck I'd appreciate any pointers.


Comments

  • #2
    Ompala
    Registered Users Posts: 906 ✭✭✭Ompala


    Ok first off the smallest volume cylinder will have a height equal to that of the tetrahedron, call this height h.
    Start by looking at the bottom, the sides of the tetrahedron lie on the edges of the circle at the base.
    Since all the length of the tetrahedron sides are of length 2a, that means this must be an equilateral triangle inside the circle, so all the angles are 60 degrees.
    Bring a line from a vertex to the centre of the circle, this will be the radius of the circle, r. This line will also bisect the angle, so we have 30 degrees between radius and the side of length 2a.
    Draw a perpendicular line from a side through the centre, which will bisect the side into equal lengths 2a.
    Those bits may seem vague, but draw it out and if you follow them correctly you should have a right angled triangle with hypotenuse of length r, adjacent of length a and the angle between 30 degrees. Use Trig to get the relationship r = 2a/(root3)
    Now you can form a right angled triangle between the radius r, the height h and a length of the tetrahedron 2a such that (2a)^2 = h^2 + r^2 , sub in r in terms of a from above to get h^2 = (8a^2)/3. Solve for h then just sub these into V = (pi)(r^2)(h) to get the desired result. Hope that helps answer your question, sorry if bits are tough to follow.


  • #3
    scallway
    Registered Users Posts: 20 ✭scallway


    Thank you so much :)


  • #4
    Ompala
    Registered Users Posts: 906 ✭✭✭Ompala


    No bother, I'm going to assume the trouble was finding the radius for the circle no?


  • #5
    scallway
    Registered Users Posts: 20 ✭scallway


    Yeah I kept getting a for it which I knew was wrong..


  • #6
    Peregrine
    Moderators, Education Moderators Posts: 26,403 Mod ✭✭✭✭Peregrine


    Alternatively, you could use the Sine rule with the base to find the radius, if you're not too comfortable with breaking down shapes into many different triangles.

    6e6d42039b9e5ff0ed13dc689645e08c.png


  • Advertisement
  • #7
    scallway
    Registered Users Posts: 20 ✭scallway


    Oh yeah! okay thanks again :)


  • #8
    Octotron
    Registered Users Posts: 19 ✭Octotron


    Hey sorry to hijack the thread, but can anyone help me out with HL Maths 2014 Q9 (c)?

    (c) A spherical snowball is melting at a rate proportional to its surface area. That is, the rate at which its volume is decreasing at any instant is proportional to its surface area at that instant.

    (i) Prove that the radius of the snowball is decreasing at a constant rate.


    (ii) If the snowball loses half of its volume in an hour, how long more will it take for it to melt completely?
    Give your answer correct to the nearest minute.

    Thanks!!


  • #9
    Ompala
    Registered Users Posts: 906 ✭✭✭Ompala


    Octotron wrote: »
    Hey sorry to hijack the thread, but can anyone help me out with HL Maths 2014 Q9 (c)?

    (c) A spherical snowball is melting at a rate proportional to its surface area. That is, the rate at which its volume is decreasing at any instant is proportional to its surface area at that instant.

    (i) Prove that the radius of the snowball is decreasing at a constant rate.


    (ii) If the snowball loses half of its volume in an hour, how long more will it take for it to melt completely?
    Give your answer correct to the nearest minute.

    Thanks!!

    http://www.boards.ie/vbulletin/showthread.php?t=2057125618
    2nd last post, still have no idea if c (ii) is right but it seems logical enough
    Then again I am quite biased here :pac:


Advertisement