Wednesday, April 8, 2009
A Recurring Problem for SATs (Functions)
SAT "Grid-In" Type
Level of Difficulty: 5 (High)
Content: Algebra 2, precalculus
The function F satisfies the condition
F(N + 6) = F(N) + 8, for all integers N.
If F(7) = -2, what is the value of F(25)?
Click on Read more to see the answer, solution, discussion.
Answer: 22
Suggested Solution:
Replace N by 7 since F(7) is known:
Therefore, F(7 + 6) = F(7) + 8 or F(13) = -2 + 8 = 6
Next, replace N by 13 since F(13) is known:
F(19) = F(13) + 8 = 6 + 8 = 14
Finally, F(25) = F(19) + 8 = 14 + 8 = 22.
Comments
(1) Too difficult for the SATs? Not really! A similar problem recently appeared. There aren't that many "hard" questions (Level 5) on the SAT but, if a student wants to score over 700 they will need exposure to these types in practice.
(2) Consider writing some variations of these function-type problems for additional practice. At first, change the constants, then consider changing the operations (from addition to multiplication for example). One could raise the bar even higher by asking the question in reverse:
If F(25) = 22, what is the value of F(7)?
(3) There is considerable advanced theory in functional equations and recurrence relations underlying these problems. However, the student needs only to feel comfortable with the function symbolism (or should I call it "Math Notations!"). Starting by "plugging in' N = 7 seems simple in retrospect but most students are too intimidated to consider it. Even the precalculus student may be able to get started, but, without experience, they will often get lost. This is all about exposure, but isn't it always?
(4) One could rewrite this problem using sequence notation:
aN+6 = aN + 8. By expressing the problem in the context of the Nth term of a sequence, students may grasp it a bit better, but, in the end, it's all about interpreting function notation.
Posted by Dave Marain at 1:51 PM 0 comments
Labels: algebra 2, functional relationships, functions, more, SAT strategies, SAT-type problems
Saturday, November 17, 2007
The Classic Cone Inscribed in the Sphere Problem: Developing Relationships Before Calculus
Update: View the series of videos here explaining the procedure for solving the cone in the sphere problem below as well as related questions.
Many Algebra 2 and Precalculus textbooks have begun to include those challenging 3-dimensional geometry questions involving 2 or more variables and/or constants. However, we know from the difficulty that calculus students continue to have with these, that we need to do more before students do their first optimization problems in calculus. You know the kind: Determine the radius of the __________ of maximum volume that can be inscribed in a _________ of radius R. These problems have fallen out of favor somewhat with the AP Development Committee, perhaps because they lack that real-world flavor or perhaps because they had become predictable or perhaps too hard. I would argue they have been part of the rites of passage for calc students for many generations for a reason - they blended the spatial reasoning of geometry with the need to identify variable relationships and reduce the number of conditions down to one function of one variable if possible. In other words, they help to develop mathematical sophistication. I 'cut my teeth' on these -- did you? Any calculus teachers reaching this topic yet in AP Calc?
Anyway here's an activity for you Algebra 2 or Precalculus students to prepare them for these challenges. As usual we proceed from the concrete (i.e., given numerical dimensions) to the abstract. Rather than attempt to draw the diagram, which is fairly challenging for me given the tools I have, I will describe the problem verbally. Good luck!
STUDENT ACTIVITY
(1) A right circular cone of height 32 is inscribed in a sphere of diameter 40.
Note: Students need to learn how to make a diagram of this problem situation.
(a) Determine the radius of the cone.
(b) Determine the volume of the cone. [Imagine asking students to memorize the formula!]
(c) Keep the diameter of the sphere at 40. This time, determine both the radius and volume of the inscribed cone whose height is 80/3. The numbers are messy but try to work in exact form (fractions, radicals) before rushing to the calculator to convert everything to decimals. Oh well, we all know what will happen here!
(d) Try another value for the height of the cone, keeping the diameter of the sphere at 40. See if you can produce a volume greater than in (c). Any conjectures?
(2) We could throw in an intermediate step by using a parameter R to denote the radius of the sphere, and use numerical values for different possible heights of the cone, but I'll leave that to the instructor. Instead, we'll jump to the abstract generalization:
A right circular cone of height h is inscribed in a sphere of radius R.
(a) Express the radius, r, of the cone in terms of R and h.
(b) Express the volume, V, of the cone as function of h alone (R is a constant here).
(c) Use your expression for r and your function for V to verify your results in (1).
(d) Calculus Students: You know what the question will be! Oh, alright:
Determine the dimensions and volume of the right circular cone of maximum volume that can be inscribed in a sphere of radius R. Anything strike you as interesting in this result?
Posted by Dave Marain at 7:36 AM 0 comments
Labels: calculus, cone, functional relationships, geometry, optimization, precalculus, sphere