Calculus countdown
Can you hit the target functions using a set of input functions and a little calculus and algebra?
Problem
In the game of Calculus Countdown you are given the following four machines into which you insert cards with functions written on them; the four machines chew up the input card(s) and spit out new cards with functions written on them. You can put any output cards back into the machines if you like. The idea of the game is to hit certain target cards given a set of initial cards.
Let's play a game. You are given the following initial seven cards (no constants of integration from the integral machine and no repeats of cards, other than the pair of $e^x$s)
Which of the following targets could you hit starting with these cards? You can use a fresh set of seven cards for each new target.
Can you make a smaller set of cards which could hit each of these targets?
Why not invent your own set of starting cards and targets?
Getting Started
To begin with, try acting on the starting functions with various operators to see the sorts of possibilities that occur.
To hit a target, think what functions would be needed to make them in one step. How might I arrive at these inputs?
Student Solutions
Game a): $\textrm{Target} = 8$
$\mathrm{D} \left( x^2 \right) = 2x$
$\mathrm{D} \left( 2x \right) = 2$
$\mathrm{P} \left( 4,2 \right) = 8$
Game b): $\textrm{Target} = x^4$
$\mathrm{P} \left( x, x^2 \right) = x^3$
$\mathrm{I} \left( x^3 \right) = \frac{x^4}{4}$
$\mathrm{P} \left( \frac{x^4}{4}, 4 \right) = x^4$
Game c): $\textrm{Target} = \frac{1}{2}$
$\mathrm{D} \left( x^2 \right) = 2x$
$\mathrm{D} \left( 2x \right) = 2$
$\mathrm{R} \left( 2 \right) = \frac{1}{2}$
Game d): $\textrm{Target} = \frac{x^6}{36}$
Method 1:
$\mathrm{I} \left( \mathrm{I}(x) \right) = \frac{x^3}{6}$
$\mathrm{P} \left( x^2,\frac{x^3}{6} \right) = \frac{x^5}{6}$
$\mathrm{I} \left( \frac{x^5}{6} \right) = \frac{x^6}{36}$
Method 2:
$\mathrm{I} \left( x^2 \right) = \frac{x^3}{3}$
$\mathrm{D} \left( \ln(x) \right) = \frac{1}{x}$
$\mathrm{R} \left( \frac{1}{x} \right) = x $
$\mathrm{P} \left( x, x \right) = x^2 $
$\mathrm{I} \left( x^2 \right) = \frac{x^3}{3}$
$\mathrm{P} \left( \frac{x^3}{3}, \frac{x^3}{3} \right) = \frac{x^6}{9}$
$\mathrm{R} \left( 4 \right) = 0.25 $
$\mathrm{P} \left( 0.25,\frac{ x^6}{9} \right) =\frac{x^6}{36}$
Game e): $\textrm{Target} = \frac{-32}{x^5}$
$\mathrm{D} \left( \mathrm{D} \left( \mathrm{D} \left( \ln(x) \right) \right) \right) = 2x^{-3}$
$\mathrm{R} \left( x \right) = \frac{1}{x}$
$\mathrm{P} \left( x^{-1}, 2x^{-3} \right) = 2x^{-4}$
$\mathrm{P} \left( 4, 2x^{-4} \right) = 8x^{-4}$
$\mathrm{D} \left( 8x^{-4} \right) = -32x^{-5}$
Game f): $\textrm{Target} = x(2 - x)$
$\mathrm{P} \left( x^2, \mathrm{R}(\exp(x)) \right) =x^2 \exp(-x))$
$\mathrm{D} \left( x^2 \exp(-x) \right) = 2x \exp(-x) - x^2 \exp(-x)$
$\mathrm{P} \left( \exp(x),2x \exp(-x) - x^2 \exp(-x) \right) = 2x-x^2$
Teachers' Resources
Why do this problem?
This problem gives a great engaging context in which to practise calculus. It is very useful for getting across important ideas concerning integration and differentiation as operations, which will be useful at university in both science and maths courses.
Possible approach
Key questions
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