A function of gradient
Find curves which have gradients of +1 or -1 at various points
Age
16 to 18
Challenge level
Problem
A continuous, differentiable curve $y = f(x)$ enters a region of the x-y plane with gradient $+1$ at the point $(1, 1)$ and emerges from the region with gradient $+1$ at either point $A=(3, 4)$, $B=(3, 3)$ or $C=(3, 2)$. Another curve $y = g(x)$ enters a region of the x-y plane with gradient $+1$ at the point $(1, 3)$ and emerges from the region with gradient $-1$ at either point $D=(3, 4)$, $E=(3, 3)$ or $F=(3, 2)$.
Image
For each of the 6 possibilities, sketch - no equations required - possible forms of the graphs of the gradients of the functions against $x$.
In which cases must the gradient function necessarily be zero at at least one place in the obscured regions?
In which cases must the gradient of the gradient function necessarily be zero at at least one place in the obscured regions? Give a clear argument.
Find equations for 6 functions which have the correct gradients at these points.
Extensions: If the gradients at points $A$ to $F$ were allowed to vary, under which circumstances might there be no zeros in the gradients and the gradients of the gradients? Explore the possibility of two different functions matching the conditions for $B$.
NOTES AND BACKGROUND
The mathematical ideas involved in this question give some insight into a university-level analytical way of viewing calculus. For example, a key result proved in first year analysis courses is the Mean Value Theorem, which states that for functions which are differentiable for $a< x< b$ we can always find a number $c$ such that $f ^{\, '}(c)=\frac{f(b)-f(a)}{b-a}$.
Getting Started
Try fitting a gradient function and then integrating.
Quadratics are the easiest to fit.
Quadratics are the easiest to fit.
Teachers' Resources
Why do this problem?
This problem involves an exploration of the gradient function
and will make or reinforce the notion that 'the gradient at each
point' is itself a function. It either paves the way for calculus
or allows those who already know about the calculus of polynomials
to apply their knowledge in an unusual context. In either case, the
reasoning is sophisticated and suitable for the highly interested
student.
The first parts of this problem are well suited for those
about to learn calculus and the second parts (involving the actual
curve fitting) well suited to those who already know about
integration and might be about to learn more involved aspects of
calculus. The problem draws together elements of coordinate
geometry, curve sketching, curve fitting and basic integration of
polynomials.
The result is both visually beautiful and perhaps
surprising.
Possible approach
This problem is suitable for quiet, reflective work,
particularly the algebraic parts.
To begin, start with case $A$ on the board. Discuss how a
curve might get smoothly from $(1,1)$ to $(3,4)$ whilst having a
gradient $1$ at each of these points. The aim of this discussion is
for students to realise that the gradient MUST increase at some
point and the MUST decrease as some other point in the hidden
region.
You might then continue with case $E$. In this case the goal
of the discussion is to realise that the gradient MUST be zero are
some point in the hidden region.
Once the problem is understood, students might wish to
continue working alone or in small groups.
For the first parts demand clear examples and
explanations.
For the second parts students can plot their functions on the
same axes to produce a pleasing mathematical image.
Key questions
What are the constraints on the curves of gradient vs
$x$?
Do the curves need to go up or down between the start and end
points? What does this imply about the gradients?
Possible extension
Extension is suggestion in the problem. You might ask the
explicit question: 'Can you find two curves which satisfy condition
$B$? If so, what is its equation? If not, what categories of curve
fail to work?
Possible support
If students are struggling to get started, suggest that they
draw curves freehand which match the conditions and then
sketch the gradient function for these examples.
For the part involving the curve fitting, suggest explicitly
that students try fitting a gradient function and then integrating.
Quadratics are the easiest to fit and all that are needed in the
problem.