Step 2

Now assume that the weight is distributed uniformly along the shape given by the answer to Step 1, that is to say, \(s\) is now the arc length of \(y=f_1(x)\).

What is \(f_2(x)\) approximately in this case, where \(f'_2(x)=s\)? Work out an answer as a polynomial, valid for small values of \(x\), including one term more than in \(f_1(x)\).

We need to work out the arc length of \(f_1(x)=1+\frac{1}{2}x^2\) to find \(s\).

But we only want an approximate answer as a polynomial, so how can we approximate the integral for arc length so that what we are integrating (the integrand) is itself a polynomial?

In the limit

  1. In the above procedure, what would Step \(n\) have been? Can you prove that Step \(n\) would have given the function \(f_n(x)\) which you predicted?

Proving that Step \(n\) follows might be tricky to do directly from the polynomial \(f_{n-1}(x)\). It might be helpful to consider what you already know about the limit function and its behaviour.

A direct approach

Can you solve the equation \(f'(x)=s\) directly?

You may find the substitution \(p=f'(x)\) helpful.