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### Hyperbolic Functions

Package of problems

## Suggestion

#### 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.