Question

What is the arc length of `f(t)=(t^2-4t,5-1/t)` over `t in [3,4]`?

Answer 1

Define `x(t), y(t)` as follows

`{ (x(t)=t^2-4t), (y(t)=5-1/t) :} => f(t) = (x(t), y(t))`

Differentiating `x(t), y(t)` wrt `t` we get:

`dx/dt = 2t-4`
`dy/dt = 1/t^2`

Then the arc length of `f(t)` over `t in [alpha,beta]` is given by;

`L = int_alpha^beta sqrt((dx/dt)^2+(dy/dt)^2)`

So over `[3,4]` we have
`L = int_3^4 sqrt((2t-4)^2+(1/t^2)^2)`

`:. L = int_3^4 sqrt(4t^2-16t+16+1/t^4)`

This definite integral does not have an intrinsic solution and would need to be solved numerically, using either a computer or estimated using the Trapezium Rule or Simpson's Rule