Why should we use derivatives in finding the rate of a ladder held up against the wall. Can't we just use average rate formula?

Question

Imagine a 5 meter long ladder held up against a wall, where the top of the ladder starts of 4m above the ground. And its slipping down the wall in such a way that the top of the ladder is dropping at 1m/s. In that initial moment what is the rate at which the bottom of the ladder is moving away from the wall. Now by pythagoras theorem the distance from the ladder to the wall is 3m. And the rate at which the top of the ladder falls is 1m/s. And i suppose this rate is constant from the time (t=0 to t=the time at which it falls to the ground) so by the formula t=(o-4)m/(-1m/s)=4s. So it takes 4 sec to fall from 4 to 0 m. And so, this much time(t=4) should take for the bottom of the ladder which is 3m away from the wall to go from -3 to -5 (i have drawn +y axis as height of wall and -× axis as the distance of the wall and ladder) so the rate(again i suppose the rate is constant becoz i did not find any other information saying otherwise) should be (-5+3)m/4s=-1m/2s. So the rate is -0.5m/s. But my answer is wrong. Where did i go wrong? I know that height of the wall as the ladder slips down that height is a function of time y(t) and the distance of the ladder and the wall as ladder moves is a also a function of time t x(t). But if the relationship is linear, then taking the average rate of change is as good as taking the derivative.

Answer 1

If $t$ is time (in seconds), then the height of the top end of the ladder is $x(t)=4-t$ (in metres). Then, the distance of the bottom end of the ladder from the wall, assuming a rigid ladder (always of length $5m$) is $y(t)=\sqrt{5^2-(4-t)^2}$ (in metres).

This is not a linear function of $t$ and so at any time $t$, if you want to know the velocity, you cannot take the average and you have to take the derivative.