If I have a ladder of length $24$ feet leaned against a wall, with the foot of the ladder being $10$ feet away from the wall; and a man pulls the foot of the ladder away from the wall at $4$ feet per second, then at what rate does the end of the ladder travel down the wall?
I modeled the problem mathematic all, first: If the leg of a right triagle is stretched (at $4$ ft/s), The hypotenuse is kept constant,and the right angle is also unchanging, then at what rate will the remaining leg change? We do know that the remaining leg should diminish as the former leg is made greater. If the rate at which the first leg grows (call this length $x$ from now on) is $dx/dt = 4$; then call the remaining leg $y$, and its rate of change with respect to the same variable y should be $dy/dt$. İt is also known that the rate at which the hypotenuse change is $0$ (call it z, so $dz/dt = 0$). Since $x^2 + y^2 = z^2$, then given the initial $x = 10$, and $z = 26$; $y$ must equal $24$. Imlplicitly differentiating the equation we have, we get $2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt)$. Substituting whats known, we get $dy/dt = - 80/48) = - 1.666...$; however, when I check if the answer is correct, I see that:
If $x$ grows its first increment, it become $14$ ($10 + 4$). By our calculations, at this increment, $y$ become $22.333...$ ($24 - 1.666...$); but, using the pythagorean theorem, this would mean that the hypotenuse is $26.358637...$; although this is quite close to $26$, in some problems, The answer deviates to as much as $2$ whole units. The answer, should have been at least as close as $\sqrt26^2 - 14^2$ giving $21.9089...$. Why does this deviation exist? İt is quite negligible here, but in other problem it makes quite the difference. Thank you in advance.
As your computations show, the rate at which the height is dropping, $\dot y$, is not a constant. Indeed you get $$\dot y =-\frac {4x}y$$ so it depends on the current position. Your "paradoxical" computation incorrectly imagines that $\dot y$ stays constant for an interval of time, but it only holds instantaneously.