A $5$-foot long ladder is resting on a wall, so that the top of the ladder is 4 feet above the ground and the bottom of the ladder is $3$ feet from the wall. At some time, the ladder is slipping so that the top of the ladder falls at a constant rate of $1 \,\frac{\text{ft}}{\text{min}}$. How fast is the bottom of the ladder slipping away from the wall?
I let $y=$ distance along the wall, and $x=$ distance along the ground, so that, $$x^2 + y^2 = 25$$
Therefore,
$$2y\frac{dy}{dt}+2x\frac{dx}{dt}=0$$
Initally,
$$2(4)(-1)+2(3)\frac{dx}{dt}=0$$
$$\therefore\frac{dx}{dt}=\frac{4}{3}$$
This is all well and good, but my confusion lies in the following analysis:
We've established $dy/dt=-1$ and $dx/dt=4/3$, and we know that initially, $$4^2 + 3^2=25$$
Let's say a minute passes, then the vertical distance should decrease by 1 ft, the horizontal distance should increase by $4/3$ ft, and the length of the ladder is always 5 ft.
$$(4-1)^2 + (3+4/3)^2 = 27.77…\neq 25$$
I would have thought the ladder would always be 5 ft long, but according to the above, it is now ~5.27 ft long. A second iteration leads to a length of ~6 ft. This ladder seems to be growing in length, which does not make sense. What is going on?
Best Answer
$4/3 \text{ ft/min}$ and $-1 \text{ ft/min}$ are the instantaneous rates of change when $x = 3$ and $y = 4$. That rate of change is constantly changing as you pass that instant, and will not stay the same for a whole minute. Thus your analysis is incorrect because it assumes constant rates of change for a whole minute.