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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?

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$4/3 \text{ ft/min}$ 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.

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The issue here is that when you substitute for $x, y, \frac{dy}{dt}$ in your differentiated expression, you are substituting those values, like you state, "at some [particular] time," say, $t_0$: If we use notation that reflects this, we have $$2 x(t) x'(t) + 2 y(t) y'(t) = 0,$$ and substituting at our particular time $t_0$, we have $$2 x(t_0) x'(t_0) + 2 y(t_0) y'(t_0) = 0.$$ Now, our given data is $x(t_0), y(t_0), y'(t_0)$ (actually, we know that $y'$ is constant, but this isn't relevant), and substituting and rearranging gives $$x'(t_0) = \frac{4}{3} \textrm{ ft/min} .$$ In particular, this tells us the speed $x'$ only at the particular time $t_0$; indeed, $x'$ varies with time.

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Your finding that $\frac{dx}{dt}=\frac{4}{3}$ relies on the fact that $y=4$ and $x=3$. This is true at $t=0$, but not true for any later time. As $x$ and $y$ change, also $\frac{dx}{dt}$ changes.

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Just to illustrate the other answers, we can calculate explicitly what $\frac{dx}{dt}$ is after a minute has passed (i.e. at $t=1$).

You've calculated that

$$ 2y\frac{dy}{dt} + 2x \frac{dx}{dt} = 0 $$

so that (using the fact that $dy / dt = -1$)

$$ \frac{dx}{dt} = \frac{y}{x}. $$

At $t=1$, $y=3$ so

$$ x^2 + 3^2 = 5^2 \implies x=4. $$

Thus, at $t = 1$,

$$ \frac{dx}{dt} = \frac{3}{4}. $$

In fact, we can calculate $\frac{dx}{dt}$ for any $t\in[0,4]$. Notice that the information you're given essentially means that $y = 4-t$, so that

$$ x = \sqrt{25 - (4-t)^2} $$

which gives us that

$$\frac{dx}{dt} = \frac{4-t}{\sqrt{25 - (4-t)^2}}. $$

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