The top of a ladder slides down a vertical wall at a rate of $0.15 \mathrm {~m} / \mathrm { s }$. At the moment when the bottom of the ladder is $1.5 \mathrm {~m}$ from the wall, it slides away from the wall at a rate of $0.3 \mathrm {~m} / \mathrm { s }$. How long is the ladder?
A) $3.9 \mathrm {~m}$
B) $2.9 \mathrm {~m}$
C) $4.4 \mathrm {~m}$
D) $3.4 \mathrm {~m}$
E) $2.4 \mathrm {~m}$

Question

Quizplus · Accepted Answer

Let the ladder have length $x$ and let the bottom of the ladder travel distance $d$. Then $\frac{d}{dt}x^2+\frac{d}{dt}d^2=0$ by the Pythagorean Theorem.

At the moment when the bottom of the ladder is $1.5\mathrm{m}$ from the wall, we have $x=\sqrt{d^2+1.5^2}$. Differentiating with respect to time, we get $$\frac{d}{dt}x=\frac{d}{dt}\sqrt{d^2+1.5^2}=\frac{d}{dt}\left((d^2+1.5^2)^{\frac{1}{2}}\right)=\frac{1}{2}(d^2+1.5^2)^{-\frac{1}{2}}\cdot 2d\left(\frac{d}{dt}\right)+0.15\mathrm{m}/\mathrm{s}$$ at that moment, using the chain rule of differentiation. We are given that $\frac{d}{dt}x=0.3\mathrm{m}/\mathrm{s}$ at this moment. We can substitute the expression for $x$ and the expression for $\frac{d}{dt}x$ into our equation to find $d$.

We get $$\frac{d}{dt}(d^2+1.5^2)+0.15^2\mathrm{m}^2/\mathrm{s}^2=0$$ at that moment, by differentiating $x^2=d^2+1.5^2$ with respect to time. Substituting $\frac{d}{dt}x=0.3\mathrm{m}/\mathrm{s}$, we get $$\frac{d}{dt}(d^2+1.5^2)=-0.09\mathrm{m}^2/\mathrm{s}^2.$$ Solving for $d$ gives $d=-1.5\mathrm{m}$ or $d=3.5\mathrm{m}$. Since the ladder is clearly getting farther away from the wall, we choose $d=3.5\mathrm{m}$. Then from $x=\sqrt{d^2+1.5^2}$, we get $x=3.4\mathrm{m},$ so the answer is $\boxed{\textbf{(D)}}$.

The Top of a Ladder Slides Down a Vertical Wall 0.15 m/s0.15 \mathrm {~m} / \mathrm { s }0.15 m/s

The Top of a Ladder Slides Down a Vertical Wall $0.15 \mathrm {~m} / \mathrm { s }$