The base of a 5-ft ladder leaning against a wall begins to slide away from the wall. At the instant of time when the base is 4 ft from the wall, the base is moving at the rate of 6 ft/sec. How fast is the top of the ladder sliding down the wall at that instant of time?
A) 5.6 ft/sec
B) 6.5 ft/sec
C) 8 ft/sec
D) 9.3 ft/sec

Question

Quizplus · Accepted Answer

We can use the Pythagorean theorem to relate the top and bottom of the ladder: $x^2 + y^2 = 25$, where $x$ is the distance from the wall to the ladder's top and $y$ is the distance from the base to the wall. 
Differentiating implicitly with respect to time, we have $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$, where $\frac{dx}{dt}$ and $\frac{dy}{dt}$ represent the rates of change of $x$ and $y$ respectively. 
At the instant when the base is 4 ft from the wall, we have $y=4$ and $\frac{dy}{dt} = 6$. We can solve for $\frac{dx}{dt}$ using the Pythagorean relation: $x = \sqrt{25-y^2} = \sqrt{25-16} = 3$. 
Plugging in the values, we get $2(3) \frac{dx}{dt} + 2(4)(6) = 0$, which simplifies to $\frac{dx}{dt} = -\frac{16}{3}$. 
Finally, we plug in $\frac{dx}{dt}$ and $y$ into the Pythagorean relation and differentiate implicitly to find $\frac{dy}{dt}$: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$, which gives $\frac{dy}{dt} = -\frac{3}{4} \cdot \frac{dx}{dt} = -\frac{3}{4} \cdot (-\frac{16}{3}) = 4.5$. Therefore, the top of the ladder is sliding down the wall at a rate of 4.5 ft/sec.

The Base of a 5-Ft Ladder Leaning Against a Wall