Question

A 13-foot ladder is leaning against a wall. If the top slips down the wall at a rate of 2 ft/s, how fast will the foot be moving away from the wall when the top is 10 feet above the ground?
The foot will be moving at ___ ft/sec

Answer 1

Answer:

$\frac{db}{dt}=\frac{20\sqrt{69}}{69}\approx2.4077\text{ ft/s}$

Step-by-step explanation:

We know that the ladder is 13 feet long.

We also know that the top of the ladder is slipping down at a rate of 2ft/s.

And we want to find the rate at which the foot of the ladder is moving away when the top is 10 feet above the ground.

We can use the Pythagorean Theorem for this problem:

$a^2+b^2=c^2$

Let's let a be the leg by the wall, and b be the leg on the ground.

Our hypotenuse is 13 and it stays 13. So, let's substitute 13 for c:

$a^2+b^2=(13)^2=169$

Now, let's take the derivative of both sides with respect to t:

$\frac{d}{dt}[a^2+b^2]=\frac{d}{dt}[169]$

Evaluate. Use implicit differentiation:

$2a\frac{da}{dt}+2b\frac{db}{dt}=0$

We know that the top of the ladder is slipping down at a rate of 2ft/s. So, da/dt is -2 (it's negative because it's slipping downwards).

We want to find db/dt when the top is 10 feet above the ground. So, substitute 10 for a.

Since a is 10 and we know that c stays constant at 13, we can figure out b using the Pythagorean Theorem:

$10^2+b^2=13^2$

Evaluate:

$100+b^2=169$

Solve for b:

$b^2=69\\b=\sqrt{69}$

So, substitute -2 for da/dt, 10 for a, and √69 for db/dt. This yields:

$2(10)(-2)+2(\sqrt{69})(\frac{db}{dt})=0$

Solve for db/dt. Multiply:

$-40+2(\sqrt{69})(\frac{db}{dt})=0$

Add 40 to both sides:

$2\sqrt{69}(\frac{db}{dt})=40$

Divide both sides by 2√69:

$\frac{db}{dt}=\frac{20}{\sqrt{69}}$

Rationalize:

$\frac{db}{dt}=\frac{20\sqrt{69}}{69}\approx2.4077\text{ ft/s}$

So, the foot will be moving at approximately 2.4077 feet per second.