Question

6 a 13 foot ladder is leaning against a wall. if the top slips down the wall at a rate of 1.25 ft/s, how fast will the bottom of the ladder be moving away from the wall when the top is 12 feet above the ground?

Explanation:

Step1: Establish the relationship

Let $y$ be the height of the top of the ladder on the wall and $x$ be the distance of the bottom of the ladder from the wall. By the Pythagorean theorem, $x^{2}+y^{2}=13^{2}=169$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of the equation $x^{2}+y^{2}=169$ with respect to $t$ gives $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. Then simplify to $x\frac{dx}{dt}+y\frac{dy}{dt}=0$.

Step3: Find the value of $x$

When $y = 12$, we can find $x$ using the Pythagorean - theorem: $x=\sqrt{169 - y^{2}}=\sqrt{169 - 144}=\sqrt{25}=5$.

Step4: Substitute the known values

We know that $\frac{dy}{dt}=- 1.25$ (negative because $y$ is decreasing), $x = 5$, and $y = 12$. Substitute these values into the equation $x\frac{dx}{dt}+y\frac{dy}{dt}=0$. We get $5\frac{dx}{dt}+12\times(-1.25)=0$.

Step5: Solve for $\frac{dx}{dt}$

First, simplify the equation: $5\frac{dx}{dt}-15 = 0$. Then add 15 to both sides: $5\frac{dx}{dt}=15$. Divide both sides by 5: $\frac{dx}{dt}=3$.

Answer:

$3$ ft/s