6. a 32 foot ladder leans against a wall. if the bottom of the ladder is pulled away from the wall at a rate…

Explanation:

Step1: Establish the relationship

Let $x$ be the distance of the bottom of the ladder from the wall and $y$ be the height of the top of the ladder on the wall. By the Pythagorean theorem, $x^{2}+y^{2}=32^{2}=1024$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of $x^{2}+y^{2}=1024$ with respect to $t$ gives $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. Then we can simplify it to $x\frac{dx}{dt}+y\frac{dy}{dt}=0$.

Step3: Find $x$ when $y = 25$

When $y = 25$, we use $x^{2}+y^{2}=1024$. So $x=\sqrt{1024 - 25^{2}}=\sqrt{1024 - 625}=\sqrt{399}\approx20$.

Step4: Substitute values and solve for $\frac{dy}{dt}$

We know that $\frac{dx}{dt}=0.4$, $x\approx20$, and $y = 25$. Substituting into $x\frac{dx}{dt}+y\frac{dy}{dt}=0$ gives $20\times0.4+25\times\frac{dy}{dt}=0$. Then $8 + 25\times\frac{dy}{dt}=0$, and $\frac{dy}{dt}=-\frac{8}{25}=- 0.32$ ft/sec.

Answer:

$-0.32$ ft/sec