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A ladder $20 \mathrm{ft}$ long leans against a vertical wall. The top end slides downwards at the rate of $2 \mathrm{ft}$ per second. The rate at which the lower end moves on a horizontal floor when it is $12 \mathrm{ft}$ from the wall is
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Verified Answer
The correct answer is:
$\frac{-8}{3}$
$$
x^{2}+y^{2}=(20)^{2}=400
$$
We have. $\frac{d y}{d t}=2 \mathrm{ft} / \mathrm{sec}$
When
$$
x=12
$$
then $\left(13^{2}+y^{2}=400\right.$
$\Rightarrow \quad 144+y^{2}=400$
$\Rightarrow \quad y^{2}=400-144=256$
$\Rightarrow \quad y=16$
Now, $2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0$
$\Rightarrow \quad x \frac{d x}{d t}=-y \frac{d y}{d t}$
$\Rightarrow \quad 12\left(\frac{d x}{d t}\right)=-16(2)$
$\Rightarrow \quad \frac{d x}{d t}=\frac{-8}{3}$
x^{2}+y^{2}=(20)^{2}=400
$$
We have. $\frac{d y}{d t}=2 \mathrm{ft} / \mathrm{sec}$
When
$$
x=12
$$
then $\left(13^{2}+y^{2}=400\right.$
$\Rightarrow \quad 144+y^{2}=400$
$\Rightarrow \quad y^{2}=400-144=256$
$\Rightarrow \quad y=16$
Now, $2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0$
$\Rightarrow \quad x \frac{d x}{d t}=-y \frac{d y}{d t}$
$\Rightarrow \quad 12\left(\frac{d x}{d t}\right)=-16(2)$
$\Rightarrow \quad \frac{d x}{d t}=\frac{-8}{3}$
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