A streetlight is at the top of an $18 \mathrm{ft}$ tall pole. A woman $6 \mathrm{ft}$ tall walks away from the pole at a speed of $4 \mathrm{ft} / \mathrm{sec}$ along a straight path. How fast (in $\mathrm{ft} / \mathrm{sec}$ ) is the tip of her shadow moving away from the pole when she is $45 \mathrm{ft}$ from the base of the pole?

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Step 1 ：A streetlight is at the top of an $18 \mathrm{ft}$ tall pole. A woman $6 \mathrm{ft}$ tall walks away from the pole at a speed of $4 \mathrm{ft} / \mathrm{sec}$ along a straight path. We want to find how fast (in $\mathrm{ft} / \mathrm{sec}$ ) is the tip of her shadow moving away from the pole when she is $45 \mathrm{ft}$ from the base of the pole.

Step 2 ：We know that $\frac{dx}{dt} = 4 \mathrm{ft} / \mathrm{sec}$, which is the rate at which the woman is moving away from the pole.

Step 3 ：We can substitute this into our equation $\frac{dy}{dt} = 3 \frac{dx}{dt}$ to find $\frac{dy}{dt}$, which is the rate at which the tip of her shadow is moving away from the pole.

Step 4 ：Substituting $\frac{dx}{dt} = 4 \mathrm{ft} / \mathrm{sec}$ into the equation, we get $\frac{dy}{dt} = 3 * 4 \mathrm{ft} / \mathrm{sec}$.

Step 5 ：Solving this, we find that $\frac{dy}{dt} = 12 \mathrm{ft} / \mathrm{sec}$.

Step 6 ：Final Answer: The tip of her shadow is moving away from the pole at a rate of $\boxed{12}$ feet per second.