A road perpendicular to a highway leads to a farmhouse located 6 mile away. (See Figure 9, page 204 of the text.) An automobile traveling on the highway passes through this intersection at a speed of 75 mph . How fast is the distance between the automobile and the farmhouse increasing when the automobile is 5 miles past the intersection of the highway and the road? The distance between the automobile and the farmhouse is increasing at a rate of $$ \square $$ miles per hour.

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To solve this problem, we can use the Pythagorean theorem and related rates. Let's define the variables involved:

- Let $$ x $$ be the distance of the automobile from the intersection along the highway.
- Let $$ y $$ be the distance from the intersection to the farmhouse, which is constant at 6 miles.
- Let $$ z $$ be the distance from the automobile to the farmhouse.

According to the Pythagorean theorem, we have:

$$
z^2 = x^2 + y^2
$$

Given that $$ y = 6 $$ miles, we can rewrite the equation as:

$$
z^2 = x^2 + 6^2
$$

Now, we need to differentiate both sides with respect to time $$ t $$:

$$
\frac{d}{dt}(z^2) = \frac{d}{dt}(x^2 + 36)
$$

Using the chain rule, we get:

$$
2z \frac{dz}{dt} = 2x \frac{dx}{dt}
$$

Now, we can simplify this to:

$$
z \frac{dz}{dt} = x \frac{dx}{dt}
$$

Next, we need to find $$ z $$ when the automobile is 5 miles past the intersection. At that moment, $$ x = 5 $$ miles. We can find $$ z $$ using the Pythagorean theorem:

$$
z^2 = 5^2 + 6^2 = 25 + 36 = 61
$$

Thus,

$$
z = \sqrt{61}
$$

The speed of the automobile is given as $$ \frac{dx}{dt} = 75 $$ mph. Now we can substitute $$ z $$, $$ x $$, and $$ \frac{dx}{dt} $$ into the differentiated equation:

$$
\sqrt{61} \frac{dz}{dt} = 5 \cdot 75
$$

Now, we can solve for $$ \frac{dz}{dt} $$:

$$
\frac{dz}{dt} = \frac{5 \cdot 75}{\sqrt{61}}
$$

Calculating this gives:

$$
\frac{dz}{dt} = \frac{375}{\sqrt{61}}
$$

Now, let's calculate the numerical value of $$ \frac{dz}{dt} $$.

First, we need to find $$ \sqrt{61} $$:

$$
\sqrt{61} \approx 7.81
$$

Now, we can calculate:

$$
\frac{dz}{dt} \approx \frac{375}{7.81} \approx 48.0
$$

Thus, the distance between the automobile and the farmhouse is increasing at a rate of approximately $$ 48.0 $$ miles per hour.

So, the final answer is:

The distance between the automobile and the farmhouse is increasing at a rate of $$ 48.0 $$ miles per hour.
The distance between the automobile and the farmhouse is increasing at a rate of approximately 48.0 miles per hour.

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