Question
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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.

Ask by Peters Hardy. in the United States
Mar 15,2025

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Tutor-Verified Answer

Answer

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

Solution

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.

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The Deep Dive

To determine how fast the distance between the automobile and the farmhouse is increasing, we can employ the Pythagorean theorem. At the moment the automobile is 5 miles past the intersection, it creates a right triangle with the farmhouse. One leg of the triangle is the distance along the highway (5 miles) and the other leg remains constant at 6 miles (the distance from the intersection to the farmhouse). The distance between the automobile and the farmhouse forms the hypotenuse. Using \( d \) to represent the distance from the automobile to the farmhouse, we have: \[ d(t) = \sqrt{(x(t))^2 + (6)^2} \] where \( x(t) \) is the distance traveled along the highway and \( dx/dt = 75 \) mph (the speed of the automobile). By substituting \( x(t) = 5 \) miles just past the intersection: \[ d = \sqrt{(5)^2 + (6)^2} = \sqrt{25 + 36} = \sqrt{61} \] Next, differentiate the equation implicitly concerning time to find \( dd/dt \) (the rate the distance is increasing). \[ \frac{dd}{dt} = \frac{1}{2\sqrt{61}}(2x \frac{dx}{dt}) \] Plugging in \( x = 5 \) and \( \frac{dx}{dt} = 75 \): \[ \frac{dd}{dt} = \frac{5 \cdot 75}{\sqrt{61}} = \frac{375}{\sqrt{61}} \approx 48.1 \text{ mph} \] The distance between the automobile and the farmhouse is increasing at a rate of approximately 48.1 miles per hour when the automobile is 5 miles past the intersection. Now, imagine if this content was transformed into a thrilling chase scene in a movie – the speeding car racing towards the farmhouse, and viewers waiting in anticipation to see how fast the distance between them grows, adding to the tension! This kind of math isn't just numbers; it's like crafting a narrative where speed and distance blur reality! For anyone dealing with similar mathematical problems, it’s essential to visualize your scenario clearly! Sketching a quick diagram can help in identifying the relationships between various elements, and make it much simpler to apply the Pythagorean theorem correctly. In the rush of calculations, it’s easy to skip some crucial parts, so remember to double-check those details for accuracy!

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