Question

Two boats leave a port at the same time, one traveling west at \(20 \mathrm{mi} /\) hr and the other traveling southwest \(\left(45^{\circ}\right.\) south of west) at \(15 \mathrm{mi} / \mathrm{hr}\). After 30 minutes, how far apart are the boats and at what rate is the distance between them changing? (Hint: Use the Law of Cosines.)

Short answer

Answer: After 30 minutes, the boats are approximately 8.435 miles apart. The rate at which the distance between them is changing at this time is approximately 12.07 mi/hr.

Step-by-step solution

Calculate the distances traveled by each boat

The boats have been traveling for \(30\) minutes which is \(0.5\,\text{hours}\). Boat A traveling west at \(20\,\text{mi/hr}\) covers a distance of: \(a = 20 \cdot 0.5 = 10\,\text{miles}\) Boat B traveling southwest at \(15\,\text{mi/hr}\) covers a distance of: \(b = 15 \cdot 0.5 = 7.5\,\text{miles}\)

Use Law of Cosines to find distance between the boats

Let \(c\) be the distance between the boats after \(30\) minutes. Using the Law of Cosines with \(\theta = 45^\circ\): \(c^2 = a^2 + b^2 - 2ab \cos\theta\) \(c^2 = 10^2 + 7.5^2 - 2 \cdot 10 \cdot 7.5 \cos(45^\circ)\) Now, we can find the value of \(c\): \(c = \sqrt{10^2 + 7.5^2 - 2 \cdot 10 \cdot 7.5 \cos(45^\circ)} \approx 8.435\,\text{miles}\) So, after \(30\) minutes, the boats are approximately \(8.435\,\text{miles}\) apart.

Calculate rates of change

Let \(d(t)\) be the distance between the boats after \(t\) hours, where \(a(t) = 20t\) and \(b(t) = 15t\). Using the Law of Cosines again: \(d(t)^2 = a(t)^2 + b(t)^2 - 2a(t)b(t) \cos(45^\circ)\) We have to find the rate of change of the distance between the boats; i.e., we need to find \(\frac{d}{dt}\big(d(t)\big)\). First, we'll differentiate both sides of the equation with respect to \(t\): \(2d(t)\cdot\frac{dd(t)}{dt} = 2a(t)\cdot\frac{da(t)}{dt} + 2b(t)\cdot\frac{db(t)}{dt} -2\,\Big(\frac{da(t)}{dt}b(t) + a(t)\frac{db(t)}{dt}\Big)\cos(45^\circ)\) Now, we'll find the derivatives: \(\frac{da(t)}{dt} = \frac{d(20t)}{dt} = 20\,\text{mi/hr}\) \(\frac{db(t)}{dt} = \frac{d(15t)}{dt} = 15\,\text{mi/hr}\) Next, plug in the values of \(a(t)\), \(b(t)\), \(d(t)\), \(\frac{da(t)}{dt}\), and \(\frac{db(t)}{dt}\) for \(t = 0.5\,\text{hours}\): \(2\cdot8.435\cdot\frac{dd(0.5)}{dt} = 2\cdot10\cdot20 + 2\cdot7.5\cdot15 -2\,\Big(20\cdot7.5 + 10\cdot15\Big)\cos(45^\circ)\) Now, solve for \(\frac{dd(0.5)}{dt}\): \(\frac{dd(0.5)}{dt} \approx 12.07\,\text{mi/hr}\)

Summary of the results

After \(30\) minutes (i.e., \(0.5\,\text{hours}\)), the boats are approximately \(8.435\,\text{miles}\) apart. The rate at which the distance between them is changing at this time is approximately \(12.07\,\text{mi/hr}\).

Key concepts

Relative Velocity

Relative velocity is the velocity of an object as observed from another moving object. This concept is crucial when analyzing problems involving multiple moving bodies, like boats in this scenario. Here, each boat's velocity is compared with respect to the other.

For example, when Boat A moves west at a speed of 20 mi/hr and Boat B moves southwest at 15 mi/hr, their relative movement affects how we compute the changing distance between them.

• The motion is not in a straight line relative to each other, making their velocities relative.
• We consider the angle between their directions (45°) in calculations.

Understanding relative velocity is crucial for determining how quickly the separation between two objects changes, especially in dynamics involving directionality.

Trigonometry

Trigonometry involves relationships between angles and lengths in triangles, and it is essential here in our example with the boats. Specifically, we use a trigonometric rule called the Law of Cosines. This law helps us determine the distance between two points given specific direction and speed.

The Law of Cosines is particularly useful for problems where angles matter, like when calculating distances at an angle. The law states:

• In a triangle with sides a, b, and c, opposite angle θ:
  \[c^2 = a^2 + b^2 - 2ab \cos(θ)\]
• Here, θ represents the 45° angle between the boats' paths.

Once you have the sides represented by the travel distances of each boat over 30 minutes (10 miles for Boat A and 7.5 miles for Boat B), this law allows us to calculate the exact distance between them by solving the equation for "c." This application highlights the importance of understanding angles in navigational problems.

Related Rates

Related rates are a method in calculus used to find the rate at which one quantity changes with respect to another. It is particularly relevant when two or more variables change over time, and you're interested in how quickly one affects another.

In our boat scenario:

• The distance between the boats changes over time.
• We are interested in determining the rate at which this distance increases, 12.07 mi/hr, at any given moment.

Using derivatives, we can relate the rates of change of functions with respect to time. With the function for distance derived via the Law of Cosines, taking its derivative gives the rate of change of distance:
\[2d(t)\cdot\frac{dd(t)}{dt} = 2a(t)\cdot\frac{da(t)}{dt} + 2b(t)\cdot\frac{db(t)}{dt} -2\,(\frac{da(t)}{dt}b(t) + a(t)\frac{db(t)}{dt})\cos(45°)\]By substituting known values like speeds and the angle, the change in distance over time is calculated. This method shows how calculus can solve complex, real-world problems by connecting different rates of change.