Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 3: Problem 14
Two ships sail from the same island port, one going north at 24 knots ( 24 nautical miles per hour) and the other east at 30 knots. The northbound ship departed at \(9: 00 \mathrm{~A} . \mathrm{M}\). and the eastbound ship left at 11:00 A.M. How fast is the distance between them increasing at 2:00 P.M.? Hint: Let \(t=0\) at 11:00 A.M.
Short Answer
Expert verified
The distance between the ships is increasing at 37.2 knots at 2:00 PM.
Step by step solution
01
Understanding the Problem
Two ships are moving from the same point at different times. The northbound ship starts at 9:00 AM and moves at 24 knots, while the eastbound ship starts at 11:00 AM at 30 knots. At 2:00 PM, we need to calculate how fast the distance between them is increasing, considering 11:00 AM as time zero (\(t=0\)).
02
Identify Variables and Functions
Let \(N(t)\) be the distance traveled by the northbound ship after \(t\) hours (starting at 11:00 AM) and \(E(t)\) be the distance traveled by the eastbound ship. Since the northbound ship had a 2-hour head start, let its position at time \(t\) be \(N(t) = 24(t + 2)\). For the eastbound ship, which starts at \(t = 0\), \(E(t) = 30t\).
03
Apply the Pythagorean Theorem
The distance, \(D(t)\), between the two ships can be modeled using the Pythagorean theorem: \[D(t) = \sqrt{N(t)^2 + E(t)^2}\]. Substituting the known functions, we get: \[D(t) = \sqrt{(24(t+2))^2 + (30t)^2}\].
04
Differentiate with Respect to Time
We need \(\frac{dD}{dt}\) to find the rate at which the distance between the ships is increasing. Differentiate \(D(t)\) implicitly: \[ \frac{dD}{dt} = \frac{d}{dt}\sqrt{(24(t+2))^2 + (30t)^2} = \frac{1}{2\sqrt{(24(t+2))^2 + (30t)^2}} \cdot (2 \cdot 24(t+2) \cdot 24 + 2 \cdot 30t \cdot 30) \].
05
Evaluate the Derivative at t = 3
At 2:00 PM, \(t = 3\). Substitute \(t = 3\) into the differentiated formula: \[ \frac{dD}{dt} = \frac{1}{2\sqrt{(24 \cdot 5)^2 + (30 \cdot 3)^2}} \cdot (2 \cdot 24 \cdot 5 \cdot 24 + 2 \cdot 30 \cdot 3 \cdot 30) \] which simplifies to \( \frac{dD}{dt} = \frac{1}{2 \cdot \sqrt{14400 + 8100}} \cdot (5760 + 5400) \).
06
Calculate the Numerical Value
Now calculate the numbers: the denominator \(2 \cdot \sqrt{22500} = 300\) and the numerator is \(11160\). Therefore, \( \frac{dD}{dt} = \frac{11160}{300} = 37.2\).
07
Conclusion
Therefore, at 2:00 PM, the distance between the two ships is increasing at a rate of 37.2 knots.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Implicit Differentiation
All calculus students eventually come across implicit differentiation. It's a technique to find the derivative of a function when it's not easily solved for one variable. In many situations, equations involve multiple variables that are intertwined with each other, making it hard to express one in terms of the other. That's where implicit differentiation comes to the rescue.
- Imagine you have a relationship like a circle's equation: \(x^2 + y^2 = r^2\). Solving for \(y\) directly and differentiating could be tricky.
- Instead, differentiate both sides with respect to, say, \(x\). You'll apply the chain rule on the \(y^2\) term, which gives you an extra factor \(2y \frac{dy}{dx}\).
- This method allows you to solve for \(\frac{dy}{dx}\) even when \(y\) is not isolated.
Pythagorean Theorem
The Pythagorean Theorem is a cornerstone of geometry, relating the sides of a right-angled triangle. It's usually phrased as \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse.
- This theorem finds use beyond simple geometry, especially when calculating distances and speeds.
- For instance, if two ships are moving at right angles to each other, their paths form a right triangle.
- The Pythagorean Theorem helps calculate their direct distance apart, making it particularly handy in related rates problems.
Calculus Problem Solving
Tackling calculus problems often involves a series of logical steps. This process can transform even the most daunting problem into manageable parts.
- First, understand what you're asked; here, it's the rate of increasing distance between two ships at a specific time.
- Next, define the variables and functions to represent distances and times. This turns abstract concepts into concrete equations.
- Apply mathematical theorems and rules like the Pythagorean Theorem to express relationships between variables.
- Finally, differentiate to find rates of change, applying techniques like implicit differentiation where direct methods fall short.
Derivatives
Derivatives are a key tool in calculus, allowing us to determine how one quantity changes with respect to another. They're used everywhere from physics to economics as they reveal the rate of change of functions.
- In problems involving movement, derivatives tell us how fast distances change between moving objects, like with our ships here.
- To calculate these rates, we differentiate the equation relating their paths. The result, \(\frac{dD}{dt}\), shows the rate at which their separation increases.
- Knowing how to find derivatives鈥攂y both simple and implicit differentiation鈥攊s crucial to tackling these kinds of questions.
