Ch. 4 - Applications of Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 4.5.53a

Chapter 4, Problem 4.5.53a

53. Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship B was sailing east at 8 knots and continued to do so all day.
a. Start counting time with t=0 at noon and express the distance s between the ships as a function of t.

Verified step by step guidance

Start by identifying the positions of the ships at noon. Ship A is 12 nautical miles north of Ship B, so the initial position of Ship A relative to Ship B is (0, 12) in a coordinate system where north is positive y and east is positive x.

Determine the position of Ship A as a function of time t. Since Ship A is moving south at 12 knots, its position at time t is given by (0, 12 - 12t).

Determine the position of Ship B as a function of time t. Since Ship B is moving east at 8 knots, its position at time t is given by (8t, 0).

Calculate the distance s between the two ships using the distance formula. The distance s is given by the formula:

\sqrt{{(8 t)}^{2} + {(12 - 12 t)}^{2}}

Simplify the expression for s to express the distance between the ships as a function of time t. This involves expanding the squares and combining like terms.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Relative Motion

Relative motion involves understanding how the position of one object changes with respect to another. In this problem, ship A moves south while ship B moves east, creating a right triangle where the legs are the distances each ship travels over time. Calculating the distance between the ships requires understanding how these movements relate to each other over time.

Pythagorean Theorem

The Pythagorean Theorem is essential for finding the distance between two points in a plane, especially when they form a right triangle. Here, the theorem helps calculate the hypotenuse, which represents the distance between the ships, using the formula s = √(x² + y²), where x and y are the distances traveled by ships B and A, respectively.

Parametric Equations

Parametric equations express the coordinates of points as functions of a parameter, often time. In this scenario, the positions of ships A and B are functions of time t, with ship A's position given by (0, 12 - 12t) and ship B's by (8t, 0). These equations allow us to express the distance between the ships as a function of time, crucial for solving the problem.

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Related Practice

Textbook Question

Finding Antiderivatives

In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.

−3x⁻⁴

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Textbook Question

Identifying Extrema

In Exercises 53–60:

a. Find the local extrema of each function on the given interval, and say where they occur.

f(x) = sin 2x, 0 ≤ x ≤ π

Textbook Question

Identifying Extrema

In Exercises 41–52:

a. Identify the function’s local extreme values in the given domain, and say where they occur.

k(x) = x³ + 3x² + 3x + 1, −∞ < x ≤ 0

Textbook Question

Finding displacement from an antiderivative of velocity

a. Suppose that the velocity of a body moving along the s-axis is

ds/dt = v = 9.8t − 3.

i. Find the body’s displacement over the time interval from t = 1 to t = 3 given that s = 5 when t = 0.

Textbook Question

Identifying Extrema

In Exercises 41–52:

a. Identify the function’s local extreme values in the given domain, and say where they occur.

f(t) = 12t − t³, −3 ≤ t < ∞

Textbook Question

25. Paper folding A rectangular sheet of 8.5-in.-by-11-in. paper is placed on a flat surface. One of the corners is placed on the opposite longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L. Try it with paper.

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a. Show that L^2=2x^3/(2x-8.5).