Distance between two ships At noon, ship A was 12 nautical...

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. \begin{equation} \begin{array}{l}{\text { a. Start counting time with } t=0 \text { at noon and express the distance }} \\ \quad{\text { } s \text { between the ships as a function of } t \text { . }} \\ {\text { b. How rapidly was the distance between the ships changing at }} \\ \quad{\text { noon? One hour later? }} \\ {\text { c. The visibility that day was } 5 \text { nautical miles. Did the ships }} \\ \quad{\text { ever sight each other? }} \\ {\text { d. Graph } s \text { and } d s / d t \text { together as functions of } t \text { for }-1 \leq t \leq 3} \\ \quad {\text { using different colors if possible. Compare the graphs and }} \\ \quad {\text { reconcile what you see with your answers in parts (b) and (c). }} \\ {\text { e. The graph of } d s / d t \text { looks as if it might have a horizontal asymptote }} \\ \quad {\text { in the first quadrant. This in turn suggests that } d s / d t} \\ \quad {\text { approaches a limiting value as } t \rightarrow \infty . \text { What is this value? }} \\ \quad {\text { What is its relation to the ships' individual speeds? }} \end{array} \end{equation}

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.

\begin{equation}
\begin{array}{l}{\text { a. Start counting time with } t=0 \text { at noon and express the distance }} \\ \quad{\text { } s \text { between the ships as a function of } t \text { . }} \\ {\text { b. How rapidly was the distance between the ships changing at }} \\ \quad{\text { noon? One hour later? }} \\ {\text { c. The visibility that day was } 5 \text { nautical miles. Did the ships }} \\ \quad{\text { ever sight each other? }} \\ {\text { d. Graph } s \text { and } d s / d t \text { together as functions of } t \text { for }-1 \leq t \leq 3} \\ \quad {\text { using different colors if possible. Compare the graphs and }} \\ \quad {\text { reconcile what you see with your answers in parts (b) and (c). }} \\ {\text { e. The graph of } d s / d t \text { looks as if it might have a horizontal asymptote }} \\ \quad {\text { in the first quadrant. This in turn suggests that } d s / d t} \\ \quad {\text { approaches a limiting value as } t \rightarrow \infty . \text { What is this value? }} \\ \quad {\text { What is its relation to the ships' individual speeds? }} \end{array}
\end{equation}

Key Concepts

Graphical Analysis

Graphical analysis involves plotting functions and their derivatives to visualize their behavior over time. It helps in interpreting how a function such as the distance between moving objects changes, how its rate of change behaves, and in identifying key features like minima, maxima, and asymptotes that provide deeper insights into the problem.

Relative Motion

Relative motion considers the movement of objects with respect to each other rather than to a fixed point. By analyzing the speeds and directions of individual objects, one can determine the effective rate at which the distance between them changes. This concept is key in problems where multiple objects are in motion, often requiring vector addition to combine their velocities.

Asymptotic Behavior

Asymptotic behavior analyzes the trends of a function as the independent variable approaches a particular limit, often infinity. This concept helps in understanding the long-term tendencies of functions, such as the rate of change of distance between moving objects, and can indicate the limiting speed or behavior as time progresses.

Differentiation

Differentiation is a fundamental tool in calculus for finding the rate at which a function changes with respect to a variable. In the context of moving objects, differentiation is used to compute how quickly the distance between objects is changing at any given moment by differentiating the distance function with respect to time.

Pythagorean Theorem

Used to relate the sides of a right triangle, the Pythagorean Theorem is essential in finding the distance between two points in a plane. When objects move along perpendicular paths, their separation can be expressed as the hypotenuse of a right triangle whose legs are the distances traveled, enabling the calculation of the distance between them.

Related Rates

This concept involves determining the rate at which one quantity changes by relating it to other quantities whose rates of change are known. In problems that involve moving objects or changing geometries, related rates allow us to connect the derivatives of various functions and determine how a change in one variable affects another over time.

Transcript

00:01 All right, so we are asked to, this is at noon, a ship a, which is here, we'll say, is 12 nautical miles north of b.

00:15 So here's b, and then it says that it's sailing at 12 knots an hour.

00:24 So a prime, or d -a -by -t, is 12 as well, and it's minus because it's going south.

00:31 And then b, it says, is going eight knots an hour east.

00:37 So b prime is 8 and it's going to be positive because it's just growing all right and then we have this d here okay so what it wants us to do is first of all in part a express the distance between the ships which they've actually called f so we'll change this to s it has a function of time so this side is going to be 12 minus 12 t okay this side is because it's 12 right now, and it's changing at 12 nautical miles per hour.

01:22 And then this side is going to be 8t.

01:26 Okay? so what we get is s is the square root of 12 minus 12t squared plus 8t all squared.

01:37 All right.

01:42 And if we expand that, we get 144 plus 144 times 104 times.

01:51 Two actually minus, not plus, because it's a negative 12t, so minus 288t, and then plus 144t squared, and then plus a further 64 t squared, which if we then put together, gives us this, 144 minus 288t, plus 208t squared.

02:20 That's part a.

02:21 Part b says, how rapidly was the distance between the ships changing at noon? okay, so for part b, at noon, it's changing at 12 knots.

02:40 Excuse me, because b and a are directly on top of each other at that point.

02:49 And then it says one hour later, one hour later, a is actually going to be at point b.

02:54 It's going to be zero in the y.

02:56 So it's going to be eight knots and it's changing at that point.

03:05 Because a becomes zero basically.

03:09 Then part c says the visibility that day was five nought of a mile.

03:15 The ships ever lose sight of each other.

03:21 So if we start out eight and 12, then they can see, oh, did they ever cite each other, not lose sight? i was going to say they're not going to see each other at the beginning...

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