(a) What quantities are given in the problem? (b) What is...

(a) What quantities are given in the problem? (b) What is the unknown? (c) Draw a picture of the situation for any time $t$ (d) Write an equation that relates the quantities. (e) Finish solving the problem. At noon, ship $\mathrm{A}$ is $150 \mathrm{~km}$ west of ship $\mathrm{B}$. Ship $\mathrm{A}$ is sailing east at $35 \mathrm{~km} / \mathrm{h}$ and ship $\mathrm{B}$ is sailing north at $25 \mathrm{~km} / \mathrm{h}$. How fast is the distance between the ships changing at 4: 00 PM?

(a) What quantities are given in the problem?
(b) What is the unknown?
(c) Draw a picture of the situation for any time $t$
(d) Write an equation that relates the quantities.
(e) Finish solving the problem.
At noon, ship $\mathrm{A}$ is $150 \mathrm{~km}$ west of ship $\mathrm{B}$. Ship $\mathrm{A}$ is sailing east at $35 \mathrm{~km} / \mathrm{h}$ and ship $\mathrm{B}$ is sailing north at $25 \mathrm{~km} / \mathrm{h}$. How fast is the distance between the ships changing at 4: 00 PM?

Key Concepts

Identification of Given and Unknown Quantities

This step involves extracting and categorizing the information provided in the problem, such as fixed quantities and known rates, versus the quantity that needs to be determined. Recognizing the knowns and unknowns is essential for setting up the correct equations and ultimately solving the problem.

Related Rates

This concept involves finding the rate at which one quantity changes with respect to time by relating it to the rate of change of another quantity. In problems like these, quantities such as distance, speed, or position often change over time, and related rates techniques help in establishing and differentiating the relationships between them.

Pythagorean Theorem

This theorem is used to relate the sides of a right triangle, especially when two quantities are perpendicular to each other. It forms the foundational equation that connects these quantities, allowing the formulation of an expression whose derivative with respect to time can be taken to determine rates of change.

Implicit Differentiation

This technique is used to differentiate equations in which the variables are not isolated. In the context of related rates, it allows one to differentiate both sides of an equation with respect to time, thereby linking the rates of change of different quantities together.

Diagrammatic Representation

Creating a visual diagram of the situation is crucial as it helps in understanding the spatial relationships and directions of the quantities involved. Diagrams simplify the process of identifying the geometric relationships and serve as a guide for setting up the necessary equations.

Transcript

00:01 So at noon, ship a is 150 kilometers west of ship b.

00:07 Ship a is sailing east at 35 kilometers, and ship b is sailing north.

00:13 So how fast is the distance between the ships changing at 4 o 'clock? so the phrase changing or rate is what tells us that this is going to involve derivatives.

00:27 It might be obvious now, but in the future, like on a find, when you have a bunch of different problem types mixed in.

00:34 This changing for the word rate is what we're looking for.

00:39 So first, what is given? so we're told that ship a is sailing east at 35 kilometers per hour.

00:50 So that means the derivative of a at dt swings with respect to time is 35.

00:58 And d .b over dt is.

01:02 25 a ship be is sailing north now it's going to be positive because east and north are positive since it's going up into the right while west and south are negative so what is the unknown the distance between the ships at four o 'clock is what we're trying to solve for so i'm going to call the distance between the ship c so dc d t at four o 'clock now this problem will make a little bit more sense once we draw the picture.

01:43 So ship a is west of ship b.

01:47 So here is a and here is b and they are 150 kilometers apart.

01:57 So ship a is going east, so it is going this way, and it is 35 km per h, kilometers per hour, and b is going north at 25 kilometers per hour.

02:15 Now, this line, the distance is 150.

02:19 What about b? we're going to say b is at 0 .0.

02:25 If we were on a coordinate plane, this would be the point 0.

02:29 And then over here for a, we have negative 150, comma, zero.

02:38 And now you can see the triangle that's being formed.

02:42 So we make the line here.

02:44 We now have the triangle and we recall this c.

02:51 Also, there's one more thing to take into account.

02:55 The at noon versus 4 o 'clock.