At 1.00 p.m. ship A is 25 km duc north of ship B. If ship A is sailing west at a rate of 16 km

step 1 For this problem, we are told that at 1 p .m. ship A is 25 kilometers due north of ship B. We th

At 1.00 p.m. ship A is 25 km duc north of ship B. If ship A...

At $1.00$ p.m. ship $A$ is $25 \mathrm{~km}$ duc north of ship $B$. If ship $A$ is sailing west at a rate of $16 \mathrm{~km} / \mathrm{h}$ and ship $B$ is sailing south at $20 \mathrm{~km} / \mathrm{h}$, at what rate is the distance between the two ships changing at $1: 30$ p.m.

Key Concepts

Related Rates

Related rates problems involve finding the rate at which one quantity changes with respect to time by relating it to another quantity whose rate of change is known. This method requires differentiating an equation that links the variables with respect to time.

Coordinate System Setup

In problems involving movement in a plane, setting up a coordinate system allows one to assign and track the positions of objects using variables for each coordinate. This is essential for determining distances and directions between moving objects.

Pythagorean Theorem

The Pythagorean theorem is used to relate the distances between points in a coordinate plane. By applying it, one can express the distance between two moving objects in terms of their horizontal and vertical separations.

Chain Rule Differentiation

The chain rule in calculus is used to differentiate composite functions with respect to time. In related rates problems, it is applied to differentiate equations that involve squared terms, ultimately linking the rates of change of these variables.

Transcript

00:01 For this problem, we are told that at 1 p .m.

00:03 Ship a is 25 kilometers due north of ship b.

00:06 We then asked that if ship a is sailing west at a rate of 16 kilometers per hour, and ship b is sailing south at a rate of 20 kilometers per hour, at what rate is the distance between the two ships changing at 1 .30.

00:20 So what we need to do first is figure out what the sort of new positions of our ships will be.

00:28 We know that essentially a -x is going to equal, let's say, just zero, and b -x will equal zero as well.

00:39 We know that a -x -prime would equal 16 kilometers an hour times half an hour, so that's going to be just eight, or specifically we could call that negative eight kilometers.

00:52 We moved eight kilometers to the left, and b -x -prime is still going to equal zero.

00:59 It's not moving east or west.

01:02 Then we'd have that a .y is going to equal, let's say, well, we started off 25 kilometers due north, so let's say we start off at 25 kilometers.

01:12 And, well, our ship a isn't moving north or south, so ay prime is going to stay 25.

01:20 And then we'd have that by is going to equal, let's say that's zero, which then means that we'd have by prime is going to equal 20 times half an hour, so 10, or negative 10 kilometers.

01:35 We went 10 kilometers south.

01:38 So now that we have that, we can figure out the distance between ship a and ship b at our time 130 by taking the square root of, we would have ax minus bx squared, so that would be just negative 8 squared, plus ay minus by squared, so that it would be 25 minus negative 10, so that would be 25 minus negative that's going to be 35 squared.

02:08 So, before i calculate that out, i'll note that more generally, we can write this as dab equals the square root of, as i was sort of saying out loud, ax minus bx, oops, ax minus bx squared, plus ay minus by squared.

02:31 We are asked, how fast is the distance changing? so we want d prime, or the derivative of that, specifically i should say with respect to t.

02:42 So d by dt of d, that is going to be 1 over 2 times the square root of what i just wrote there...

Please give Ace some feedback