(a) One car leaves a given point and travels north at 30 mph . Another car leaves the same point

step 1 Okay, so we have a problem here. One car leaves a given point traveling north at 30 miles per ho

(a) One car leaves a given point and travels north at 30 mph...

(a) One car leaves a given point and travels north at 30 $\mathrm{mph}$ . Another car leaves the same point at the same time and travels west at 40 $\mathrm{mph}$ . At what rate is the distance between the two cars changing at the instant when the cars have traveled 2 hours?(b) Suppose that, in part (a), the second car left 1 hour later than the first car. At what rate is the distance between the two cars changing at the instant when the second car has traveled 1 hour?

Key Concepts

Related Rates

This concept involves finding the rate at which one quantity changes by relating it to other quantities whose rates of change are known. In problems like these, the quantities (distances, in this case) are connected through an equation, and by differentiating that equation with respect to time, one can determine the unknown rate of change.

Pythagorean Theorem in Motion

The Pythagorean theorem plays a central role in relating distances in problems involving perpendicular motion. When two objects move along perpendicular paths, the direct distance between them can be expressed as the hypotenuse of a right triangle formed by their individual displacements. This geometric relation forms the basis for applying calculus techniques to find the rate of change of the overall distance.

Implicit Differentiation and Chain Rule

Implicit differentiation is used when the equation relating the quantities is not explicitly solved for one variable. The chain rule facilitates the differentiation of composite functions with respect to time, accounting for how each variable changes in time. Together, they allow for the differentiation of the distance equation derived from the Pythagorean theorem, thereby linking rates of changes of different dimensions.

Time Offsets in Motion

When different objects begin their motion at different times, it is important to account for these time shifts in formulation. Adjusting the time variable ensures that the object which started later is properly represented in the equations, allowing for accurate calculation of distances and their rates of change at specific instants.

Transcript

00:01 Okay, so we have a problem here.

00:03 One car leaves a given point traveling north at 30 miles per hour.

00:09 So let's draw a picture.

00:10 So here's our point.

00:12 One car leaves going 30 miles per hour.

00:19 And then another car leaves the same point and travels west to 40 miles per hour.

00:32 And then we want to know at what rate is the distance between the two cars changing at the instant when the cars have traveled two hours.

00:42 So the second car is out here and the first car is out here.

00:49 Here's the distance between them.

00:53 Okay.

00:55 So we'll call that s and this is y and we'll say this is x.

01:03 So y is the distance that the first cars traveled traveling north.

01:08 X is the distance that the second cars travel traveling west.

01:11 S is the distance between them.

01:15 And so the relationship between x, y, and s is, of course, west and north are perpendicular.

01:23 So we have the pythagorean theorem, x squared plus y squared is s squared.

01:31 And of course, 30 and 40 are the rates of change of x and y.

01:38 So dx, dt is 40, d, d, y, d, t is 30.

01:47 And so if we differentiate this with respect to t, we get 2x, dx, dt, plus 2y, dy, d y, dt, equals 2s, d, d, s, d, d, d, t.

02:06 Okay, and so what we need is, okay, so we have dx, dx, dx, d, d, d, you're looking for d s, d, d, d, d, d, d, d, d, d, d, d, d, d, d, d, d, d, d, d, we need x and y and s.

02:18 Well, it says after two hours.

02:21 Okay, so after two hours, so after two hours, well, x, so the second car has been traveling 40 miles per hour, so it's gone 80 miles.

02:42 Y has been traveling 30 miles for two hours, so that's 60 miles.

02:48 And then if we come up here and just do 80, square, plus 60 squared is s squared.

02:59 Well, you do 80 squared plus 60 squared.

03:01 You're going to get, let's see, 10 ,000.

03:07 And so then square of 10 ,000 is 100.

03:09 So s is 100 miles...