# Calculus application of rate of change problem

I am not exactly sure how to go about solving the following question. I made a drawing to help visualize the problem and thought about assuming just for the sake of this problem that one car is stationary while the other is driving, so I will be able to fit the two cars onto the same drawing, and thus make the problem somewhat more clear.

However, I am really lost, and I don't really know how to approach this question as I find there's too much information that needs to be used, which is quite overwhelming.

Here's the question:

Two cars travel on parallel “north-south” roads that are 2km apart from each other. The first car is traveling north on road #1 at a speed of 20km/h. While the second car is traveling south at a speed of 60km/h. How fast is the distance between them changing, when the distance between them is 8km?

Furthermore, are there any general strategies that can be applied to questions of such nature? What would be an effective way to become comfortable with solving such questions.

Any help would be greatly appreciated!

• Have you written down a formula for the distance between the cars? Commented Dec 1, 2019 at 3:58
• Would this be valid? d = sqrt(20^2 + 60^2)? Commented Dec 1, 2019 at 4:01
• I see. That's why I thought of assuming that the second car is going at a speed of 40km/h and the first one is stationary (0 km/h) just to make it easier to tackle, however, I'm not sure if it's even a valid approach. Commented Dec 1, 2019 at 4:04

Using vectors make it easier to handle.

First car:
$$(x,y) = (0, 20t)$$

Second car:
$$(x,y) = (2, -60t)$$

Distance between them is
$$d(t) = \sqrt{2^2 + (80t)^2}$$

• Thanks a lot! I do really appreciate your answer. However, I was wondering if there's also a way to solve this question using derivatives? Commented Dec 1, 2019 at 4:12
• Hey $d(t)$ is just an expression for the distance. You still have to differentiate it to get the rate of change in distance $d'(t)$ Commented Dec 1, 2019 at 4:14
• I'm sorry, I misunderstood it at first! So, would I need to differentiate d, and set it equal to 8? Commented Dec 1, 2019 at 4:17
• Exactly! There is a better way though. Try differentiating $d^2$ instead Commented Dec 1, 2019 at 4:17
• $d^2 = 2^2 + (80t)^2 \Rightarrow 2dd' = 0 + 80^2\times 2t$ Commented Dec 1, 2019 at 4:18

Let $$y$$ and $$l$$ be the vertical and direct distances between the cars, respectively. Then,

$$y^2+2^2 = l^2$$

Take the time derivatives,

$$\frac {dl}{dt}=\frac yl \frac {dy}{dt} = \frac {\sqrt{l^2-2^2}}l \frac {dy}{dt}$$

Plug in $$l=8km$$ and $$\frac {dy}{dt}= (20+60)=80kmh$$ to obtain the rate of change

$$\frac {dl}{dt}= 20\sqrt{15}kmh$$

• Thanks a lot, Quanto! Commented Dec 1, 2019 at 4:20
• @Harry Battersby - No problem.... Commented Dec 1, 2019 at 4:22