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### Course: AP®︎/College Calculus AB>Unit 4

Lesson 5: Solving related rates problems

# Related rates: Approaching cars

In this video, we explore the fascinating world of related rates with two cars approaching an intersection. We'll figure out how the rate of change of the distance between the two cars changes as they move. It's a real-world application of math that shows how calculus helps us understand motion and rates of change. Created by Sal Khan.

## Want to join the conversation?

• Did anyone notice that around that the triangle was a 3-4-5 triangle?
• I knew the numbers were so conveniently chosen! :D
• 2 caveats: what if we redid the problem, but instead the intersection of the roads wasn't perpendicular, say 60º, and the cars were moving away from the intersection...at different times?
• What happens when he two cars go beyond the interception?
Do I have to change only the sign of the rate of change found or do I have to recalculate everything?
• you only have to change the signs because all though the speeds are the same the cars are now moving AWAY from the intersection, so dy/dt = 60, not -60, and dx/dt = 30, not -30. This means that instead of 1.2 * -30 + 1.6 * -60 = 2 ds/dt, it would be 1.2 * 30 + 1.6 * 60 = 2 ds/dt. Therefore if you simplify you get a ds/dt of 66 miles/hour, which means that the cars are moving away from each other which does indeed occur.
• shouldn't the answer be -66 miles^2/hour instead of -66 mph?...[1.2 miles * (-30mph) + 1.6 miles * (-60mph)] = -66 miles^2/hour.
• I don't think so, because we also divided by 2S, which is also a distance. S happened to be 1 mile. So we had:
[1.2 miles * (-30mph) + 1.6 miles * (-60mph)] / 2 miles.
• Great video!
It have that impact on me to raise some questions!

Do we already know that two cars will "intersect"? What if they just drive from each other at a certain positive distance?
Let call that distance G (from gap). Then G can be called a global minimum of a function representing the distance between two cars.

The main question is how I can calculate that G?
I guess I can still use the benefits of geometry tools, but kinda got confused...
• You're right that it can be calculated in this way. See my answer to @creationmak's question for the calculations - they won't intersect and their closest approach is 0.18 miles. But your general idea is right - expressing their distance as a function of time, using pythagoras, that's how I came up with the definition of s(t) in that answer. Then to find the global minimum of that function, I differentiate and solve for s'(t)=0.
• what a beautiful car and a track
• at , during the calculation of X^2+Y^2=S^2, are we using the implicit differentiation? since there are multiple variables, I wondered if we should use either partial derivative or implicit derivative techniques...
• Implicit differentiation. This is single-variable calculus and doesn't require you to use any partial derivatives. I hope I have helped you.
• does it matter what in interval you use when you find the rate of change of a proportional relationship
• Now how do you determine how much time until the cars crash? I have an air traffic control question similar to this and the second part of the question says: "How much time does the air traffic controller have to get one of the planes on a different flight path?"
• what will be the least value of 's' possible? How will we calculate that?
• To find the smallest value of s, you can express s as a function of t:

s(t) = sqrt( (0.8-60t)^2 + (0.6-30t)^2 )

Then take the derivative, using the chain rule and simplifying I get:

s'(t) = ( -60(0.8-60t)-30(0.6-30t) ) / sqrt(0.8-60t)^2 + (0.6-30t)^2)

Now we already know that as t increases then s(t) will keep decreasing until the closest pass, and then it starts to increase again. So at this point, s'(t) will be 0, at which point we'll have the minimum. So we set our expression for s'(t) = 0 and solve for t. I won't show the steps here but you can do it for yourself, the result is t = 11/750. So the closest approach occurs after 11/750 of an hour (or approximately 53 seconds, by multiplying 11/750 with 3600) . You can insert this into the s(t) to find the distance at that point, and you'll see that s(11/750) is approximately 0.18 miles (to 2 significant figures).