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

Lesson 4: Introduction to related rates- Related rates intro
- Analyzing problems involving related rates
- Analyzing related rates problems: expressions
- Analyzing related rates problems: expressions
- Analyzing related rates problems: equations (Pythagoras)
- Analyzing related rates problems: equations (trig)
- Analyzing related rates problems: equations
- Differentiating related functions intro
- Worked example: Differentiating related functions
- Differentiate related functions

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# Analyzing related rates problems: equations (Pythagoras)

A crucial part of solving related rates problems is picking an equation that correctly relates the quantities. We recommend making a diagram before doing that.

## Want to join the conversation?

- What does the answer end up being?(12 votes)
- I used two methods to calculate it and got -102.307 (approximated) twice(8 votes)

- why is this so hard(8 votes)
- If you do not feel ready for this, I encourage you to watch the previous videos in the Calculus 1 playlist.

If you still face difficulty, go to the Precalculus playlist.

If you still don't feel ready, go to the Get ready for precalculus playlist.

I hope this helps.(2 votes)

- Why is distance of car to intersection relevant? can this be solved if we just put respective car speeds into pythagorean theorem (as they denote rate of change of car distance to the intersection at point t0). If you do that you get similar result ~102.9km/h(3 votes)
- To get the answer you have to find the instantaneous rate of change of function d(t) at instant t0. To get this value, you would find what the function of d(t) is, get it's derivative, then plug in the values to get your answer. To do this you need the values, d, x(t), and y(t). X(t) and Y(t) are the distances to the intersection, while d can be found using the pythagorean theorem. As you found out, doing it the way you described does not give you an exact answer and if you put that on a test you would most likely get the question wrong.(3 votes)

- I was wondering if you could set it up as d(t) = √(x(t)^2 +y(t)^2) and then take the derivative. d'(t) = 1/2(x(t)^2+y(t)^2)^-1/2 (2x'(t)+2y'(t)

and then plug in all of your info. Does that work?(3 votes)- I had the same idea.

I think you are almost correct. I think it should be:

d'(t) = (1/2(x(t)^2+y(t)^2)^(-1/2))(2xx'(t)+2yy'(t))

that is, you forgot to apply the chain rule to (x(t))^2 and (y(t))^2(1 vote)

- Is this like some sort of function that is defined by its derivative?(2 votes)
- I'm not 100% sure what you mean, but you take the first equation ---> a^2+b^2=c^2 and then find the derivative of that ---> 2a(aPrime) + 2b(bPrime) = 0(3 votes)

- Is t sub zero just the time at zero seconds?(3 votes)
- I think the question needs to be more elaborated. The solution is using Euclidean distance. Another accepted answer would be using Manhattan distance.(1 vote)

## Video transcript

- [Instructor] Two cars are driving towards an intersection from
perpendicular directions. The first car's velocity is 50 km/h and the second car's velocity is 90 km/h. At a certain instant, T sub zero, the first car is a
distance, X sub T sub zero or X of T sub zero, of half a kilometer from the intersection and the second car is a distance, Y of T sub zero, of 1.2 kilometers from the intersection. What is the rate of
change of the distance, D of T, between the cars at that instant? So, T sub zero. Which equation should be
used to solve the problem? And they give us a
choice of four equations, right over here, so you
could pause the video and try to work through it on your own but I'm about to do it, as well. So, let's just draw what's going on. That's always a healthy thing to do. So, two cars are driving towards an intersection from
perpendicular directions. So, let's say that this
is one car right over here and it is moving in the X direction towards that intersection,
which is right over there. And then you have another car that is moving in the Y direction. So, let's say it's moving like this. So, this is the other car. I should have maybe done a top view. Well, here we go. This square represents the car and it is moving in that direction. Now, they say at a certain instance. T sub zero, so let's draw that instant. So, the first car is a distance, X of T sub zero, of 0.5 kilometers. So, this distance, right over here, let's just call this X of T and let's call this distance, right over here, Y of T. Now, how does the distance between the cars relate to X of T and Y of T? Well, we could just use
the distance formula, which is essentially just
the Pythagorean theorem, to say, well, the
distance between the cars would be the hypotenuse
of this right triangle. Remember, they're traveling
from perpendicular directions. So, that's a right triangle, there. So, this distance, right over
here, would be X of T squared plus Y of T squared, and
the square root of that, and that's just the Pythagorean
theorem, right over here. This would be D of T, or we could say that D of T squared is equal to X of T squared plus Y,
too many parentheses. Plus Y of T squared. So, that's the relationship between D of T, X of T, and Y
of T, and it's useful for solving this problem because, now, we can take the derivative of both sides of this equation with respect to T. We'd be using various derivative rules, including the chain
rule, in order to do it, and then that would give us a relationship between the rate of change of D of T, which would be D prime of T, and the rate of change of X of T, Y of T, and X of T and Y of T, themselves. And so, if we look at these
choices, right over here, we indeed see that D sets up that exact same relationship that
we just did, ourselves. That it shows that the distance squared between the cars is equal to that X distance from the
intersection squared, plus the Y distance from
the intersection squared, and then we can take the
derivative of both sides to actually figure out this
related, rates question.