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## Calculus 1

### Course: Calculus 1 > 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: expressions

When we have a related rates problem on our hands, it's best to first make sure we understand all the involved quantities.

## Want to join the conversation?

- What does A'(t₀) end up being in this problem?(19 votes)
- I think it should be A'(t0) = 8.5 m^2/hour(13 votes)

- When we take the derivative of both sides, wouldn't the derivative of 1/2 = 0 ?(5 votes)
- No. When you take the derivative of both sides, only a constant added onto either side would = 0. If 1/2 was added to the right-hand side of the equation, it would = 0 in the derivative. However, because the 1/2 is a coefficient (and is being multiplied, not added), the 1/2 remains.

This is shown in a derivative rule:

d/dx[A * f(x)] = A * f'(x)

As you can see, the coefficient is kept when the derivative is taken.(11 votes)

- Why is the derivative of the area at that instant "not given"? Khan showed us that we can calculate the derivative at that instant by plugging in.(1 vote)
- Yes, we can calculate 𝐴'(𝑡₀), but its value is not given in the problem.(5 votes)

- For these problems, is the rate of change that we are looking for always going to be constant? When we solved a problem in the last video, we used measurements from a specific moment to generalize the rate of change of the area of a circle to all times (I think we did this at least).

Can we use the specific quantities at one moment like they usually give us coupled with our given rates of change to make a general rate of change for whatever we're looking for?

Or are there times when, for example, the rate of change of the area of a circle with respect to time is 1 m/s at t=3 and then 3 m/s at t=10?

Sorry about the confusing wording.(1 vote)- The rate of change the we're looking for isn't constant, if it were, there would be no reason for the problem to specify the rate
*at this instant*. Sal derived the general rate of change of the area of the triangle at6:42. If he had plugged in his known values in a different order, you would have seen that the rate of change of the area of the triangle was 6b-13h, not constant.

The answer was a single number because the answer was that function 6b-13h evaluated at a point.(1 vote)

## Video transcript

- [Instructor] The base b of
t of a triangle is decreasing at a rate of 13 meters per hour, and the height h of t of
the triangle is increasing at a rate of six meters per hour. At a certain instant t sub zero, the base is five meters and
the height is one meter. What is the rate of
change of the area A of t, so our area is going to be a function t, what is the rate of change of the area of the triangle at that instant? And so what we're going
to do in this exercise, instead of going straight
and trying to solve it, what we need to do here is
to identify the various units of different expressions
and then try to think about what information is given and what's not. And then that will actually equip us to actually solve this
rate of change problem. So let's just do this first part. Let's match each
expression with its units. And like always, pause the video, and see if you can do it on your own. All right, so the first
one is b prime of t. So this is the rate of change at which, of which the base is changing
with respect to time. So if we think about it,
b of t, that is the base. That is going to be in meters. So this is going to be in meters. If we say b prime of t, this is going to be how
much our base is changing with respect to time. So this is going to be meters per, and they give us right over here, they say it's decreasing at
a rate of 13 meters per hour, so the units here are meters per hour. And so b prime of t, that is
going to be in meters per hour. A at time t sub zero, remember, A is the area of our triangle. And we're measuring everything in meters. You can tell from the
information they've given us. And so area is going
to be in square units, and so it's going to be in square meters. Now, the height at time t sub zero, well, both the base and the
height, those are lengths. They're gonna be measured in meters, and so our height at
time t sub zero is going to be in meters. And then here we have the rate of change of our area with respect to time. So our area, we already
know is in meters squared. But we want know this here, that this is going to
be the rate of change of our area with respect to time. So it's going to be an
amount of area per unit time. And time here, we're using
hours, as you can see from some of the information
they've given us. So this is going to be area per unit time, or meters squared per hour. So it's going to be right over here. That's area per unit time. And the length we're
using in this is meters, and time is hours. All right. Now they say match each
expression with its given value. So what is the base of the triangle at time t sub zero? Do they give that to us? Well, let's see. They say, at a certain time, at a certain instant t sub zero, the base, I'm gonna underline this
in a different color, the base at a certain instant t sub zero, the base is five meters. So they say the base at time t sub zero, the base is a function of time, but they tell us that it is five meters. So this is five meters right over here. Now what about the rate of change of the base with respect to time? Do they tell us that? Well, look right over here. That's actually the first piece
of information they gave us. The base b of t of a
triangle is decreasing at a rate of 13 meters per hour. So the rate of change of the base, that it b prime of t, which is equal to db/dt, and they tell us that that is, it's decreasing at a rate
of 13 meters per hour. So that would be negative
13 meters per hour. And so the rate of change of
the base with respect to time is going to be negative 13. They gave us that. Now A prime of t, this is the
rate of change of the area at time t sub zero. Did they give us this? Well, they ask us that. What is the rate of
change of the area A of t of the triangle at that instant? So this is what we actually
need to figure out, but they haven't given it to us, otherwise there is no problem to solve. So this one right over here is not given. In fact, this is what we
are trying to solve for. And then finally we have the first derivative of the
height with respect to time. So you could view this as dh/dt. What is this going to be? Do they give it to us? Well, look right over here. They say the height of
the triangle is increasing at a rate of six meters per hour. So if they're saying h of t is increasing, they're telling us the rate of change of h of t with respect to time, so that's h prime of t. And they're telling us
that it is increasing at six meters per hour, so it's gonna be positive
six meters per hour. So they did indeed give us that. Now why is all of this a
useful exercise to go through? Well, now we are really
ready to solve the question. Because, in general, if we're
talking about any triangle, we know that area is equal to 1/2 base times height. Now in this situation, area
and our base and our height, they're all going to be functions of t, so we could write A of t is equal to 1/2 times b of t times h of t. And if we want to find the rate of change of our area at that instant, and the instant that they're talking about is at time t sub zero, well, then what we would want
to do is take the derivative of both sides with respect to t. So the derivative on the left-hand side with respect to t would be A prime of t, and then on the right-hand
side it would be 1/2 times, and we would actually
use a combination of, well, it's really just the
product rule right over here. The derivative of the first
function with respect to t, so it's b prime of t
times the second function, this is just the product rule here, plus the first function b of t times the derivative
of the second function with respect to time. And we need to figure out not
just the general expression, they want us to know what the
rate of change of the area, so A prime of t at that
instant, at t sub zero. So what we want to figure out, we want to figure out A
prime at time t sub zero. Well, that's just going to
be equal to 1/2 times b prime of t sub zero times h of t sub zero, plus b of t sub zero, times h prime of t sub zero. Now this might seem daunting, except they've given us a
lot of this information. What is b prime of t sub zero? Well, they tell us the rate of change of b with respect to time, and it seems like it's just gonna stay at negative 13 meters per hour. So they gave us this. H, what is the height at time t sub zero? Well, they tell us right over here. At a certain instant,
the base is five meters and the height is one meter. So they give us both
b and h at t sub zero. So they gave us this. They gave us this. And what is the rate
of change of the height at time t sub zero? Well, they tell us. The height of the triangle is increasing at a rate of six meters per hour. So they tell us that as well. All of that stuff is given, and so you just have to
plug it in to figure out what is the rate of change
of the area at t sub zero, at that instant.