How can you tell whether a modeling function describes an amount or a rate?

Read the wording of the description, and then check the units . Gallons on their own describe an amount, whereas gallons per second describe a rate, and that one word changes what the derivative of the function means.

Define a particle in a motion problem.

A particle is the general term for the moving object, treated as though it were the size of a single point. Modeling it that way means its three-dimensional shape can be ignored entirely.

What is the difference between displacement and distance?

Displacement is measured relative to a fixed point and carries a sign, so it can be positive, negative or zero. Distance is the magnitude of displacement and is never negative, and it may instead mean the total length traveled.

True or False? When a cone drains but keeps its shape, its radius can be treated as a constant.

False. Both the radius and the height change as the cone shrinks, so both have to be differentiated. What stays constant is the ratio of radius to height, and that is what lets one of them be written in terms of the other.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 4: Contextual Applications of DifferentiationRates of Change & Related Rates

Rates of Change & Related Rates (College Board AP® Calculus BC): Flashcards

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FrontMeaning of a Derivative in Context
What are the units of $\frac{d y}{d x}$ ?

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Cards in this collection (25)

What are the units of $\frac{d y}{d x}$ ?
They are the units of $y$ divided by the units of $x$ .
So if a tank's volume $v$ is in liters and time $t$ is in seconds, $\frac{d v}{d t}$ is in liters per second.
True or False?
If $f (t)$ is described as the rate of flow of gasoline, then $f^{'} (t)$ is the rate at which gasoline is being pumped.
False.
Here $f$ is already a rate, so $f^{'}$ is the rate of change of that rate, measured in gallons per second per second.
It says whether the flow is speeding up or slowing down, not how fast the gasoline is arriving.
How can you tell whether a modeling function describes an amount or a rate?
Read the wording of the description, and then check the units.
Gallons on their own describe an amount, whereas gallons per second describe a rate, and that one word changes what the derivative of the function means.
A function $r$ gives the rate of change of the volume of water in a container. Fill in the two missing words for what $r (t) > 0$ together with $r^{'} (t) < 0$ tells you:
the volume is still $______$ even though it is doing so at a $______$ rate overall.
The completed interpretation is: the volume is still increasing even though it is doing so at a decreasing rate overall.
A positive $r$ makes the volume rise, and a negative $r^{'}$ makes that rise slow down.
True or False?
$f^{'} (t)$ can be negative at a moment when $f (t)$ itself is positive.
True.
The two say quite different things: $f$ gives the amount, and $f^{'}$ gives whether that amount is rising or falling.
A tank holding $24$ gallons and losing $3$ gallons a minute has $f > 0$ and $f^{'} < 0$ at the same instant.
A tank has $W^{'} (7) = - 3$ gallons per minute. Interpret that in the context of the problem.
At $t = 7$ minutes the amount of water in the tank is decreasing at a rate of $3$ gallons per minute.
The minus sign is carried by the word "decreasing", so it is not repeated on the number as well.
A height is increasing at a rate of $4$ inches per second. How would you write that rate as a derivative?
As $\frac{d h}{d t} = 4$ , where $h$ is the height in inches and $t$ is the time in seconds.
The units name the two variables for you: inches measure the height and seconds measure the time, so the units settle which goes on top.
For $y = 3 x$ the rate of change is $3$ , while for $v = \frac{1}{3} t^{3}$ it is $t^{2}$ . What is the difference?
For $y = 3 x$ the rate of change is the same at every point, because the graph is a straight line.
For $v = \frac{1}{3} t^{3}$ the rate depends on $t$ , so every point has its own rate: $4$ at $t = 2$ and $25$ at $t = 5$ .
The depth of water in a harbor has $f^{'} (12) = - 2.721$ and $f^{''} (12) = - 0.822$ , in feet and hours. What does each one mean?
The first says that at $t = 12$ the depth is decreasing at a rate of $2.721$ feet per hour.
The second says that the rate at which the depth is changing is itself decreasing, at $0.822$ feet per hour per hour.
Define a particle in a motion problem.
A particle is the general term for the moving object, treated as though it were the size of a single point.
Modeling it that way means its three-dimensional shape can be ignored entirely.
What is the difference between displacement and distance?
Displacement is measured relative to a fixed point and carries a sign, so it can be positive, negative or zero.
Distance is the magnitude of displacement and is never negative, and it may instead mean the total length traveled.
True or False?
A particle with negative acceleration is always slowing down.
False.
What matters is whether the velocity and the acceleration have the same sign, not the sign of the acceleration on its own.
A particle with velocity $- 4$ and acceleration $- 2$ is speeding up, since both are negative.
What is the difference between speed and velocity?
Velocity carries a sign, which gives the direction of motion along the line.
Speed is the magnitude of velocity, so a velocity of $- 6$ corresponds to a speed of $6$ .
Fill in the two missing derivatives for motion in a straight line:
$v =______$
$a =______= \frac{d^{2} s}{d t^{2}}$
The completed results are $v = \frac{d s}{d t}$ and $a = \frac{d v}{d t} = \frac{d^{2} s}{d t^{2}}$ .
Velocity is the slope of a displacement-time graph, and acceleration is the slope of a velocity-time graph.
At $t = 2$ a particle has velocity $- 1$ and acceleration $12$ . Is its speed increasing or decreasing?
Decreasing.
The velocity is negative and the acceleration is positive, so the two have opposite signs and the particle is slowing down.
True or False?
A particle can be instantaneously at rest and accelerating at the same moment.
True.
Being at rest means only that $v = 0$ at that instant, and it says nothing at all about $a$ .
A particle whose velocity is changing sign passes through $v = 0$ with $a \neq 0$ , which is exactly the moment it turns round.
What does $a = 0$ tell you about a particle's motion?
At that instant the velocity is neither increasing nor decreasing.
Constant velocity is a different claim: it needs $a = 0$ right across an interval, not just at a single moment.
A differentiable velocity function has $v (2) = 2$ and $v (4) = - 3$ . Justify that the particle is at rest somewhere between those times.
Since $v$ is differentiable it is continuous, and it changes sign between the two times.
By the intermediate value theorem there is therefore at least one time in that interval where $v = 0$ , which is when the particle is at rest.
What links several rates of change together in a related rates problem?
The chain rule, in the form $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ , adapted to whichever variables the context uses.
The chain can be extended to more than two factors when more variables are involved.
You know $\frac{d h}{d t}$ and can work out $\frac{d v}{d h}$ . Fill in the two gaps to reach $\frac{d v}{d t}$ :
$\frac{d v}{d t} = \frac{d v}{______} \cdot \frac{______}{d t}$
The completed equation is $\frac{d v}{d t} = \frac{d v}{d h} \cdot \frac{d h}{d t}$ .
Write the rate you want first, fill the outer numerator and denominator to match it, and then every remaining term must look as though it cancels.
True or False?
When a cone drains but keeps its shape, its radius can be treated as a constant.
False.
Both the radius and the height change as the cone shrinks, so both have to be differentiated.
What stays constant is the ratio of radius to height, and that is what lets one of them be written in terms of the other.
You can work out $\frac{d V}{d r}$ but the chain you are building needs $\frac{d r}{d V}$ . What do you do?
Take the reciprocal, since $\frac{d r}{d V} = \frac{1}{(\frac{d V}{d r})}$ .
That step is only valid where $\frac{d V}{d r}$ is not zero.
A shape has $V = \frac{1}{3} π r^{3}$ with $\frac{d V}{d t} = 7$ . Find $\frac{d r}{d t}$ when $r = 2$ .
It is $\frac{7}{4 π}$ .
Since $\frac{d V}{d r} = π r^{2}$ , the chain gives $\frac{d r}{d t} = \frac{1}{π r^{2}} \cdot 7$ , which at $r = 2$ comes to $\frac{7}{4 π}$ .
In a related rates problem, what is the alternative to building a chain of derivatives?
Differentiate the equation linking the quantities implicitly with respect to time.
For $V = \frac{1}{3} π r^{3}$ that gives $\frac{d V}{d t} = π r^{2} \frac{d r}{d t}$ directly, and the known values can then be substituted in.
A $17$ meter ladder slides down a wall with its top falling at $6$ meters per second. How fast is the foot moving when it is $15$ meters from the wall?
At $3.2$ meters per second.
Pythagoras gives $x^{2} + y^{2} = 289$ , so $y = 8$ there, and differentiating with respect to time gives $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$ with $\frac{d y}{d t} = - 6$ .

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