CEGEP Calculus I — Related Rates Worksheet

All 5 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Setting Up a Related Rates Relation

   A sealed bicycle pump holds a fixed quantity of air. While the plunger is pushed in slowly, the pressure $P$ (in kilopascals) and the volume $V$ (in cubic centimetres) of the trapped air satisfy $$PV=7200.$$ At the instant the trapped air occupies $60{\text{cm}}^3$, the volume is decreasing at $8{\text{cm}}^3/\text{s}$.

   1. Name the two quantities that vary with time, and state which rate is given and which is wanted, using Leibniz notation.
   2. Differentiate the relation with respect to time.
   3. Find the rate of change of the pressure at that instant, with units.
2. Q2Related Rates in Geometric Figures

   Road salt is poured into a hopper shaped like an inverted right circular cone. The hopper is $6$ metres deep and its circular top has radius $3$ metres. Salt enters at a constant $2{\text{m}}^3/\text{min}$, and the pile in the hopper always fills the cone up to a depth $h$.

   Recall $V=\frac{1}{3}\pi r^{2}h$ for a cone of base radius $r$ and height $h$.

   1. Express $V$ in terms of $h$ alone. Explain why the relation $r=h/2$ may be substituted before differentiating, while $h=4$ may not.
   2. Find the rate at which the depth of salt is rising when $h=4$ metres. Give an exact value and a decimal to three places, with units.
3. Q3Related Rates in Motion Problems

   Two straight service roads leave a depot, one due east and one due north. A van drives away from the depot along the east road at a constant $60\text{km}/\text{h}$, and a truck drives away along the north road at a constant $45\text{km}/\text{h}$.

   Consider the instant at which the van is $12$ km from the depot and the truck is $9$ km from the depot.

   1. Let $x$ and $y$ be the two distances from the depot and $z$ the distance between the vehicles. Write the relation connecting them and differentiate it with respect to time.
   2. Find ${\displaystyle \frac{dz}{dt}}$ at that instant, with units, and say in one sentence what is happening to the gap between the vehicles.
4. Q4Interpreting a Related Rates Answer

   A settling tank is being drained. An engineer models the tank correctly and, for one particular instant, obtains $$\frac{dV}{dt}=-0.25{\text{m}}^3/\text{min},\frac{dh}{dt}=-0.05\text{m}/\text{min},\frac{dA}{dt}=0{\text{m}}^2/\text{min},$$ where $V$ is the volume of liquid, $h$ is the depth and $A$ is the area of the free surface of the liquid. No further computation of derivatives is needed below.

   1. Write one sentence for each rate, saying what it tells you about the tank at that instant, with units, and whether the quantity is growing or shrinking.
   2. These quantities satisfy ${\displaystyle \frac{dV}{dt}}=A{\displaystyle \frac{dh}{dt}}$. Use this to state the free-surface area at that instant, and explain why ${\displaystyle \frac{dA}{dt}}=0$ does not mean the surface has area zero. What does it say about the shape of the tank at that depth?
   3. A colleague concludes: “${\displaystyle \frac{dh}{dt}}$ is only $-0.05$, so the tank will still be draining an hour from now.” Explain why the computed rate does not justify that claim.
5. Q5Synthesis — drawing on several topics in this set

   A water trough is $4$ metres long. Its cross-section is an isosceles triangle standing point down: $1$ metre deep, and $1.2$ metres across at the top. Water runs in at a constant $0.06{\text{m}}^3/\text{min}$. Let $h$ be the depth of water, in metres.

   1. Express the volume of water in terms of $h$ alone, stating which substitution is legitimate before differentiating and why.
   2. Find the rate at which the water level is rising when $h=0.5$ metres, with units.
   3. Without computing a second value, explain whether the level rises faster or more slowly as the trough continues to fill, and justify your answer from the relation you differentiated.

The 3 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

Every letter in the picture is a function of time

Q1 opens by asking you to name the two quantities that vary with time and to state, in Leibniz notation, which rate you are given and which one you want. That looks like bookkeeping. It is the problem. The pump's pressure and volume are not two numbers, they are two functions of the clock, and the moment you write dV/dt and dP/dt you have committed to that reading — after which differentiating the relation with respect to t is the only move available.

Error one: substituting before differentiating

In Q1 the trapped air occupies 60 cm³ at one instant. Put that 60 into the relation before you differentiate and the left-hand side becomes a number; a number has derivative zero, so you have just told the calculus that nothing in the pump is changing — which is the one thing you know is false. The same trap is set out in the open in Q2, where part (a) asks you to say why one substitution may be made before differentiating and the other may not.

Error two: a quantity that really is constant — or one you failed to notice

The mirror-image mistake is to differentiate something that was never varying, or to miss that a phrase in the question has quietly handed you a rate. The relation in Q1 has a fixed number on its right-hand side, and that side genuinely is the same at every instant, because the pump is sealed. Its derivative is zero, and that zero is not a technicality — it is the entire reason the two remaining rates are tied to one another.

Geometry: getting down to one variable

A cone or a trough hands you a relation with two varying lengths in it at once — Q2 has a radius and a depth, Q5 has a width and a depth — and you cannot solve for one rate while a second unknown rate is still sitting in the equation. The way out is similar triangles: the surface of the pile or of the water is always a scaled copy of the top of the container, so the ratio between the two lengths is fixed by the shape.

That ratio is exactly the kind of fact step 3 lets in early: it is true throughout the filling, not only at the moment you are asked about. Substitute it, get the volume as a function of the depth alone, and differentiate after that. Both Q2 and Q5 ask you to state which substitution is legitimate and why, so the reasoning earns marks in its own right and not merely as a route to a number.

Motion: the relation is usually Pythagoras

Q3 sends two vehicles down perpendicular roads and asks how fast the gap between them changes. The relation connecting the three distances is the right-triangle one, and it is worth differentiating in its squared form rather than solving for the hypotenuse first:

A square root would demand a chain rule inside a chain rule for no gain. Differentiate the squares, and divide by what is left over on the very last line. Two sanity checks on an answer of this kind: the gap between two vehicles can never open faster than their two speeds added together, and a positive rate means separating while a negative one means closing. If your number fails either check, the error is upstream, in the relation.

Interpreting the answer is a criterion, not a courtesy

Q4 is unusual, and deliberately so: the modelling has been done for you, three rates arrive already computed, and you are asked only what they mean. No derivative has to be found anywhere in it. That is not a soft question — the devis lists a just interpretation of the results as a performance criterion in its own right, which means a correct number reported without its sign, its units and a direction is an incomplete answer, not a stylistically plain one.

You can write the units down before you have solved anything at all: a depth rate is metres per minute, a volume rate is cubic metres per minute, an area rate is square centimetres per second. Doing it in that order means the units are never what gets dropped in the last thirty seconds of an exam — and a rate whose units you cannot name is nearly always a rate you have set up wrongly.

Q5 closes the set by asking for a trend rather than a value: whether the level in the trough rises faster or more slowly as it goes on filling, justified from the relation you differentiated and without computing a second value. Rearrange that relation until the rate you care about stands alone, then look at what it still depends on. Reading the behaviour of a quantity off the form of an expression, rather than off a table of values, is the habit this whole course is built to teach.

Getting the most out of it

Split your labelling into "always true" and "true now"

Two lists down the side of the page. On the left, the facts about the setup: the shape ratio, the fixed length of the trough, the sealed product, the constant inflow. On the right, the readings for this one instant: the depth, the volume, the distance. Everything on the left may enter the equation before you differentiate; nothing on the right may. That one division prevents both of the errors above, and it costs about fifteen seconds.

Write the units before the number

You know a depth rate is in metres per minute the moment you have read the question, long before you have differentiated anything. Fill them in first and they can never become the thing you forget.

Do the whole differentiation in symbols

Keep every varying quantity a letter through the d/dt line, and let numbers appear for the first time on the line after it. It feels slower and it is far quicker to repair: if a numerical value shows up in your working above the derivative line, ask at once whether the quantity it stands for was really constant.

Say the answer back as a sentence, then test it against the story

"The pressure is rising, by so many kilopascals each second, at the instant the air fills sixty cubic centimetres." Then ask whether the story agrees: squeezing air into a smaller space should raise its pressure, salt poured into a hopper should raise the pile, two vehicles driving away from the same depot should be separating. A sign that contradicts the physical situation has just found you an error for free.

Common questions