Vanier 201-SN2-RE Calculus Tutoring

Your Vanier differential calculus course

Vanier’s Pure and Applied Science course list places 201-SN2-RE Differential Calculus in semester 1. The course develops limits, continuity and derivatives, including trigonometric, exponential and logarithmic functions, with applications to graphing, related rates and optimization.

Vanier lists Social Science Calculus 1 separately as 201-107-VA. Use the full code on your outline when requesting help.

Topics to work on

Limits and continuity
Derivatives of algebraic, trigonometric, exponential and logarithmic functions
Graphs and rates of change
Related rates and optimization

Before you start

Review factoring, function notation and the product and chain rules. For an application, write the permitted values of the variable before solving the derivative equation.

Choose a starting point

A slope on an implicit curve Track the dependent variable through each derivative.
Classify stationary points Use a sign chart to justify a local maximum or minimum.
Build and optimize an open box Start with the dimensions and physical domain, then try the changed-size question.

Differential calculus: implicit slopes and turning points

Differentiate an implicit equation, then use a derivative sign chart to classify stationary points.

Before you start: Allow 20–25 minutes to try both questions, then 15 minutes to check and retry any missed steps.

1. A slope on an implicit curve

On the curve $x^2+xy+y^2=7$ , find $dy/dx$ at $(1,2)$ . Check that the point belongs to the curve.

Show the explained solution

The point satisfies $1+2+4=7$ . Treat $y$ as a function of $x$ .
Differentiation gives $2x+y+xy^\prime+2yy^\prime=0$ . In particular, the derivative of $xy$ needs both product-rule terms.
$y^\prime=-\frac{2x+y}{x+2y}.$ At $(1,2)$ , the slope is $-4/5$ .

Review implicit differentiation step by step

2. Classify stationary points

For $h(x)=x^3-3x^2+2$ , find the stationary points and classify them using the sign of $h^\prime$ .

Show the explained solution

$h^\prime(x)=3x(x-2)$ , so the stationary inputs are $x=0$ and $x=2$ .
The derivative is positive for $x<0$ , negative for $0<x<2$ , and positive for $x>2$ .
Therefore $(0,2)$ is a local maximum and $(2,-2)$ a local minimum. They are not global extrema on the whole real line.

Use derivative signs to classify an extremum

Use your answers to plan your revision

Implicit differentiation
Underline every occurrence of the dependent variable. Explain where each factor of its derivative comes from.
Classification
Make a sign table before using the words maximum or minimum; identify whether your conclusion is local or global.

For Vanier calculus help, send the rule or sign-chart step you could not explain. Keep your section’s outline beside the practice so the review matches your assessment.

Request 201-SN2-RE exam preparation help

Original Nablio practice, not an official examination.

Course topic sources

Which cut makes the largest open box?

A square sheet of card is 24 cm on each side. Cut a square of side $x$ cm from each corner, then fold up the sides to make an open-top box. Ignore the thickness of the card. Write the volume $V(x)$ , state the physical domain (the allowed values of $x$ that make a box) and find the cut that gives the largest volume.

Show a hint

The box height is $x$ . Each base side loses a length $x$ at both ends, so it becomes $24-2x$ .

Show the reasoning

1. Name the dimensions and domain
The height is $x$ cm and each base side is $24-2x$ cm. A box must have a positive height and a positive base side, so $0<x<12.$
Volume is height times base area: $V(x)=x(24-2x)^2.$ Here $V$ is measured in cubic centimetres.
2. Differentiate both factors
The function is a product of $x$ and $(24-2x)^2$ . Apply the product rule, and use the chain rule on the square: $\begin{gathered}V^\prime(x)=(24-2x)^2\\\quad-4x(24-2x)\\=(24-2x)(24-6x).\end{gathered}$
The factor $-4$ comes from $2(24-2x)$ multiplied by the inner derivative $-2$ . Keeping only $(24-2x)^2$ would omit the change in the base area.
3. Check the candidates against the domain
The equation $V^\prime(x)=0$ gives $x=12$ or $x=4$ . The value $x=12$ makes the base side zero and lies outside $0<x<12$ . The only interior candidate is $x=4$ .
4. Show why it is a maximum
Within $0<x<12$ , the factor $24-2x$ is positive. The other factor, $24-6x$ , is positive for $0<x<4$ and negative for $4<x<12$ . Thus the volume increases up to $x=4$ and decreases after it.
This sign change establishes the maximum over the physical domain. Solving the derivative equation alone would only identify candidates.
5. State the dimensions and volume
Cut out 4 cm squares. The resulting box has height 4 cm and base dimensions $16\times16$ cm. Its maximum volume is $\begin{aligned}V(4)&=4(24-8)^2\\&=1024\text{ cm}^3.\end{aligned}$ A length answer of 4 cm and a volume answer of 1024 cm³ describe different quantities.

Your turn

For the same construction using a square card with side length $L>0$ , the volume is $V(x)=x(L-2x)^2$ . What cut gives the maximum? Identify the domain and explain which derivative root must be excluded.

Check your answer

The domain is $0<x<L/2$ and $V^\prime(x)=(L-2x)(L-6x)$ . The root $x=L/2$ collapses the base; the interior root is $x=L/6$ . The derivative is positive before $L/6$ and negative afterward, so that cut maximizes the volume. For $L=24$ , this returns $x=4$ .

If the dimensions were wrong, sketch the sheet and label both corner cuts on one edge. If the model was right but the derivative was wrong, work through the derivative-rule check before trying another application.

What to work through with a tutor

Your tutor can help you set up the problem or review what you have started, label the variables with you and check which values make sense in the situation.

Your tutor can help you choose the first derivative rule and explain each product, quotient or chain-rule step. If you have already tried the derivative, you can look together for a missing factor or step.

Your tutor can check your derivative sign analysis with you, help you justify a maximum or minimum and connect the answer to the problem’s quantities and units.

Build a mixed practice set

Pair a limit or continuity question with a derivative-rule question and one application from your outline. For every optimization problem, write the domain, find the candidates and justify the maximum or minimum. Redo the setup without the solution before changing the numbers.

Independent practice

Continue with these calculus lessons

10 min

Derivative Rules: A Three-Question Check

For choosing the first rule and spotting a missing product-rule term.

18 questions

Calculus I Practice for Limits and Derivatives

For limits, continuity, derivative definitions and mixed derivative methods.

30 min

Before Calculus: 10 Algebra Skills You Should Know

For factoring and rearranging before or after differentiation.

35 min

Sketch a rational function without crossing its domain breaks

Extend polynomial sign analysis to a rational function with excluded inputs. Curve-sketching scope is supported by the historical Fall 2024 outline; use your current section outline.

Official course information

Lesson questions

How do lessons work?

Lessons are currently taught in English, online through Zoom or Google Meet. Send your course code, topic and preferred time; we confirm course support, availability, lesson details and price before payment.

What should I bring?

Bring your current course outline and a question you are allowed to share, if you have them. Any work you have started is useful, but you can also ask for help with the first step. Your outline helps us choose topics that match your section and follow its assessment rules.

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201-SN2-RE Differential Calculus tutoring