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Collège André-Grasset · 201-SN2-RE

Differential calculus tutoring for André-Grasset 201-SN2-RE

A curve can keep rising while its slope gets smaller. Learn to separate those two ideas, work through the example, then try a model whose horizontal tangent is neither a maximum nor a minimum.

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Your differential calculus course at André-Grasset

André-Grasset’s 201-SN2-RE, Calcul différentiel, uses limits and derivatives to analyse functions and science models. This guide focuses on reading first- and second-derivative signs to explain growth, extrema and concavity.

Use your current course outline to choose the rest of your revision topics.

Course topics

  • Limits and derivatives
  • Rates of change and function growth
  • Extrema and concavity
  • Asymptotic behaviour and science models

Give each derivative a job

Let tt be the time input and B(t)B(t) the quantity being modelled. If the model is twice differentiable, B′(t)B^{\prime}(t) measures how the quantity changes. On an interval where this derivative is positive, the quantity increases; on an interval where it is negative, the quantity decreases. The second derivative B′′(t)B^{\prime\prime}(t) measures how the first derivative changes. A negative second derivative means the graph is concave down, even on an interval where the quantity is still increasing.

Two conclusions to justify

  • A negative B′′(t)B^{\prime\prime}(t) does not tell you that B(t)B(t) is decreasing. Check the sign of B′(t)B^{\prime}(t) for that.
  • A zero first derivative gives a stationary point. Check the signs on both sides before calling it a local maximum or minimum.

A growth rate can fall while biomass rises

A simplified biomass model is B(t)=20+12t−2t2B(t)=20+12t-2t^2 for 0≤t≤40\le t\le4. Here tt is the numerical time in days and B(t)B(t) is the numerical biomass in grams. Find when the biomass increases and decreases, explain the concavity, and find the greatest biomass on this interval. At day 2, is the biomass already decreasing?

Show a hint

Find B′(t)B^{\prime}(t) and B′′(t)B^{\prime\prime}(t) separately. Solve B′(t)=0B^{\prime}(t)=0, then test the first derivative on each side. For the greatest value on the full interval, compare the interior candidate with both endpoints.

Show the worked example

  1. 1. Separate the two rates

    Differentiating once and then again gives B′(t)=12−4tB^{\prime}(t)=12-4t and B′′(t)=−4.B^{\prime\prime}(t)=-4. The first derivative is a biomass rate in grams per day. The second is the change in that rate per day, in grams per day squared.

  2. 2. Use the first-derivative signs

    The first derivative is zero at t=3t=3. It is positive for 0<t<30<t<3 and negative for 3<t<43<t<4. Thus biomass increases up to day 3, then decreases. The change from positive to negative establishes a local maximum at day 3.

  3. 3. Interpret the concavity

    The second derivative is negative throughout 0<t<40<t<4, so the graph is concave down throughout the interior of the modelled interval. At day 2, B′(2)=4B^{\prime}(2)=4: biomass is still increasing at 4 g/day. Its rate of change is decreasing, which is different from the biomass itself decreasing. There is no inflection point here because the concavity does not change.

  4. 4. Compare the actual biomass values

    The polynomial is continuous on the closed interval. Compare its endpoint values with the interior stationary value: B(0)=20,B(3)=38,B(4)=36.\begin{aligned}B(0)&=20,\\B(3)&=38,\\B(4)&=36.\end{aligned} The greatest biomass in this model is 38 g at day 3. The value B′(3)=0B^{\prime}(3)=0 describes a zero rate at that instant; it does not mean zero biomass.

Your turn: must a horizontal tangent be a turning point?

Consider a different simplified biomass model, C(t)=30+(t−2)3C(t)=30+(t-2)^3, for 0≤t≤40\le t\le4. Again, tt is the numerical time in days and C(t)C(t) is the numerical biomass in grams. Find both derivatives. At day 1, can biomass be increasing while the graph is concave down? At day 2, is there a local extremum, an inflection point, both or neither? Finally, locate the greatest biomass on the full interval.

Show a hint

Factor the first derivative and compare its sign on either side of 2. Then check whether the second derivative changes sign. A zero value of either derivative by itself does not settle the classification.

Check each step

  1. 1. Differentiate and interpret day 1

    The derivatives are C′(t)=3(t−2)2C^{\prime}(t)=3(t-2)^2 and C′′(t)=6(t−2).C^{\prime\prime}(t)=6(t-2). At day 1, C′(1)=3C^{\prime}(1)=3 g/day and C′′(1)=−6C^{\prime\prime}(1)=-6 g/day². Biomass is increasing while its rate is decreasing, so the graph is concave down there.

  2. 2. Classify the stationary point

    At day 2, C′(2)=0C^{\prime}(2)=0. On both sides of 2, the square makes C′(t)>0C^{\prime}(t)>0. Biomass increases before and after this stationary point, so it is neither a local maximum nor a local minimum. A horizontal tangent alone does not establish a turn.

  3. 3. Check for a change of concavity

    For 0<t<20<t<2, the second derivative is negative; for 2<t<42<t<4, it is positive. The graph changes from concave down to concave up at day 2. Since C(2)=30C(2)=30, the graph has a stationary inflection point at (2,30)(2,30). Here “stationary” means the tangent is horizontal, and “inflection” means the concavity changes.

  4. 4. Include the endpoints

    The model is increasing across the interval, with just the one horizontal tangent. The candidate values are C(0)=22,C(2)=30,C(4)=38.\begin{aligned}C(0)&=22,\\C(2)&=30,\\C(4)&=38.\end{aligned} The greatest biomass is 38 g at day 4, an endpoint. Solving only C′(t)=0C^{\prime}(t)=0 would miss it.

If the signs were the obstacle, mark the derivative’s zeros and test one input in each resulting interval within the model’s domain. If the interpretation was the obstacle, write two separate sentences: what the biomass is doing and what its rate of change is doing.

Work through the signs with a tutor

Bring a practice question you are allowed to share, even if you have not started. You can work on the first step together or review any calculation you have tried. Your tutor can help you separate the function, first derivative and second derivative, label their units, and build a sign chart with one test value in each interval.

Your tutor can ask you to explain the curve aloud: where it rises, where its slope decreases and why a stationary point is or is not an extremum. Together, change an interval endpoint and discuss whether the greatest value changes before calculating again.

Build a focused revision set

Use your section’s current topics and assessment instructions. Practise one first-derivative sign chart, one growth-versus-concavity interpretation and one absolute-extremum question with endpoints. For each answer, include the sign evidence that supports it. Then redo a question with a different interval and compare your conclusions.

Independent practice

Choose your next practice step

André-Grasset course information

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