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Collège Ahuntsic · 201-SN2-RE

Differential calculus tutoring for Ahuntsic 201-SN2-RE

Build a clear connection between a rate over an interval and a derivative at one instant. Try the worked example or ask a tutor to explain the connection and help you start a question.

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Your differential calculus course at Ahuntsic

Ahuntsic’s 201-SN2-RE, Calcul différentiel, introduces limits and derivatives for science applications. This guide works through the derivative definition and the difference between continuity and differentiability.

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

Course topics

  • Limits and continuity
  • Derivative definition and rules
  • Function analysis
  • Rates of change, related rates and optimization

Before calculating a rate

Label the input, output and their units. Mark the two input values used for an average rate. To find an instantaneous rate, let the second input approach the first and check what happens from both sides.

Two steps worth checking

  • Keep the time increment nonzero while simplifying the difference quotient, then take its limit.
  • At a change of formula, check the function values and the one-sided slopes separately.

From an average filling rate to a derivative

A tank’s water volume is modelled by V(t)=20+2t+t2V(t)=20+2t+t^2 for 0≤t≤20\le t\le2. Here tt is the numerical time in minutes and V(t)V(t) is the numerical volume in litres. Find the average filling rate from 1 to 2 minutes, then use the derivative definition to find the instantaneous rate at 1 minute. Why can the two answers differ?

Show a hint

Start with volume change divided by elapsed time. For the derivative, replace the second time by 1+h1+h, where h≠0h\ne0 is a time increment measured in minutes. Expand (1+h)2(1+h)^2 before simplifying.

Show the worked example

  1. 1. Find the volume change

    The model gives V(1)=23V(1)=23 L and V(2)=28V(2)=28 L. The tank gains 28−23=528-23=5 L over a time interval of 2−1=12-1=1 minute. A volume in litres and a filling rate in litres per minute describe different quantities.

  2. 2. Calculate the average rate

    Let ravgr_{\mathrm{avg}} be the average filling rate over this interval. Then ravg=(28−23) L(2−1) min=5 L/min.\begin{aligned}r_{\mathrm{avg}}&=\frac{(28-23)\,\mathrm{L}}{(2-1)\,\mathrm{min}}\\&=5\,\mathrm{L/min}.\end{aligned} On a volume–time graph, this is the slope of the secant through the points at the ends of the interval.

  3. 3. Shorten the interval without dividing by zero

    Using the model’s minute and litre values, V(1+h)=20+2(1+h)+(1+h)2=23+4h+h2.\begin{aligned}&V(1+h)\\&=20+2(1+h)\\&\quad +(1+h)^2\\&=23+4h+h^2.\end{aligned} Subtract V(1)=23V(1)=23 before dividing. For 0<∣h∣≤10<|h|\le1, V(1+h)−V(1)h=4h+h2h=4+h.\begin{aligned}&\frac{V(1+h)-V(1)}{h}\\&=\frac{4h+h^2}{h}\\&=4+h.\end{aligned} This quotient is an average rate in L/min. For h=0.1h=0.1, it gives 4.1 L/min; for h=−0.1h=-0.1, it gives 3.9 L/min. A negative increment uses an earlier time.

  4. 4. Take the limit and interpret it

    As hh approaches zero from either side, 4+h4+h approaches 4. Thus V′(1)=lim⁡h→0V(1+h)−V(1)h=4.\begin{aligned}&V^{\prime}(1)\\&=\lim_{h\to0}\frac{V(1+h)-V(1)}{h}\\&=4.\end{aligned} The instantaneous filling rate is therefore 4 L/min. It is the tangent slope at 1 minute, measured in litres per minute. The average rate over the following whole minute is higher because the filling rate increases during that interval. The derivative is a limiting rate; it does not describe an extra 4 L arriving all at once.

Your turn: is a continuous function always differentiable?

An idealized filling model changes at 1 minute. Let W(t)W(t) be its numerical volume in litres, with tt in minutes. For 0≤t≤10\le t\le1, use W(t)=20+2t+t2W(t)=20+2t+t^2. For 1<t≤21<t\le2, use W(t)=23+6(t−1)W(t)=23+6(t-1). First check the volume at the join. Next find the left-hand and right-hand rates there. Does W′(1)W^{\prime}(1) exist? Finally, what number could replace 6 to make the two rates match?

Show a hint

For continuity, compare W(1)W(1) with the volume limits from each side. For the derivative, use (W(1+h)−W(1))/h\bigl(W(1+h)-W(1)\bigr)/h: choose the first formula when h<0h<0 and the second when h>0h>0.

Check each step

  1. 1. Check that the volumes meet

    The first formula gives W(1)=23W(1)=23 L. Both one-sided volume limits are also 23 L, so WW is continuous at 1 minute. There is no jump in volume.

  2. 2. Check the left-hand rate

    For h<0h<0 near zero, the first formula applies and the difference quotient is 4+h4+h, just as in the worked example. Its limit as h→0−h\to0^- is 4 L/min.

  3. 3. Check the right-hand rate

    For h>0h>0 near zero, W(1+h)=23+6hW(1+h)=23+6h. Hence W(1+h)−W(1)h=6hh=6.\begin{aligned}&\frac{W(1+h)-W(1)}{h}\\&=\frac{6h}{h}=6.\end{aligned} The right-hand rate is 6 L/min. Since 4≠64\ne6, the two-sided derivative W′(1)W^{\prime}(1) does not exist. The volume graph has a corner even though it is continuous.

  4. 4. Make the slopes match

    Replace 6 by 4 in the second formula: W(t)=23+4(t−1)W(t)=23+4(t-1) for 1<t≤21<t\le2. The join remains at 23 L and both one-sided rates are now 4 L/min. The modified model is differentiable at 1 minute.

If the algebra blocked you, expand the square and subtract the entire starting value on separate lines. If you treated a matching pair of volumes as matching slopes, compare the two difference quotients in the self-check.

Work through the difficult step 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 ask you to identify the interval, label the units and explain which quantity the question asks for. Together, you can trace a missing subtraction or an expansion error before returning to the limit.

For a piecewise function, your tutor can help you choose the correct formula on each side and sketch the join. Then change one coefficient yourself and explain whether the value, the continuity or the derivative changes.

Build a focused revision set

Use your section’s current topics and assessment instructions. Start with one average-rate question, one derivative from the definition and one piecewise continuity question. Write what each limit measures, check the result, then redo the difficult step without the solution. Add derivative-rule and application problems from the topics your class has covered.

Independent practice

Choose your next practice step

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