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 February 15th, 2016, 06:29 AM #1 Member   Joined: Mar 2015 From: Los Angeles Posts: 73 Thanks: 7 suggestions for how to get student unconfused I'm tutoring a calculus student, and he is struggling. I was wondering if other teachers here have ideas about how to clarify this. He gets confused about the relationship between a function and its derivative. Say the problem is $f(x) = x^2 - x + 3$. Find $f'(x)$. Then find the equation of the tangent line at $x = 3$. So this requires keeping three concepts straight in your mind. First you have the equation defining $f(x)$. Second, you work out $f'(x) = 2x - 1$ and that's a different equation. Then you take $f'(3)$ to get the derivative, i.e. slope of the tangent line, which is 5. But THEN you have to plug 3 into $f$, NOT into $f'$, in order to get the point on the line, which is $(3,9)$. So to do this you have to keep the meaning of $f(x)$ and $f'(x)$ straight. But then you have to work out $y = 5x - 6$, a THIRD equation. So he easily gets all these things confused. I try as much as possible to graph these things. Obviously this is a very visual problem. But he has difficulty making the connection between the algebra and the graph. So when I draw a graph, it's almost like I'm making it even harder for him. Just wondering if anyone has teaching tips. February 15th, 2016, 07:01 AM #2 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 You didn't say how our suggestions worked out in your last thread, only a few days ago. This was about basically the same subject. helping to improve the phsyicality of the concepts February 15th, 2016, 07:06 AM   #3
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 Originally Posted by mike1127 So this requires keeping three concepts straight in your mind. First you have the equation defining $f(x)$. Second, you work out $f'(x) = 2x - 1$ and that's a different equation. Then you take $f'(3)$ to get the derivative, i.e. slope of the tangent line, which is 5. But THEN you have to plug 3 into $f$, NOT into $f'$, in order to get the point on the line, which is $(3,9)$. So to do this you have to keep the meaning of $f(x)$ and $f'(x)$ straight. But then you have to work out $y = 5x - 6$, a THIRD equation.
To be honest, your description sounds a little confusing to me too.

I might start the last part from the end. We seek the equation of the tangent which is a straight line. There are various ways to frame the equation of a straight line: $y=mx+c$ and $y-y_0 = m(x-x_0)$ are respectively the most common and the most useful for this problem, but I'd work with whichever the student comes up with.

In each case, you focus the student on determining one value at a time out of the target equation: and it will be best to start with $m$ the gradient of the straight line.

The straight line is tangent to the curve at the point $x=3$. This means that it has the same gradient as the curve does at the point $x=3$. The gradient of the curve anywhere is given by $f'(x)$. And the point we want is $x=3$, so substitute that into the equation. That is our $m$.

Now, if necessary, you can highlight that there are (infinitely) many straight lines with that gradient. But we want the one that just touches the curve at $x=3$. When we say that the curves touch, we mean that they share a point. So for our straight line, the $y$ value we get at $x=$ is the same as the $y$ value we get for the curve $y=f(x)$ at $x=3$. And this latter is $y=f(3)$.

We can then use these values ($x=3$ and $y=f(3)$) to determine the equation of our straight line (either by $x_0=3$, $y_0=f(3)$ or by setting $x=3$ and $y=f(3)$ and determining the value of $c$ algebraically.

Last edited by skipjack; February 15th, 2016 at 10:06 PM. February 15th, 2016, 09:59 PM   #4
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 Originally Posted by studiot You didn't say how our suggestions worked out in your last thread, only a few days ago. This was about basically the same subject. helping to improve the phsyicality of the concepts

Hi Studiot, yes thank you so much for your references. I love "calculus made easy." However, my student resisted reading it. He is anxious about math, but his response to that is to hurry, hurry, hurry. I have pointed out to him several times that it's no good to hurry if you get the wrong answer, and I am slowly making an impression on him. He is changing his tune gradually.

Also, the class is moving slowly and has mostly focused on algebra review so far. There have been some calculus problems actually, but it's a class for non-math majors, and the teacher wants to make it easy, and so she is spending 90% of the time reviewing algebra so far in the first three weeks. She sometimes provides a review problem set prior to a test, and my student says that he only wants to work the type of problems on the review set, because it seems faster to him. I keep pointing out that broad knowledge supports understanding, and taking shortcuts now will only make the class harder later in the semester. He is an athlete so I tried to use an athletic example: I said that his coaches want him to cross-train. "You wouldn't think of doing only a single activity if you are an athlete, right?"

So I really appreciate your suggestions and I will continue to make use of them as the class progresses.

Mike February 15th, 2016, 11:30 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 The derivative of a function can be thought of as its slope. Your student should be aware of the $df/dx$ or $dy/dx$ notation, which makes this slightly easier to understand. The tangent line isn't "at $x = 3$". It's the straight line through the point $(3,\,f(3))$ with slope (i.e. derivative) equal to $f\,'(3)$. Hence it's convenient to obtain the tangent's equation in point-slope form: $\displaystyle y - y_1 = m( x - x_1 )$, where $(x_1,\,y_1)$ is the point alluded to in the question and $m$ is the common slope. The task then reduces to writing that line equation and the values of $x_1$, $y_1$ and $m$. Point out that $x_1=3$, $y_1=f(3)$ and $m=f\,'(3)$. At this level, many problems require the student to (a) recognize what the problem means, (b) write down the relevant equations, and (c) carry out the appropriate calculations. Note that to tackle this kind of problem, the student must previously have learned how to differentiate a function (at least for simple functions, such as polynomials). Thanks from mike1127 February 16th, 2016, 02:29 AM #6 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 Mike, thank you for the clarification, but I have to say that it was skipjack, not I that provided the references. Further my offering was towards a different simpler approach from that given by all the big brains here. I did not give much detail because I hoped a maths tutor would pick up on that and ask for more if needed. Unfortunately you are tutoring someone who doesn't want to learn maths, just pass a minor hurdle to his (her) career and so will simply try to make specific learned responses to specific test questions, hence the concentration on the review. For this person KISS and learning by drill rote is the appropriate teaching method. February 16th, 2016, 06:51 AM   #7
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 Originally Posted by studiot Mike, thank you for the clarification, but I have to say that it was skipjack, not I that provided the references. Further my offering was towards a different simpler approach from that given by all the big brains here. I did not give much detail because I hoped a maths tutor would pick up on that and ask for more if needed. Unfortunately you are tutoring someone who doesn't want to learn maths, just pass a minor hurdle to his (her) career and so will simply try to make specific learned responses to specific test questions, hence the concentration on the review. For this person KISS and learning by drill rote is the appropriate teaching method.
Oh yes, I was responding to the wrong person. Yes, your suggestions are right on--you are right in the heart of what's going on, and where he struggles with an appreciation of the real meaning of "variables," "independent variables," "functions." And you mention that a graph of a line can demonstrate what happens when you double x (y doubles).

He really has a hard time making connections between graphs and equations. He also seems to have a hard time imagining a graph as a living thing. I try to say "imagine yourself running along the curve... what value are you approaching?" But he just looks blank.

I think you are right in the heart of what needs to be explained, but I am not sure how to do it.

But I had an idea yesterday.

Let me actually address this entire thread. The explanations here are very nice. I like how you guys have "organized" the explanation so that it appears to be simpler. I think that keeping things simple and obvious, to the greatest extent possible, is really important at all levels of math (even category theory).

However, I still think this explanation would be beyond my student's comprehension. Yet I don't want to give up the idea he can really understand at least some of this. Here is my idea: I could ask him to organize his own explanation. Basically I would introduce the ideas one at a time, and ask him to make notes, or a graph, or a cheat sheet, in a form that appeals to him. I would direct his attention to how he organizes the paper and ask him to consider what would help his understanding.

The goal is not to get a beautiful set of notes. They probably won't be very organized. The goal is to shift the process from just feeding him information to getting him to be an active participant.

Lots of general learning strategies are helpful for him, like the ones in the book "How to Solve It." Sometimes it helps to ask him to write out the statement of the problem himself. That kind of thing.

Mike February 16th, 2016, 06:55 AM   #8
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 Originally Posted by skipjack ... At this level, many problems require the student to (a) recognize what the problem means, (b) write down the relevant equations, and (c) carry out the appropriate calculations. Note that to tackle this kind of problem, the student must previously have learned how to differentiate a function (at least for simple functions, such as polynomials).
Yes, your explanation is clearer than mine. He struggles with recognizing what the problem means. Also general attention to detail. But I don't want to give up the idea he will come to understand what the problem really means. As mentioned in my prior post, I'm going to try to get him to be more actively involved in the way he constructs concepts.

Mike February 16th, 2016, 08:19 AM #9 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,161 Thanks: 734 Math Focus: Physics, mathematical modelling, numerical and computational solutions The above posts make lots of good points! Here's a couple of exercises that might help if you need something more specific to follow. I would personally leave at least a few hours or a day between the two. Required knowledge: - basic algebraic manipulation - polynomial differentiation - basic 2D sketching (coordinates, axes, limits) - equation of a line ($\displaystyle y = mx + c$; $\displaystyle m$ = gradient, $\displaystyle c = y$-intercept) Make sure the student has a ruler and works in pencil. Also, try and identify if any of the above are difficult. If so, go back and study those again. The student should not attempt the following until the above topics are understood. ------------------------------------------------------------------------- Exercise 1: gradient concept: Start with a polynomial function $\displaystyle f(x) = x^3 - 2x + 5$ 1. Make a table with four columns: $\displaystyle x$, $\displaystyle f(x)$, "slope" and $\displaystyle f'(x)$ on a left-hand page of a workbook (or piece of paper) and two sets of axes on a right-hand page (or a separate piece of paper), one directly above the other. The x-axis limits should be -3 to 3 and the scale should be identical on both axes. This is important for this exercise to work. The y-axis limits for the upper graph should be between 10 and -6. The y-axis limits on the lower axes can be left for now 2. Put the following values into the x column: -2.5, -2.0, -1.5, -1.0, -0.5, 0.0, 0.5, 1.0, 1.5 and 2.0 3. Ask the student to evaluate the function for each x-value and put the answer in the second column. Working out should be shown. Use this opportunity to see if he can substitute properly and check correct use of notation. e.g. $\displaystyle f(-2) = (-2)^3 - 2(-2) + 5 = -8 +4 +5 = 1$ 4. Ask the student to plot $\displaystyle y = f(x)$ on the upper set of axes. Use this opportunity to see if the graph is of decent quality, axes labels included, correct points, etc. The line of best fit should be a curve going through all the points, but it doesn't need to be super accurate. 5. Ask the student to look at each point and describe how steep the line is at each point. The answers should go in the third column. i) is it sloping upwards (positive) or downwards (negative) ii) is it very steep, steep, a bit steep, shallow, very shallow, horizontal. Answers here need not be amazing, but each one should have a sign (+ or -) and a label. You'll want to watch out for obvious mistakes, like saying it is really steep when the function is close to a turning point 6. Ask the student to work out $\displaystyle f'(x)$ using the usual differentiation method. Check the result. 7. Ask the student to fill out column 4 by substituting $\displaystyle x$ into $\displaystyle f'(x)$ for each value. The student should notice that the numbers in this column differ 8. Ask the student to plot $\displaystyle y = f'(x)$ on the lower set of axes by going through each point. The x-axis limits should be the same, but the y-axis limits can be different. The student should notice that the graph looks totally different. 9. Ask the student to compare the numbers in the f'(x) column with the predictions for steepness made earlier. Are all the negative slopes negative in the results? Are the shallow descriptions matching with smaller numbers and the steep predictions matching with big numbers? 10. Ask the student to identify turning points on the top graph ($\displaystyle y=f(x)$). 11. Ask the student to, for each turning point, draw a dotted line (MUST be dotted.. this is not a tangent line!) vertically down the page until it hits the other curve. Now ask the student to write what the coordinate is for each point on the lower curve. If the student doesn't notice it already, point out that the y-value is 0 (or very close to 0 depending on the quality of the line of best fit) ------------------------------------------------------------------------ Exercise 2: tangent concept (requires more care than the previous one) Use the same function as exercise 1. 1. Pick points $\displaystyle x = -1.5, x = 0$ and $\displaystyle x = 1.5$. For each point ask the student to work out $\displaystyle f(x)$ without looking at the previous exercise. 2. Ask the student to work out$\displaystyle f'(x)$ using differentiation again without looking at the previous exercise. 3. Ask the student to calculate $\displaystyle f'(x)$ for the those three points without looking at the previous exercise. 4. Go back to the top figure of the previous exercise. Explain that a tangent line is a line that has the same slope (or steepness) at a point. Ask the student to draw a tangent line for the curve at $\displaystyle x = 0$. Make sure a ruler is used and make sure this tangent is good! Using a ruler to double-check that the steepness is the same before drawing the line is a good idea... the ruler should kinda follow the line's course in this particular case because it is in the middle of the polynomial curve. If all else fails, ask the student to redraw the original plot (there is a table of values from the previous exercise after all) and start again with a fresh set of axes. 5. Point out where the tangent line crosses the y-axis. Note down this coordinate ($\displaystyle x_1, y_1$) and ask what the y-intercept is. 6. Point out where the tangent crosses the x-axis. Note down this coordinate ($\displaystyle x_2, y_2$). 7. Ask the student to work out the equation of the straight line from the two points. (i.e. draw a triangle, obtain m and c using $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$ and $\displaystyle c = y$-intercept). Double-check the answer carefully to make sure the answer is correct (a common mistake is to accidentally swap $\displaystyle x_2$ and $\displaystyle x_1$ around). The final equation should be in the form $\displaystyle y = mx + c$ 8. Ask the student what the slope is (it is $\displaystyle m$). Is the slope negative? Is $\displaystyle m$ negative? 9. Ask the student to compare their value of $\displaystyle m$ with the $\displaystyle f'(x)$ value they calculated earlier. Are they similar? They should be! If they are different signs or very different in magnitude, a mistake has been made, so check over the previous work. 10. Ask the student to repeat steps 4-9 for the other two points ($\displaystyle x=1.5$ and $\displaystyle x = -1.5$). Watch to make sure the tangent lines are good. If they are not, ask the student to rub out the line and try again. After these exercises, I'd maybe retry the one you are currently trying to do. It depends on how much progress you can make. Thanks from greg1313 and mike1127 Last edited by Benit13; February 16th, 2016 at 08:27 AM. February 17th, 2016, 01:59 AM #10 Member   Joined: Mar 2015 From: Los Angeles Posts: 73 Thanks: 7 Thanks, Benit. Your exercise idea is very useful. As an aside, there is a lot of "psychology" in my work with this student. I mean that he has counterproductive habits that hamper his learning. One thing is that he doesn't like doing exercises that he thinks he understands already, because he regards them as a waste of time. I am coaching him to regard exercises as good training that deepen one's understanding. I.e. understanding something is not all or nothing. It's a question of how deep and confident you can get. Mike Tags student, suggestions, unconfused Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post cherrie New Users 1 August 1st, 2014 12:04 AM Liwayway New Users 2 February 20th, 2014 04:53 AM ejreams New Users 3 November 1st, 2011 04:30 PM

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