Slopes of tangent lines

By | July 26, 2011

One of the most difficult items for the Philippine sample in the Trends and Issues in Science and Mathematics Education Study (TIMSS) for Advanced Mathematics and Science students conducted in 2008, is about comparing the slopes of the tangent at a point on a curve. The question is constructed so that it assesses not only the students understanding of tangent lines to the graph of a trigonometric function but also students’ skill to use mathematics to explain their thinking. The question is one of the released items of TIMSS Advanced 2008 so I can share it here. The graph actually extends beyond point B in the original item.

Sophia is studying the graph of the function y=x+cos x. She says that the slope at point A is the same as the slope at point B. Explain why she is correct.

I don’t have information  if the students’ difficulty has to do with their mathematical understanding or it is the way the question is asked. I have a feeling that had the question been ‘What is the derivative of the function y = x + cos x?’, the students would have been able to answer it. But of course, the item is also assessing students’ conceptual understanding of derivative as the slope of the tangent line at a point on a curve.

The TIMSS Advanced tests were given to Year 11/12 populations. Because the country does not have senior high schools, the Philippines sample were Year 10 students from Science High Schools where calculus is a required subject. The group of teachers we were discussing this question with said they are only able to cover up to the derivative of polynomial functions although the syllabus cover derivative of trigonometric functions. Indeed, the problem should not be difficult to those who have taken calculus or at least have reached the topic about the derivative of trigonometric functions. The solution is pretty straight forward. The derivative of the function y = x + cos x is 1+-sin x so the slope of the tangent  at Π and 2Π is 1.

Covering the syllabus is really a problem because of lack of time. Even if the students are well selected, I think it is still a tall order to cover topics what other countries would cover with an additional two years in high school. Quality of teaching suffers when teachers will teach math at lightning speed. One is forced to do chalk and talk.

The TIMSS item shown above can still be solved with basic knowledge of trigonometric function and slopes of tangent lines. The function y = x + cos x is a sum of the function y = x and y = cos x. The slope of y = x is 1. That slope is constant. The function y = cos x has turning points at Π and 2Π hence the slope of the tangents at these points is 0. So, Sophia is correct in saying that the slope of the tangents at Π and 2Π in y = x + cos x are the same. Students are more likely to analyze the problem this way if they have a conceptual understanding of the functions under consideration and if they are exposed to similar way of thinking, especially of expressing representations in equivalent and more familiar form. This way of thinking need to be developed early on. For example, learners need to be exposed to tasks such

1) Find as many ways of  expressing the number 8.

2) What number goes to the blanks in 14 + ___ = 15 + ____.

3) Solve 3x = 2x – 1 graphically.

You may want to read my other posts to items based on TIMSS framework here and proposed framework for analysing understanding of function in equation form and sample problem on sketching the graph of the derivative function.


References for understanding the idea of derivatives

1.Students’ conceptual understanding of a function and its derivative in an experimental calculus course [An article from: Journal of Mathematical Behavior]
2. Calculus: An Intuitive and Physical Approach (Second Edition)


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3 thoughts on “Slopes of tangent lines

  1. Shaun

    Hi Erlina,

    I just found your site, and I think it teaches very similar concepts that my site does as well. Would you like to link my site in your blogroll? It is Math Concepts Explained at I am going to link to your site right now and hopefully some of my readers will come pay you a visit. 🙂



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