I personally stick to the analytic definition, and in fact usually assume "curves" are continuously differentiable. These are not, however, the appropriate levels of generality to foist on calculus students. The preceding examples are, respectively, calculus-level articulations of these two viewpoints. I have given this serious thought and can find no argument to counter my colleagues observation of the inadequacy of the informal definition in the case when the curve in question is already a straight line.Īlso, if I do this again in future with another colleague how can I avoid embarrassment again? At what point did I go wrong here with my explanations? Should I have avoided the geometric view completely and gone with rate of changes instead? I am not a teacher but have taught calculus from first principles to many people over the years and would be very interested in how it should be done properly. Now this whole exchange left me feeling rather stupid as I hold a Phd in Maths myself and I could not adequately define a tangent without using the notion of differential calculus - and yet when I was taught calculus at school it was shown as a tool to calculate the gradient of a tangent line and so this becomes a circular argument. I had no comeback to this argument at all and had to concede that I should have just defined the tangent as the line passing through the point on the curve that has gradient equal to the derivative at that point. A secant line is one which intersects a curve at two points. He argued that in this case the definition of the tangent line as "just touching the curve at that point" is simply not true as it is coincident with the line itself and so touches at all points. The slope of this tangent line is f(c) ( the derivative of the function f(x) at xc). He then came back with the argument of this definition when the "curve" in question is a straight line. My colleague then said that for a cubic curve, the line can touch the curve again at other points so I explained the concept again but restricted to a neighbourhood about the point in question. I was describing the tangent line to a curve at a specific point in the same way that I was taught at school - that it is a line that just touches the curve at that point and has gradient equal to the derivative of the curve at that point. While doing this, my colleague came back at me with an argument for which I had no satisfactory reply. It is the line that looks most like the function f ( x) near x = 1.I was recently explaining differentiation from first principles to a colleague and how differentiation can be used to obtain the tangent line to a curve at any point. The green line is a a tangent line that passes through (1, 2). It passes through (1, 2) and (5, 18) with a slope of 4. The red line on the graph is the secant line. The lines and function are graphed below. The equation of the tangent line is y = 2. The equation of the tangent line is y = 0 x + b. We have already found that f (1) = 2 and the other function value is The derivative is found by applying the definition of the derivative at a point, The equation of the secant line is y = 4 x – 2.įor part b, the tangent line must pass through (1, 2) with a slope given by f ′(1). To find the value of b, substitute one of the points for x and y: This gives us the secant line equation y = 4 x + b. It is the same as the instantaneous rate of change or the derivative. For part b, we’ll find the slope of the line using the derivative of f ( x) at x = 1 or f ′(1).įor part a, the line must pass through the point (1, 2) and (5,18) since A tangent line is a line that touches a graph at only one point and is practically parallel to the graph at that point. For part a, we’ll find the slope of the line using two points on the function. Solution For each of these parts, we’ll find the equation of a line using y = mx + b. Tangent is a cofunction of cotangent A cofunction is a function in which f (A) g (B) given that A and B are complementary angles.
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