2Differential calculus is concerned with the rate at which things change. For example, the speed of a car is the rate at which the distance it travels changes with time.First we shall review the gradient of a straight line graph, which represents a rate of change.
3Gradient of a straight line graph The gradient of the line between two points (x1, y1) and (x2, y2) iswhere m is a fixed number called a constant.A gradient can be thought of as the rate of change of y with respect to x.
4Gradient of a curveA curve does not have a constant gradient. Its direction is continuously changing, so its gradient will continuously change too.y = f(x)The gradient of a curve at any point on the curve is defined as being the gradient of the tangent to the curve at this point.
5We cannot calculate the gradient of a tangent directly A tangent is a straight line, which touches, but does not cut, the curve.xyOATangent tothe curveat A., as we know only one point on the tangent and we require two points to calculate the gradient of a line.We cannot calculate the gradient of a tangent directly
6Using geometry to approximate to a gradient Look at this curve. B1B2Tangent tothe curveat A.B3xOLook at the chords AB1, AB2, AB3, . . .For points B1, B2, B3, that are closer and closer to A the sequence of chords AB1, AB2, AB3, move closer to becoming the tangent at A.The gradients of the chords AB1, AB2, AB3, move closer to becoming the gradient of the tangent at A.
7A numerical approach to rates of change Here is how the idea can be applied to a real example. Look at the section of the graph of y = x2 for 2 > x > 3. We want to find the gradient of the curve at A(2, 4).ChordxchangesfromygradientAB1AB2AB3AB4AB5B1 (3, 9)2 to 34 to 9= 5B2 (2.5, 6.25)2 to 2.54 to 6.25= 4.5B3 (2.1, 4.41)B4 (2.001, )2 to 2.14 to 4.41= 4.1A (2, 4)Complete the tableThe gradient of the chord AB1 is2 to 2.0014 to= 4.0012 to4 to
8As the points B1, B2, B3, . . . get closer and closer to A the gradient is getting closer to 4. This suggests that the gradient of the curve y = x2 at the point (2, 4) is 4.xy24y = x2
9The gradient of the chord is Example (1)Find the gradient of the chord joining the two points with x-coordinates 1 and on the graph of y = x2. Make a guess about the gradient of the tangent at the point x = 1.The gradient of the chord is(1.001, )1.0012= 2.001I’d guess 2.(1, )1)
10The gradient of the chord is Example (2)Find the gradient of the chord joining the two points with x-coordinates 8 and on the graph of y = x2. Make a guess about the gradient of the tangent at the point x = 8.The gradient of the chord is(8.0001, )=I’d guess 16.(8, )64
11Let’s make a table of the results so far: x-coordinategradient1243567816You’re probably noticing a pattern here. But can we prove it mathematically?
12I will call it ∆x.I need to consider what happens when I increase x by a general increment. I will call it h.(2 + h, (2 + h)2)(2, 4)h
13Let y = x2 and let A be the point (2, 4) Let B be the point (2 + h, (2 + h)2)xyA(2, 4)y = x2OHere we have increased x by a very small amount h. In the early days of calculus h was referred to as an infinitesimal.B(2 + h, (2 + h)2)Draw the chord AB.Gradient of ABIf h ≠ 0 we can cancel the h’s.= 4 + hUse a similar method to find the gradient of y = x2 at the points(i) (3, 9)(ii) (4, 16)As h approaches zero, 4 + h approaches 4.So the gradient of the curve at the point (2, 4) is 4.
14We can now add to our table: x-coordinategradient1243567816It looks like the gradient is simply 2x.68