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Differential calculus

Differential calculus. Differential 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.

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Differential calculus

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  1. Differential calculus

  2. Differential 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.

  3. Gradient of a straight line graph The gradient of the line between two points (x1, y1) and (x2, y2) is where 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.

  4. Gradient of a curve A 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.

  5. Tangent to the curve at A. y A O x A tangent is a straight line, which touches, but does not cut, the curve. We cannot calculate the gradient of a tangent directly , as we know only one point on the tangent and we require two points to calculate the gradient of a line.

  6. y Tangent to the curve at A. A O x Using geometry to approximate to a gradient Look at this curve. B1 B2 B3 Look 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.

  7. B4 (2.001, 4.004001) B2 (2.5, 6.25) B3 (2.1, 4.41) A 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). B1(3, 9) 4 to 9 = 5 2 to 3 4 to 6.25 = 4.5 2 to 2.5 = 4.1 2 to 2.1 4 to 4.41 Complete the table A (2, 4) The gradient of the chord AB1 is 4 to 4.004001 = 4.001 2 to 2.001 2 to 2.00001 4 to 4.0000400001 4.00001

  8. y y = x2 4 2 x As 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.

  9. I’d guess 2. (1, ) Example (1) Find the gradient of the chord joining the two points with x-coordinates 1 and 1.001 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 = 2.001 1)

  10. I’d guess 16. (8, ) Example (2) Find the gradient of the chord joining the two points with x-coordinates 8 and 8.0001 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, ) 8.00012 = 16.0001 64

  11. Let’s make a table of the results so far: You’re probably noticing a pattern here. But can we prove it mathematically?

  12. I will call it ∆x. (2, 4) h I need to consider what happens when I increase x by a general increment. I will call it h. (2 + h, (2 + h)2)

  13. y y = x2 B(2 + h, (2 + h)2) A(2, 4) O x If h≠ 0 we can cancel the h’s. Let y = x2 and let A be the point (2, 4) Let B be the point (2 + h, (2 + h)2) Here we have increased x by a very small amount h. In the early days of calculus h was referred to as an infinitesimal. Draw the chord AB. Gradient of AB Use a similar method to find the gradient of y = x2 at the points (i) (3, 9) (ii) (4, 16) = 4 + h As h approaches zero, 4 + h approaches 4. So the gradient of the curve at the point (2, 4) is 4.

  14. It looks like the gradient is simply 2x. We can now add to our table: 6 8

  15. Let’s check this result. y = x2

  16. Let’s check this result. y = x2 Gradient at (3, 9) = 6

  17. Let’s check this result. y = x2 Gradient at (2, 4) = 4

  18. Let’s check this result. y = x2 Gradient at (1, 1) = 2

  19. Let’s check this result. y = x2 Gradient at (0, 0) = 0

  20. Let’s check this result. y = x2 Gradient at (–1, 1) = –2

  21. Let’s check this result. y = x2 Gradient at (–2, 4) = –4

  22. Let’s check this result. y = x2 Gradient at (–3, 9) = –6

  23. ZOOM IN Another way of seeing what the gradient is at the point (2, 4) is to plot an accurate graph and ‘zoom in’. y = x2

  24. When we zoom in the curve starts to look like a straight line which makes it easy to estimate the gradient. 0.8 0.2

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