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Maths IB standard

Maths IB standard. Calculus Yay!. Starter activity. How do we know what the gradient of a line is? Y = mx + c If we are given two points how do we find the gradient?. Gradient/ Slope. The gradient or slope of a line describes the steepness. To find the gradient we use the phrase rise

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Maths IB standard

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  1. Maths IB standard Calculus Yay!

  2. Starter activity • How do we know what the gradient of a line is? • Y = mx + c • If we are given two points how do we find the gradient?

  3. Gradient/ Slope • The gradient or slope of a line describes the steepness. To find the gradient we use the phrase rise run • Remember you rise up in the morning before you go for a run • Click here for a summary of everything

  4. Gradient/ Slope • To find the gradient/slope between two points you have to find the rise (difference between the y’s) over the run (difference between the x’s) • Look at this applet MathsNet: A Level Pure C1 Module Click here for examples of finding the gradient between two points

  5. Formula for the gradient/slope • Can you think of a formula to find the gradient between two points (x1,y1) and (x2,y2)? • M = y2-y1 • x2-x1 • This is the differences of the y’s over the differences of the x’s

  6. Example 1 • Find the gradient between the following points: • 1) (4,2) (6,3) • 2) (-1,3) (5,4) • 3) (-4,5) (1,2) • 4) (2,-3), (5,6) • 5) (-3,4) (7,-6)

  7. More practice • Find the gradient between these points • 1) (-2,-4) (10,2) • 2) (-12,3) ( -2,8) • 3) (-2,7) (4,5) • 4) (2,3) (5,7) • 5) (3b,-2b) (7b,2b)

  8. Example 2 • Work out the gradients of the following equations: • 1) y = -2x+5 • 2) y = -x+7 • 3) –x+2y -4 = 0 • 4) -3x+6y+7 = 0 • 5) 9x+6y+2 = 0

  9. Looking at the gradient on a curve • Let's think the slope of the secant line that goes through two points (a,f(a)) and (a+h,f(a+h)) on y = f(x). • The slope is (f(a+h)-f(a))/h. • When we make h smaller, what does (f(a+h)-f(a))/h approach ?We call this taking a limit when we investigate h becoming smaller.

  10. Gradients of y = x2 • Complete this table by looking at this applet

  11. Gradients of y= x3 • Complete this table by using the applet

  12. Gradients of y= x4 • Complete this table by using the applet

  13. So far • y= x2 has a gradient function of • Y = x3 the gradient function is • Y = x4 the gradient function is

  14. The magic phrase Put the power in front and then drop the power by one!!!!!! • We call the gradient function the derivative of the function i.e. dy DON’T WRITE IT LIKE • dx THIS dy/dx

  15. The Derivative • So if y = x n • Then dy dx = n x n-1

  16. F(x) or y • This is also called the derivative noun • Another way to express dy dx the gradient function is to say f(x) or y  • This is called f dash x or y dash • The verb is differentiate • These are all names for the GRADIENT FUNCTION

  17. The derivative of a number • What is the gradient of the line y = 10? • What is the gradient of the line y = 5? • What does this line look like? • So what is the derivative of y = x5 + 10?

  18. Derivative • So if y = x n where n is a number • Then dy dx = n x n-1 • Example 1 • Find derivative of • A) y = 4x2 • dy • dx = 8x • b) y= 5x3 • dy dx= 15x2

  19. More examples • Find the derivative of the following:

  20. Answers • Find the gradient function (dy when y is: • dx) • A) 2x2 – 6x +3 • B) ½ x2 +12x • C) 4x2 – 6 • D) 8x2+7x+12 • E) 5+4x-5x2

  21. Some rules to follow • When y = kxn and k is a constant • Then dy • dx = nkx n-1 • Example 1 • If y= 4x3 then dy dx = 12x2 • When y = u + v • Then dy du dv • dx = dx + dx • Example 2 • If y = 3x3 – 5x2 then dy/dx = 9x2 – 10x • Ditto for subtraction

  22. Some rules to follow • When y = u + v • Then dy = du + dv • dx dx dx

  23. Example 2 • If y = 3x3 – 5x4 then • dy • dx = 9x2 –20x3 • Ditto for addition

  24. Example 3 • (i) What about the derivative of y = 1 x3 • (ii) Find the derivative of Y= 20 – 1/3x4+10x • dy = - 4/3 x3 +10 • dx • Rules: 1) derivative of a number (constant) is zero • 2) derivative of 10x is 10 why?

  25. Example 4 • Find the derivative of Y= 3x - 1+1/5 x2 +x4 • The derivative is3 + 2/5 x + 4x3

  26. Example 5 • Find the gradient of the curve with equation y = (2x-1)(3x+2) at the point x= 3 • Here we multiply the brackets out first by using the moon! • Y= 6x2+x-2 • dy dx = 12x+1 • Now the gradient at x=3 is 12(3)+1= 37

  27. Starter • Quick questions • Write down the magic phrase to find the derivative of a function. • Write down at least three different types of notation used for the derivative of a function. • What is another name for derivative?

  28. Quick questions- find the derivative

  29. In the next few lessons we will… • A) Find out how to find the second derivative. • B) Use first derivative as a rate of change for real life situations • C) Review: Finding gradients at particular points on a curve • D) Find the equation of a tangent and normal to a curve

  30. Applications of the first derivative • Some movies but got to get correct format

  31. An online quiz • Rates of change • Distance, velocity and acceleration

  32. Second derivative • The second derivative means just differentiate dy dx again! • The second derivative looks like this: • d2y dx2 or y ‘’ or f ‘’(x) • Download livemath plugin now to view inter active pages

  33. An example of second derivative • Find the second derivative:

  34. Quiz • A MC quiz on differentiating polynomials

  35. Gradient as a rate of change • Remember that the first derivative is the gradient function or the rate of change of one variable y with respect to x. • What if we were given an equation of displacement s with respect to time? • For example s = t2+3t • If I differentiate this I get a rate of change of distance with respect to time- this is called • Velocity! So v = ds dt

  36. The derivative as a rate of change • The derivative tells us the gradient function of any curve. If we have a distance time graph what does the gradient tell us? • The gradient tells us the rate of change of the distance with respect to time which is also called speed! • Click here for the surfer • Click here for AS guru Game

  37. Example 1 • The displacement of a body at a time t seconds is given in metres by s = t2+3t • Find a) the velocity of the body at time t • b) the initial velocity of the body • Solution • A)The velocity = ds dt = 2t+3 • B) The initial velocity is when t=0 • so v = 2(0)+3 = 3 m/s • So the initial velocity is 3 m/s

  38. The rate of change- gradient • Example The volume of an expanding spherical balloon is related to its radius, r cm by the formula V = 4/3 r3. Find the rate of change of volume with respect to the radius at the instant when the radius is 5 cm. • Ex 7G

  39. Example 1 • Find the gradient of the curve f(x) = x(x+1) at the point P(0,0).

  40. Example 2 • Find the gradient of the curve y = 3x2 +x –3 at the point (1,1)

  41. Example 3 • Find the gradient of the curve y = ½ x2 + 3/2 x at the point (1,2)

  42. Drag and drop • Put these tangents onto the curve

  43. Equations of tangents and normals • Some definitions first:

  44. Finding the equation of a Tangent • What does the curve y = (x-3)(x+2) look like? • At the point x = 2 we can draw a tangent to the curve. • How do we find the equation of this line? • What information do we know already?

  45. Tangents • If we know the gradient, m at any point on a curve and we have the coordinates of that point (x ,y) we can easily find the equation of the tangent line at that point. • What is the equation of a straight line?

  46. The steps • Example 1 • Find the equation of the tangent of the curve y = (x-3)(x+2) at the point x = 1 y = - 6

  47. The steps • Complete these sentences: • 1) Find the gradient of the curve at the point x = 1 by……. • 2) The equation of a straight line is…… • 3) Now substituting x = 1 and y = -6 and m = ……… to obtain the equation of the tangent.

  48. An example • Find the equation of the tangent on the curve Y= (x+3)(x-1) at the point (1,5). • How do I find the equation of a line given a gradient and point? • Y = mx + c • How do I find the gradient? • Y = x2+3x-x-3 • Y= x2+2x-3 dy dx = 2x+2 • Gradient at x = 1 dy • dx = 2(1) + 2= 4 • Y-5 = 4(x-1) • Y = 4x+1 is the equation of the tangent at the point x = 1

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