1. Rectilinear Motion and Tangent Lines Katie Faith Laura Woodlee

2. Let’s Review some Calculus! ARF

3. What is Rectilinear Motion? Definition: It is any motion along a straight line. Application to Calculus: –In the AP Calculus AB course, this method is used to solve problems dealing with particle motion. Three equations are found and utilized to find the position, velocity, and acceleration of the particle at different points of time.

4. Rectilinear Motion (Continued) The position of the particle is found using the position function, which is the original function. This is usually written as x(t) or p(t).It describes the particle’s location at a specific point in relation to time (t). If given velocity, this function is found by taking the integral of the velocity function and then plugging in a known point to find the constant (C).

5. Rectilinear Motion (Continued) The velocity of the particle is found by taking the derivative of the original function (position function). It can also be found by taking the integral of the acceleration function and then plugging in a known point to find the value of the constant (C). The function is written as v(t). The velocity describes the movement of the particle, either left, right, or stopped. This direction is determined using a sign line.

6. STOP!!! SIGN LINE REVIEW!!! In this application of calculus, sign lines are used to show when the velocity of the particle is moving to the right (positively), moving to the left (negatively), or changing direction (the point when it switches from positive to negative). To make a sign line, first the zeroes of the function must be found. For example: v(t)= t²-3t + 2 v(t)= (t-1)(t-2) Then the sign line is set up: + - + The zeroes are written on the (t-1) Side (t-2) 1 2 a time line with the numbers is written on the bottom This sign line shows that the function is increasing from negative infinity to 1, resting at 1, decreasing from 1 to 2, resting at 2, and increasing from 2 to infinity. Dots represent the zeroes (this is where it is at rest) Things to remember: Negative (dotted) lines are always to the left of the zero (point) unless there is a negative in front of t Positive (solid) lines are always to the right unless there is a negative in front of t V(t)

7. Rectilinear Motion (Continued) The acceleration of the particle is found by taking the second derivative of the position function, or the first derivative of the velocity function. It is written as a(t). The acceleration of the particle describes the absolute value of distance that the particle travels during a certain interval of time. The zeroes of this function describe when the particle is at rest.

8. Example Problem 1 1) 1983 AB 2: A particle moves along the x-axis so that at time t its position is given by: x(t)= t³ - 6t²+ 9t + 11 a)What is the velocity of the particle at t=0? b)During what time intervals is the particle moving to the left? c)What is the total distance traveled by the particle from t=0 to t=5?

9. Step-by-Step Solution 1 a)To solve, you must first find the formula for velocity, which is done by taking the derivative of the position formula. Using chain rule, this would be: v(t)= 3t² - 12t + 9 To find the velocity when t=0, plug in 0 for t: v(0)= 3(0)²- 12 (0) + 9 v(0)=9 b)To find the time intervals that the particle is moving to the left, you must make a sign line of the velocity function. First find the zeroes of the velocity function: v(t)= 3t² - 12t +9 v(t)= 3(t-3)(t-1) Then make the sign line: + - + 3 (t-3) (t-1) 1 3 The sign line shows the particle is moving to the left from 1 to 3. V(t)

10. Step-by-Step Solution 1 c)To find the total distance traveled, you will be looking at the position that the particle is at different times. The sign line, in the last part of the problem, showed that the particle was at rest at t=1, and changed direction at t=3. These are two important points that the position must be found at, along with the endpoints, which are t=0 and t=5. First we must find the position of the particle at these critical points using the position formula: p(t)= t³ - 6t² + 9t + 11 Then find the total distance traveled at each interval and add to find total distance: p(0)=11 p(1)=15 + p(3)=-43 + p(5)=143 698 units total 4 58 186

11. Example Problem 2 2) 1982 AB 1: A particle moves along the x-axis in such a way that its acceleration at time t for t > 0 is given by: a(t)= 3/t² When t=1, the position of the particle is 6 and the velocity is 2. a)Write an equation for the velocity, v(t), of the particle for all t>0. b)Write an equation for the position, x(t), of the particle for all t>0. c)Find the position of the particle when t=e.

12. Step-by-Step Solution 2 a)To find the equation for velocity, you must first take the integral of the acceleration equation and then plug in the known point v(1) = 2 to find the value for C: a(t)= 3 t² v(t)= -6 + C t 2= -6 + C C= 8 1 Final Velocity Equation: v(t)= -6 + 8 t

13. Step-by-Step Solution 2 b) To find the equation for position, you must take the integral of the velocity equation and then plug in the known point x(1)=6 to find the value for C: v(t) = -6 + 8 t x(t) = -6 ln t + 8t + C 6 = -6 ln 1 + 8(1) + C 6 = -6(0) + 8 + C 6 = 8 + C C = -2 c) To find the position when t =e, plug in e for t in the position formula: x(e) = -6 ln e + 8e –2 = -6 + 8e –2 = 8e – 8 or 8(e – 1) Final Position Equation: x(t) = -6lnt + 8t - 2

14. Try Me Problem 3) 1987 AB 1: A particle moves along the x-axis so that its acceleration at any time t is given by a(t)= 6t – 18. At time t=0 the velocity of the particle is v(0)=24, and at time t=1, its position is x(1)= 20. a)Write an expression for the velocity v(t) of the particle at any time t. b)For what values of t is the particle at rest? c)Write an expression for the position x(t) of that particle at any time t. d)Find the total distance traveled by the particle from t=1 to t=3.

15. Solution a)v(t) = 3t² - 18t + 24 b)t = 2, t = 4 c)x(t) = t³ - 9t² + 24t + 4 d)x(1) = 20 x(2) = 24 x(3) = 22 4 2 6 total units

16. Work (If Needed)

17. What are Tangent Lines? Definition: A line that corresponds to a function and only intersects it at one specific point. General Equation: y-y 1 =m(x-x 1 )

18. Tangent Lines (Continued) To solve tangent line problems, certain components must be given or solved for: –The original function –The derivative of that function –X-coordinate (x 1 ) –Y-coordinate(y 1 ) –The solution of the derivative by plugging in the original x- and y- coordinates Once these components are solved for, plug them into the general equation.

19. Example Problem 1 1978 AB 1: given the function f defined by f(x)= x³ - x² - 4x + 4. Write an equation of the line tangent to the graph of f at x= -1

20. Step-by-Step Solution 2 To find the equation of the line tangent, you must first find the y- coordinate and the slope: f(x)= x 3 -x 2 -4x+4 f(-1) = (-1)³ - (-1)² - 4(-1) + 4 f(-1) = -1 – 1 + 4 +4 f(-1) = 6 f’(x) = 3x² - 2x – 4 f’(-1) = 3(-1)² - 2(-1) –4 f’(-1) = 3 + 2 –4 = 1 When the elements are plugged into the general equation, the tangent line is: y – 6 = 1 ( x + 1) or y = x + 7

21. Example Problem 2 1985 AB 1: Let f be the function given by : f(x)= 2x – 5 x² - 4 Write an equation for the line tangent to the graph of f at the point (0, f(0)).

22. Step-by-Step Solution 2 To find the equation of the line tangent, you must first find the y- coordinate and the slope: f(x)= 2x – 5 x² - 4 f(0) = 2(0) –5 (0)² - 4 f(0) = 5 4 f’(x) = (x² -4)(2) – (2t – 5)(2t) (t² -4)² f’(0) = (0² - 4)(2) – (2(0) – 5)(2(0)) (0² - 4)² f’(0) = (-4)(2) 16 = -8 When the elements are plugged into the general equation, the tangent line is: y-(5/4) = -8x or y = -8x + (5/4)

23. Try Me Problem 1986 AB 1: Let f be defined by: f(x) = 7 – 15x + 9x² - x³ Write an equation of the line tangent to the graph of f at x=2.

24. Solution y- 5 = 9(x – 2) or y =9x –13 or y – 9x = -13

25. Work (If Needed)

26. What are Normal Lines? Normal lines are similar to tangent lines except their slope is perpendicular to the tangent lines’ slope. The general equation is the same: y-y 1 =m(x-x 1 ) except the slope is the opposite reciprocal

27. Normal Lines (Continued) To solve normal line problems, certain components must be given or solved for: –The original function –The derivative of that function –X-coordinate (x 1 ) –Y-coordinate(y 1 ) –The solution of the derivative by plugging in the original x- and y- coordinates, then finding the opposite reciprocal of the solution

28. STOP!!! WHAT DOES PERPENDICULAR MEAN?!?!? A perpendicular line has the opposite reciprocal slope of the line it is perpendicular to. This means it crosses the line at a 90º angle. Opposite reciprocal means that the slope is “flipped” and the sign is changed. For example: 1) m= -2 m=½ 2) m= 7 m= -12 12 7

29. Example Problem Find the equation of the normal line to the curve: f(x)=x 2 -2x+1 at the point x=3.

30. Step-by-Step Solution To find the equation of the line normal to the graph, you must first find the y- coordinate and the slope: f(x)=x 2 -2x+1 f(3)=(3) 2 -2(3)+1 f(3)=4 f’(x)=2x-2 f’(3)=2(3)-2 f’(3)=4 This is the slope of the tangent line and needs to be made perpendicular for the line to be normal to the graph. m= -¼ When the elements are plugged into the general equation, the normal line is: y-4=-¼(x-3) y-4=-¼x+(3/4) y=-¼x+ 19 / 4

31. Try Me Problem Find the equation of the normal line to the graph of: f(x)=(x 2 +3) ½ at the point (−1, 2).

32. Solution y-2=2(x+1) or 2x-y=-4

33. Work (If Needed)

34. Bibliography "Rectilinear motion." Connexions - Sharing Knowledge and Building Communities. N.p., n.d. Web. 1 Mar. 2011..."Rectilinear motion." Connexions - Sharing Knowledge and Building Communities. N.p., n.d. Web. 1 Mar. 2011... Picture. 1 Mar. 2011..Picture. 1 Mar. 2011.. Picture. 2 Mar. 2011..Picture. 2 Mar. 2011.. "Tangents and Normals." Series Math Study. N.p., n.d. Web. 6 Mar. 2011.."Tangents and Normals." Series Math Study. N.p., n.d. Web. 6 Mar. 2011.. KAYE AUTREYKAYE AUTREY © Laura Woodlee and Katie Faith, 3/7/11