IB Math SL Year 2 Name: Date: 8-1 Rate of Change and Motion

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1 Name: Date: 8-1 Rate of Change and Motion Today s Goals: How can I calculate and interpret constant rate of change? How can I calculate and interpret instantaneous rate of change? How can we use derivatives to model displacement, velocity, and acceleration for a moving object? Important notes before we begin: A derivative, we know, is the of a tangent line of a function. It also tells us the instantaneous rate of change of one variable with respect to another. In this lesson we will study instantaneous rates of change and motion. Instantaneous means the rate of change at a particular moment. Let s Apply It! 1. During one month, the temperature of the water in a pond is modeled by the function C(t) = e t + x 2-9x, where t is measured in days and C is measured in degrees Celsius. Question Notes a. Find the rate of change in temperature on day 15. *Note, question was modified for C (t). b. At what time will the temperature reach its maximum? c. Calculate the maximum temperature reached.

2 Let s try another 2. Let x be the number of thousands of units of an item produced. The revenue for selling x units is r(x) = and the cost of producing x units is c(x) = 2x 2. Question Notes a. The profit p(x) = r(x) c(x). Write an expression for p(x). b. Find and. c. Hence, find the number of units that should be produced in order to maximize profit.

3 Using Derivatives to Analyze Motion an object is moving, a ball in the air for example, we can use functions and calculus to analyze its displacement, velocity, and acceleration. Displacement- position (distance from original location) Velocity - speed in a given direction over time; change in positioning over time Acceleration - change in the rate of speed (or the change in velocity) *Key Notes Given that s(t) is a displacement function, The velocity function is the derivate of the displacement function. The acceleration function is the second derivative of the displacement function (or derivate of velocity function). 3. A diver jump from a platform at time t = 0 seconds. The distance of the diver above water level at time t is given by s(t) = -4.9t t + 10, where s in in meters. Note, velocity is the rate of change for a Find the instantaneous velocity of the diver at t =1 second. distance function. What direction is the diver moving? Explain how your calculations justify your answer.

4 Practice 4. a stone is thrown vertically upwards, its height above the ground (in meters) is given by, where time is measured in seconds. a. Find the height of the stone when t = 1.2 seconds. b. Find the rate of change in height at 4.5 seconds. c. At what time will the stone reach its maximum height? d. Calculate the maximum height reached by the stone.

5 5. If a ball is thrown vertically upwards. Its height in metres above the ground t seconds after it is thrown is modeled by the function h(t) = -4.9t t a. Find the height of the ball when t = 0 seconds and when t = 2 seconds. b. Find the instantaneous rate of change of the height of the ball when t = 1 second, t = 2 seconds, and t = 3 seconds. Explain what each of these values tells you about the motion of the ball. 6. A particle moves on a vertical line so that its coordinate at time t is y = t 3 12 t + 3, t > 0. a. Find its velocity function. b. Find its acceleration function. c. During what intervals is the particle is moving upward and when is it moving downward?

6 7. The displacement s metres at time t seconds is given by s = 5 cos 3t + t , for 0 < t < 2. (a) Write down the minimum value for t. Thus, find the minimum value of s. (b) Find the velocity, v, at time t. (c) Find the velocity at 1.5 seconds. 8. The velocity, v, in ms 1 of a particle moving in a straight line is given by v = e 3t 2, where t is the time in seconds. (a) Find the acceleration of the particle at t =1. (Note! You will need to apply chain rule when deriving!) (b) At what value of t does the particle have a velocity of 22.3 m s 1? (c) Was the particle moving right or left at 1 second? Justify your answer.