3.4 Velocity, Speed, and Rates of Change

Consider a graph of displacement (distance traveled) vs. time. time (hours) distance (miles) Average velocity can be found by taking: A B The speedometer in your car does not measure average velocity, but instantaneous velocity. (The velocity at one moment in time.)

Velocity is the first derivative of position.

Example: Free Fall Equation Gravitational Constants: Speed is the absolute value of velocity.

A fly ball is hit vertically upward (a pop-fly). It’s position is given by the equation s(t) = -5t 2 +30t, where the origin is at ground level and the positive direction is up. a) Find the max height of the baseball. Grade 10: Complete the Square s(t)= -5(t 2 – 6t) = -5(t 2 – 6t +9 – 9) = -5[(t – 3) 2 – 9] = -5(t – 3) Calculus: Use derivatives Find where v(t)= s’(t) = 0 v(t)= -10t + 30 Set v(t) = 0 and solve -10t + 30 = 0 t = 3 Sub t = 3 into s(t) to find height: s(3) = -5(3) (3) = 45 The max height of the ball is 45m

b) When will the ball hit the ground? Set s(t) = 0 and solve for t -5t 2 +30t = 0 -5t(t - 6) = 0 t = 0 or t = 6 c) What is the velocity when it hits the ground? v(6) = -10(6) + 30 = -30 It was travelling 30m/s when it hit the ground. It took 6 seconds to return to the ground.

The Cheetah Problem The position function of a cheetah moving across level ground in a straight line chasing after prey is given by the equation s(t) = t 3 -15t 2 +63t, where t is in seconds and s(t) is in metres. a)What is the cheetah’s velocity at 1s?4s?8s? b)When is it momentarily stopped? Did it turn around? c)When is its motion positive? negative? d)Find the position of the cheetah after 10s? e)Find the total distance travelled after 10s.

The position function of an object moving in a straight line is s(t) = 3t 2 – 0.25t 4. At time t = 3, is the object moving towards the origin or away from the origin? Solution: The velocity of the object is: v(t) = s’(t) = 6t – t 3 At time t = 3, s(3) = and v(3) = - 9 In the diagram, let the positive direction be to the right. Thus, s(3) >0 implies the object is to the right of the origin and v(3) < 0 implies the object is moving to the left (in this case, to the origin) when t = 3. 0 s Direction of motion at t = 3

Acceleration is the derivative of velocity. example: If distance is in: Velocity would be in: Acceleration would be in: metre

Starting at time t = 0, a dragster accelerates down a strip and then brakes and comes to a stop. Its position function s(t) is s(t) = 6t 2 – 0.2t 3. a)After how many seconds does the dragster stop? b)What distance does the dragster travel? c)At what time does the braking commence? Solutions: a)v(t) = 12t – 0.6t 2. To stop v(t) = 0 12t – 0.6t 2 =0 so that t = 0 or t = 20. It stops after 20s. b)It travels from s(0) = 0 to s(20)= 800. Since it did not reverse direction, the distance travelled was 800m. c)a(t) = v’(t) = 12 – 1.2t = 1.2(10 – t) Note: when t 0 and that if t>10 then a(t) <0. Thus the dragster accelerates from 0s to 10s and brakes from 10s to 20s. The braking begins after 10s.

A ball thrown in the air slows as it rises and speeds up as it falls Negative acceleration means that the ball slows down as it rises. It indicates that the velocity is decreasing. Positive acceleration means that velocity is increasing. An object is accelerating if a(t) x v(t) > 0 decelerating if a(t) x v(t) < 0

time distance acc pos vel pos & increasing acc zero vel pos & constant acc neg vel pos & decreasing velocity zero acc neg vel neg & decreasing acc zero vel neg & constant acc pos vel neg & increasing acc zero, velocity zero It is important to understand the relationship between a position graph, velocity and acceleration:

Rates of Change: Average rate of change = Instantaneous rate of change = These definitions are true for any function. ( x does not have to represent time. )

Example 1: For a circle: Instantaneous rate of change of the area with respect to the radius. For tree ring growth, if the change in area is constant then dr must get smaller as r gets larger.

from Economics: Marginal cost is the first derivative of the cost function, and represents an approximation of the cost of producing one more unit. Marginal revenue is the first derivative of the revenue function, sometimes given as R(x) = xp(x) where p(x) is the price per unit (p(x) is sometimes called the price function or demand function.) and x is the number of items. The instantaneous rate of change of cost with respect to the number of items produced is called the marginal cost.

Example: Suppose you are running a company that assembles stoves. You have been keeping Track of your expenses, but it is a small company and you only have partial figures. marginal cost # Stoves Cost If you are currently producing 10 stoves, how much will the 11 th stove cost? If we use the slope of the secant around 10 we can use it to estimate the cost of making the 11 th stove. Slope==$195

Marginal cost is a linear approximation of a curved function. For large values it gives a good approximation of the cost of producing the next item. This approximation is based on the idea of local linearization. It is the process of using a tangent or a secant to approximate a value of a function in close proximity to a known point marginal cost

Example: Suppose it costs: to produce x stoves. If you are currently producing 10 stoves, the 11 th stove will have a marginal cost of approximately: marginal cost The actual cost is: actual cost Note that this is not a great approximation – Don’t let that bother you.

Guidelines for Successful Problem Solving with Related Rates 1.Make a sketch and label known quantities 2.Introduce variables to rep. quantities that change (let x rep…) 3.Identify what you are to find 4.Find an equation (area, volume, pythag…) 5.Find the derivative 6.Substitute and solve 7.Conclude in the context of the problem 