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with upper and lower limits it mean f(x) continuous on closed interval [a, b] and F(x) is the antiderivative (indefinite integral) of f(x) on [a,b], then
[ ]∫ −=b
a
b
aaFbFxFdxxf )()()()( (Fundamental Theorems of Calculus)
4.5 Applications of Differentiation and Integration
4.5.1 Displacement, Velocity, and Acceleration
Given an equation for the displacement of a moving object, find an equation for its velocity and an equation for its acceleration, and use the equations to analyze the motion.
Example 1
Suppose a football is punted into the air. As it rises and falls, its displacement (directed distance) from the ground is a function of the number of seconds since it was punted.
y = -16t2 + 37t + 37 where y is the football’s displacement in feet and t is the number of seconds since it was punted. The velocity of the ball gives its speed and the direction in which it’s going. Because velocity is the instantaneous rate of change, it is a derivative.
velocity = dy/dt = y’ = -32t + 37 Find velocity at t = 1 at t = 2 The dy/dt symbol reminds you of the units for velocity (ft/sec). Speed is the absolute value of velocity. Speed tells how fast an object is going without regard to its direction. Describe the speed and velocity at t = 1 and t = 2 sec. Note that the velocity changes from t = 1 to t = 2 sec. The instantaneous rate of change in velocity is called acceleration. Using v for velocity, v = -32t + 37. Find acceleration. The dv/dt symbol for the derivative gives the units of acceleration. dv/dt is in (feet/second)/sec and written “ft/sec2” Interpret the idea of negative acceleration.
PROPERTIES: Velocity, Speed, and Acceleration If x is the displacement of a moving object from a fixed plane (such as the ground), and t is time, then Velocity: v = x’ = dx/dt
Note that the acceleration is constant, -32 ft/sec2, for an object acted on only by gravity. To tell quickly whether an object is speeding up or slowing down, compare the signs of the velocity and acceleration.
Example 2
The position of a ball thrown into the air is observed to be described by the equation
216t12t20y(t) −+=
where y is in feet. We wish to determine the equation of its velocity, and its acceleration.
Solution:
The word indefinite is used because an antiderivative always has an unspecified constant, C, added. This constant is called the constant of integration. The velocity of a moving object is the antiderivative of the acceleration, and the displacement of the object is the antiderivative of the velocity.
TECHNIQUE: Speeding Up or Slowing Down • If velocity and acceleration have the same sign, the object is speeding up. • If velocity and acceleration have different signs, the object is slowing
down.
DEFINITION: Antiderivative, or Indefinite Integral
Function g is an antiderivative (or indefinite integral) of function f iff g’(x) = f(x).
The term rate implies a change in a quantity with respect to another quantity, often time. A car’s rate, for example, is its speed and is the distance traveled in a unit of time. If two rates are related, the rate of change of one quantity is tied to the rate of change of another quantity. The Chain Rule offers ways to treat such problems.
Example 3
Suppose air is blown into a spherical balloon at a rate of 36 in3/sec. How fast is the radius of the balloon increasing at the instant that the radius is 3 inches?
Solution:
A procedure for handling a related rates problem like the one above is as follows:
1 Draw a picture of the situation.
2 Introduce variables for quantities that are
changing. Let V be volume and r the radius of the balloon.
3 Write down explicitly what rates are involved. We know that
dt
dV = 36 in3/sec.
We want dt
dr when r = 3 in.
Note: decreasing quantity means a negative derivative.
4 Find an equation relating the quantities involved. V = (
3
4) π r3.
5 Differentiate both sides with respect to t (like implicit differentiation), considering the changing quantities as functions of t.
V = (3
4) π r3, so
dt
dV = (
3
4) π (3 r2
dt
dr)
= 4 π r2 dt
dr
Warning: don't forget the chain rule!
6 Substitute the given rates and other constants. 36 = 4 π 32
1. The position of an object is observed to be described by the equation,
10358)( 23 −+−= tttts
where s is in meters. We wish to determine the equation of its velocity, and its acceleration.
2. The position of an object is observed to be described by the equation
2)43( −= ts
where s is in meters. We wish to determine the equation of its velocity, and its acceleration.
3. At time t, the volume V 3cm of water leaking tank is V, where 23100 ttV −−= .
Find the rate of water flow from the tank at t = 6 seconds.
4. The radius, r , of a spherical balloon at time t is given by ttr += 2 . Express the volume if the balloon 3Vcm in terms of t and find the rate of change of the volume at 4=t seconds.
5. The radius r cm of a circle increases at a constant rate of 0.5 1−cms . If the
initial radius is 3.5cm, find the radius of the circle after 10 seconds.
6. The radius of the circle is increasing at the rate of 5cm per minute. Find (a) the rate of change of the area of the circle when its radius is 12 cm. (b) the radius of the circle when its area is increasing at a rate of 50π 12 −scm . 7. The volume of a constant height cone is decreasing at a rate of 4 13 −scm . Find
the rate of change in its cross sectional radius when the radius is 5 cm and the height is 8 cm.