2.
Derivatives as it Relates to Velocity, Position, and Acceleration
Velocity = the rate of change of displacement
Displacement is change in position (x(t)=position)
Instantaneous Velocity= Derivative of Position(x’(t)= position)
x(t)= position
x ‘(t)= v(t)= velocity
x “(t)= v (t) = a (t)= acceleration
Velocity is speed with direction
3.
Derivatives as it Relates to Velocity, Position, and Acceleration (cont’d)
Example 1
The position x of a particle at time t is given by the function x ( t ) = kt ^2 where k = −5 m s −2. Find (a) the velocity as a function of time; (b) the velocity at time t = 3 s.
(a) The velocity vx is the derivative with respect to time of the position function x ( t ) which is of the form At^n , with A = k and n = 2.
(b) Remembering that k is given as −5 m s−2, the velocity at time t = 3 s is now easily obtained from the equation above, as follows
vx (3 s) = 2 kt = 2 × (−5 m s−2) × (3 s) = −30 m s−1
Note that we have multiplied m s^−2 by s to give m s^−1
A spot light is on the ground 20 ft away from a wall and a 6 ft tall person is walking towards the wall at a rate of 2.5 ft/sec. How fast the height of the shadow changing when the person is 8 feet from the wall? Is the shadow increasing or decreasing in height at this time?
In this case we want to determine y’ when the person is 8 ft from wall or x=12 ft. Also, if the person is moving towards the wall at 2.5 ft/sec then the person must be moving away from the spotlight at 2.5 ft/sec and so we also know that x’=2.5.
In this case the equation we’re going to work with is, y/6=20/x
Y=120/x
Now all that we need to do is differentiate and plug values into solve to get y’.
Two people are 50 feet apart. One of them starts walking north at a rate so that the angle shown in the diagram below is changing at a constant rate of 0.01 rad/min. At what rate is distance between the two people changing when theta=0.5 radians?
We want to find x’ and we could find x if we wanted to at the point in question using cosine since we also know the angle at that point in time. However, if we use the second formula we won’t need to know x as you’ll see. So, let’s differentiate that formula.
Air is being pumped into a spherical balloon at a rate of 5 cm3/min. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm.
We know that air is being pumped into the balloon at a rate of 5 cm3/min. This is the rate at which the volume is increasing.
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