Applied Derivatives

Introduction

A common applied derivative example is distance, speed and acceleration. Speed is the derivative of distance; meaning the rate at which is distance is accumulated is speed. The derivative of speed is acceleration; meaning the rate at which speed is accumulated is acceleration.

When talking about moving objects such as projectiles shot in the air, these will often be min/max problems. The maximum or minimum will occur when the object changes directions. This happens at the instant velocity is equal to zero, velocity being the derivative of position.

For example:

1.1   A model rocket blasts off from the ground with an initial velocity of 160 ft/s. The formula for height off the ground (position) is f(t) = 160t – 16t²
a)   Find the height and velocity of the rocket at 1 second.
       Height = f(t)
       = 160t – 16t²
       f(1) = 144 ft.

       Velocity = f`(t)
       = 160 – 32t
       f`(1) = 128 ft/s.

b)   Find the time of the maximum height and the maximum height reached by the rocket.
       f`(t) = 0 at the maximum height
       0 = 160 – 32t
       32t = 160
       t = 5
       5 seconds

       f(t) = 160t – 16t²
       f(5) = 160(5) – 16(5)²
       f(5) = 800 – 400
       f(5) = 400
       400 ft

c)   Find the time the rocket hits the ground and find its velocity at impact.
       f(t) = 160t – 16t²
       0 = 160t – 16t²
       16t² = 160t
       t = 10
       10 seconds

       f`(t) = 160 – 32t
       f`(10) = 160 – 32(10)
       f`(10) = -160
       -160 ft/s

       Note this is double the five seconds it took to reach maximum height. It took five seconds to go up and five to come down. Also the initial velocity was 160 ft/s at launch and at impact it was -160 ft/s. This is always the case when objects are launched from the ground or height 0. If the object is launched from any height this symmetry will not hold true.

1.2   Given,
a)   Find