D - Rates of Change Lesson

Rates of Change

The derivative is an important tool for understanding and modeling the way things change in the world around us. Although it is natural to think of a quantity changing with respect to time, other variables can be treated in the same way. Derivatives are applied when chemists monitor reaction rates (the rate of change in the concentration of a reactant with respect to time), when geologists examine the rate at which a piece of molten rock cools by heat conduction into surrounding rocks, when physicists investigate temperature gradient (the rate of change of temperature with respect to position), and when economists scrutinize marginal demand, marginal revenue, and marginal profit, all of which are the derivatives of the demand, revenue, and profit functions. In each of these applications the rates of change are actually slopes of tangents, although their interpretations are different in each area.

Average Velocity

One of the most common applications of rate of change is velocity. Average velocity is the ratio of distance traveled (change in distance) to time elapsed (change in time) for a position function s = f(t), and is denoted by $v_{avg}=\frac{\Delta s}{\Delta t}$. Average velocity can be thought of as the slope of the line joining the points on the graph of s(t) corresponding to the endpoints of the interval. View the presentation on average rate of change.

Instantaneous Velocity and Speed

To find the velocity of a moving body at an exact instant t, take the limit of the average velocity over the interval from t to t + ∆t as ∆t approaches zero. The instantaneous velocity is the derivative of the position (distance) function s = f(t) with respect to time, and is denoted by

$v\left(t\right)=\frac{ds}{dt}=\lim_{t \to 0}\frac{f\left(t+\Delta t\right)-f\left(t\right)}{\Delta t}$

Velocity can be either positive or negative. Velocity is often confused with speed. The absolute value of velocity is speed (speed = |v|), which means speed is always nonnegative. Think about the speedometer in a car. It always denotes a zero or positive reading no matter whether the car is at rest or moving forward or backward.

View the presentation below on using the Power Rule to find velocity.

Acceleration

Acceleration is the derivative of velocity with respect to time. If a body's position at time t is s = f(t), then the body's acceleration at time t is

$a\left(t\right)=\frac{dv}{dt}=\frac{d^2s}{dt^2}$

The rate at which a body's velocity changes is its acceleration. Acceleration measures how quickly a body picks up or loses speed. View the presentation on finding acceleration using the Quotient Rule.

View the presentation below to see how higher order derivatives are applied to finding average velocity, instantaneous velocity, and acceleration.

A sudden change in acceleration is called a jerk. When you experience a jerky ride in a car, it is not that the accelerations are necessarily large but that the changes in acceleration are abrupt. Jerk is the derivative of acceleration with respect to time and is denoted by $j\left(t\right)=\frac{da}{dt}=\frac{d^3s}{dt^3}$.

The figure below represents the graphs of four functions. One is the position of a car, one is the velocity of the car, one is the car's acceleration, and one is the jerk. Note that whenever s = s(t) has a horizontal tangent, the velocity function v(t) = s'(t) = 0. Whenever v(t) has a horizontal tangent, the acceleration function a(t) = v'(t) = s''(t) = 0. Likewise, when a(t) has a horizontal tangent, the jerk function j(t)= a'(t) = v''(t) = s'''(t) = 0.

Gravity

Near the surface of the Earth all bodies fall with the same constant acceleration, assuming no air resistance.  Galileo's experiments with free fall led to the equation s = (-1/2) gt², where s is the distance fallen and g is the acceleration due to Earth's gravity. The value of the gravitational acceleration g is approximately 32 ft/sec² or 9.8 m/sec².

Problem One: Consider a free fall situation where an object is dropped from rest at time t = 0. To determine its velocity, speed, and acceleration in terms of metric measurements at t = 4 seconds, we use s = 4.9 t².

$s=4.9t^2\\ v\left(t\right)=s'\left(t\right)=9.8t\\ v\left(4\right)=9.8\left(4\right)=39.2m/sec\\ speed=|v(4)|=39.2m/sec\\ a(t)=v'(t)=s''(t)=9.8\frac{m}{sec^{2} }$

Note that the acceleration is a constant.

Problem Two: If a particle is projected straight upward from a height s₀ above the ground at time t = 0 and with initial upward velocity v₀ , then its height above the ground at time t is s = (-1/2) gt² + v₀t + s₀. Differentiating s produces the velocity of the particle at time

t: v = ds/dt = -gt + v₀. The derivative of the velocity is the acceleration of the particle at time t: a = dv/dt = -g. When you consider that a particle is projected upward, it follows that it is moving against rather than with gravity. Therefore, the negative value of g represents acceleration in the opposite direction of gravity.

Rates of Change Practice

A rocket is propelled upward and reaches a height of $h\left(t\right)=4.9t^2$ in t seconds. Use this information to complete the questions below.

Rates of Change: Even More Problems!

Complete problems from your textbook and/or online resources as needed to ensure your complete understanding of rates of change.

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