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Rates of Change

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Rate of Change
A lot of real world relationships can be described by functions of the form y=f(x)y=f\left(x\right)y=f(x), where
xxx is the independent variable
y or fy\ or\ fy or f is the dependent variable that changes with the independent variable
How quickly the dependent variable changes when there is a change in the independent variable is called the      rate of change     

Average Rate of Change
The average rate of change between x=ax=ax=a and x=bx=bx=b is the slope of the secant line of the curve between the points x=ax=ax=a and x=bx=bx=b

Average Rate of Change=f(b)−f(a)b−a or ΔyΔx=y2−y1x2−x1\displaystyle \boxed{\text{Average Rate of Change}=\frac{f\left(b\right)-f\left(a\right)}{b-a}}\ \text{or}\ \frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}Average Rate of Change=b−af(b)−f(a)​​ or ΔxΔy​=x2​−x1​y2​−y1​​

Instantaneous Rate of Change
The instantaneous rate of change at a point x=ax=ax=a is the slope of the tangent line of the curve at the point x=ax=ax=a
Instantaneous Rate of Change=lim⁡h→0 f(a+h)−f(a)h\displaystyle \boxed{\text{Instantaneous Rate of Change}=\lim_{h\to0}\ \frac{f\left(a+h\right)-f\left(a\right)}{h}}Instantaneous Rate of Change=h→0lim​ hf(a+h)−f(a)​​

Common Rates of Change Questions
Velocity: the speed with a direction → this is the rate of change of the position relative to the initial position
Cost, revenue, & profit: Profit=Revenue−Cost\text{Profit}=\text{Revenue}-\text{Cost}Profit=Revenue−Cost
Volume & Surface Area: The rate of change of volume is surface area

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Example: Velocity
A rock is dropped from a cliff with a height of 45 m. After ttt seconds, the rock is sss meters above the ground, where s(t)=45−5t2s(t)=45-5t^2s(t)=45−5t2 where 0≤t≤30\le t\le30≤t≤3

a) Find the average velocity of the rock between the times t=1t=1t=1 and t=2t=2t=2.
b) Find the instantaneous velocity of the rock at the time t=1t=1t=1.

a) 
Since velocity is the rate of change of position, the average velocity is the average rate of change of position.
Avg Velocity=s(2)−s(1)2−1\displaystyle\text{Avg Velocity}=\frac{s\left(2\right)-s\left(1\right)}{2-1}Avg Velocity=2−1s(2)−s(1)​
Avg Velocity=(45−5(2)2)−(45−5(1)2)2−1\displaystyle\text{Avg Velocity}=\frac{\left(45-5\left(2\right)^2\right)-\left(45-5\left(1\right)^2\right)}{2-1}Avg Velocity=2−1(45−5(2)2)−(45−5(1)2)​
Avg Velocity=(45−20)−(40)1\displaystyle\text{Avg Velocity}=\frac{\left(45-20\right)-\left(40\right)}{1}Avg Velocity=1(45−20)−(40)​
Avg Velocity=−15\text{Avg Velocity}=-15Avg Velocity=−15
Therefore, the average velocity between t=1t=1t=1 and t=2t=2
t=2 is −15-15−15 m/s

b)
The instantaneous velocity of the rock is the instantaneous rate of change of the position.
Instantaneous Rate of Change=lim⁡h→0 s(1+h)−s(1)h\displaystyle \text{Instantaneous Rate of Change}=\lim_{h\to0}\ \frac{s\left(1+h\right)-s\left(1\right)}{h}Instantaneous Rate of Change=h→0lim​ hs(1+h)−s(1)​
Instantaneous Rate of Change=lim⁡h→0 [45−5(1+h)2]−[45−5(1)]h\displaystyle \text{Instantaneous Rate of Change}=\lim_{h\to0}\ \frac{\left[45-5\left(1+h\right)^2\right]-\left[45-5\left(1\right)\right]}{h}Instantaneous Rate of Change=h→0lim​ h[45−5(1+h)2]−[45−5(1)]​
Instantaneous Rate of Change=lim⁡h→0 [40−10h−5h2]−[40]h\displaystyle \text{Instantaneous Rate of Change}=\lim_{h\to0}\ \frac{\left[40-10h-5h^2\right]-\left[40\right]}{h}Instantaneous Rate of Change=h→0lim​ h[40−10h−5h2]−[40]​
Instantaneous Rate of Change=lim⁡h→0 [−10h−5h2]h\displaystyle \text{Instantaneous Rate of Change}=\lim_{h\to0}\ \frac{\left[-10h-5h^2\right]}{h}Instantaneous Rate of Change=h→0lim​ h[−10h−5h2]​
Instantaneous Rate of Change=lim⁡h→0 h(−10−5h)h\displaystyle \text{Instantaneous Rate of Change}=\lim_{h\to0}\ \frac{h\left(-10-5h\right)}{h}Instantaneous Rate of Change=h→0lim​ hh(−10−5h)​
Instantaneous Rate of Change=lim⁡h→0 (−10−5h)\displaystyle \text{Instantaneous Rate of Change}=\lim_{h\to0}\ \left(-10-5h\right)Instantaneous Rate of Change=h→0lim​ (−10−5h)
Instantaneous Rate of Change=−10\text{Instantaneous Rate of Change}=-10Instantaneous Rate of Change=−10
Therefore, the instantaneous velocity of the rock at the time t=1st=1st=1s is −10-10−10m/s

Practice: Rate of Change of Revenue
The revnue of a local theater is given by R(t)=640+120t−10t2R\left(t\right)=640+120t-10t^2R(t)=640+120t−10t2 where ttt is the number of months from January (for example, t=1t=1t=1 corresponds to the revenue in January, t=4t=4t=4 corresponds to the revenue if April, etc.)

a) Find the theather's revenue in June.
b) How fast is the revenue increasing or decreasing in June? (i.e. find the instantaneous rate of change at t=6t=6t=6

a) Enter the exact revenue in $

b) Enter how fast the revenue is changing in June

I don't know

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Example: Rate of Change of Medicine
The amount of drug measured in milligrams in a person's body ttt hours after it has been administered is M(t)=−12t2+2M(t)=-\frac{1}{2}t^2+2M(t)=−21​t2+2 where 0≤t≤20\le t\le20≤t≤2. Find the rate of change of the amount of drug in the body half an hour after it has been administered. Interpret your answer.

The rate of change after half an hour is the slope of the tangent line at the point t=12t=\frac{1}{2}t=21​:
m=lim⁡h→0 M(12+h)−M(12)h\displaystyle m=\lim_{h\to0}\ \frac{M\left(\frac{1}{2}+h\right)-M\left(\frac{1}{2}\right)}{h}m=h→0lim​ hM(21​+h)−M(21​)​
m=lim⁡h→0 [−12(12+h)2+2]−[−12(12)2+2]h\displaystyle m=\lim_{h\to0}\ \frac{\left[-\frac{1}{2}\left(\frac{1}{2}+h\right)^2+2\right]-\left[-\frac{1}{2}\left(\frac{1}{2}\right)^2+2\right]}{h}m=h→0lim​ h[−21​(21​+h)2+2]−[−21​(21​)2+2]​
m=lim⁡h→0 [−12(14+h+h2)+2]−[158]h\displaystyle m=\lim_{h\to0}\ \frac{\left[-\frac{1}{2}\left(\frac{1}{4}+h+h^2\right)+2\right]-\left[\frac{15}{8}\right]}{h}m=h→0lim​ h[−21​(41​+h+h2)+2]−[815​]​
m=lim⁡h→0 [−18−12h−12h2+2]−[158]h\displaystyle m=\lim_{h\to0}\ \frac{\left[-\frac{1}{8}-\frac{1}{2}h-\frac{1}{2}h^2+2\right]-\left[\frac{15}{8}\right]}{h}m=h→0lim​ h[−81​−21​h−21​h2+2]−[815​]​
m=lim⁡h→0 [−12h−12h2]h\displaystyle m=\lim_{h\to0}\ \frac{\left[-\frac{1}{2}h-\frac{1}{2}h^2\right]}{h}m=h→0lim​ h[−21​h−21​h2]​
m=lim⁡h→0 h(−12−12h)h\displaystyle m=\lim_{h\to0}\ \frac{h\left(-\frac{1}{2}-\frac{1}{2}h\right)}{h}m=h→0lim​ hh(−21​−21​h)​
m=lim⁡h→0 −12−12h\displaystyle m=\lim_{h\to0}\ -\frac{1}{2}-\frac{1}{2}hm=h→0lim​ −21​−21​h
m=−12\displaystyle m=-\frac{1}{2}m=−21​

Therefore, the rate of change of the amount of drug in the body half an hour after it's been administered is −12-\frac{1}{2}−21​ mg/hr.

This rate of change is negative because the amount of drug in the body is decreasing at this time.

The profit of a company measured in dollars is modelled by the function P(x)=400x−x2P(x)=400x-x^2P(x)=400x−x2, where xx
x is the amount of product sold in a week.
a) Find the profit if 10 products are sold in a week.
b) Find the rate of change of the profit if 100 products are sold per week.

a) Enter the profit in dollars

b) Enter the rate of change in dollars/product sold

I don't know

Practice: Rate of Change of Volume & Surface Area
Determine the instantaneous rate of change of the following volumes.