3.11 Related Rates Flashcards
What are related rates?
Related rates are finding the rate in what property changes in relation to the other property’s quantity changes in respective to time.
What are the steps when solving a related rates problem?
First you would identify the variables that are given in the problem
Then assign variables to these quantities
After which you would write an equation that involve these variables or what the question asks.
An example would be like finding the radius of a circle πr^2 etc.
Now differentiate the equation in respective to time.
Then substitute known values and solve for the unknown rate
Solve this Example:
Epidemiologists study factors and track movements of diseases in the human populations. One epidemiologist is studying a virus outbreak in New York City and notices the outbreak area is decreasing in a circular patch centered around one particular neighborhood at a rate of 250pi km^2/day. How fast is the radius of the virus outbreak decreasing when the radius is 25km?
First I would understand what formula and variables would be would be used first.
So I’d consider..
a = πr^2
r = 25 km
dA/dt = 250π km^2/day
Then following this format and plugging in the variables/function
dA/dt=2π r^2 dr/dt
250π = 2π (25) dr/dt
250π = 50π dr/dt
Divide by 50 pi
250π / 50π = 50π /50π dr/dt
The result is dr/dt= 250pi/50pi = 5 km/day
Solve this Example:
A balloon is being filled with air at a rate of 60 ft^3/sec. How fast is the radius of the balloon changing when the radius is 1 ft (assuming the balloon is a perfect sphere)?
First I would identify the given variables/functions as well as the formula needed to find the answer
V= 4/3πr^3
r = 1 ft
dA/dt = 60ft^3/sec
Then following the format
dV/dt=4/3πr^3 dr/dt
But first I need to break down 4/3 pi r^3 by applying the chain rule
4/3πr^3 dr/dt
4πr^2 dr/dt
Now going back to the problem then plug in the new terms
dV/dt=4πr^2 dr/dt
Plug in
60=4π(1) dr/dt
60/4π=4π/4π dr/dt
dr/dt=15/4π
dr/dt=4.77 ft/sec
The result is 4.77 ft/sec
