3.10 Related Rates
This section has one of the most interesting and most involved of the topics we have covered so far; therefore, a systematic method is suggested for solving related rates problems:
1) Read the problem and take note of units and given values.
2) Draw a picture.
3) Develop a static equation relating all variables.
4) Implicitly differentiate the static equation, usually with respect to time.
5) Substitute the given and/or implied values stated in the problem and solve the equation for the requested unknown.
A television camera is positioned 4000 ft from the base of a rocket launching pad. The mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Assuming that the rocket rises vertically, and that its speed is 600 ft/s when it has risen 3000 ft, how fast is the distance from the television camera to the rocket changing at that moment?
Step 1) All units are in feet and seconds, no conversion necessary. Horizontal distance to rocket is 4000 ft. The height of the rocket = 600T. The time= 5 sec, the height = 3000 ft.
Step 3) (360000T)(T) + (16000000) = N*N
Step 4) 2(600T)*600 = 2N*(dN/dT)
Step 5) (720000T)/2N = dN/dT
3600000/2N = dN/dT
N*N= (3000*3000) + (4000*4000)
N*N = 25000000
N = 5000
36000000/10000 = dN/dT
dN/dT = 360 ft/s
These are two sites which have explanations and examples for related rate problems.
If Noah's flood was caused by rain over 40 days and nights, what was the rate at which the rain fell?
SAe = 5.096*10^8 sq km
SA(land) = 1.48*10^8 sq km
SA(water) = 3.616*10^8 sq km
Elevation of Mt. Everest = 8.85 km
Average land elevation = 0.5 km
Volume of water = (3.616*10^8)8.85 + 0.5*(1.48*10^8)
Volume of water = 3.27*10^9 cubic km
average depth rained = V/SA
average depth rained = (3.27*10^9)/(5.096*10^8)
average depth rained = 6.42 km
6.13 km/960 hrs = 6.69 m/hr or .186 cm/s