course Mth 271
The thing that I dont understand about asst 1.5 17-22 was it gives you a graph for each of the problems and it has the same equation for each y= f(x) and then it gives you a c= 3 for # 17 and a point (3,1) amd then it asks for the lim....I am not sure what I need to do here ....thanks!
If a sand pile 4 meters high has a mass of 146000 kg, then what would we expect to be the mass of a geometrically similar sand pile 14 meters high? Using the differential estimate the mass of sand required to increase the height of the pile from 4 meters to 4.03 meters.
Y= k x^3
146000= k 4^3
2281.25=k
y= 2281.25* 14^3
y=6259750 kg
y=2281.25* 4^3
=146000kg
y= 2281.25 4.03^3
=149309.7kg
149309.7- 146000= 3309.7
"
The differential estimate uses the derivative at x = 14 to estimate the change.
The derivative of y = k x^3 is y ' = 3 k x^2. This is the rate of change of y with respect to x.
When multiplied by the change in x it gives you the approximatechange in y.
What is y ' when x = 14, and how much change does it predict if x changes by .03?