Which operations are the most efficient to use to solve this problem? Mario hauled 64 containers of oil that each held 73 gallon
1 answer:
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Answer:
h=0.4--> 15
h=0.2 --> 14.06
h=0.1 --> 13.71
Step-by-step explanation:
This is a numerical solution using Euler's method. Euler's method enables us to numerically approch a solution with a suitable step size. As the step size gets smaller, the approximation will be more accurate. Euler's method is as the following.
Here, h is the step size. The reason why first derivation used here is to make an appriximation with using the rate of increase or decrease. As the step size is smaller, the icreases or decreases are followed more accurately. Now, let's solve the question:
for h= 0.4:
y(0.4)=y(0)+0.4*y'
The trick here is y' is equal to y, thus we can write that y'=y(0.4) and our starting point y(0) is given as 9 in the question and the equation becomes:
y(0.4)=9+0.4*y(0.4) and this is easy to solve.By replacing y(0.4) functions to be at the same side of the equation, we get:
0.6*y(0.4)=9 and by solving this equation, y(0.4) is found to be 15.
for h=0.2:
This will be similar to the previous question, but since the step size is 0.2, we will first calculate y(0.2) and then y(0.4).
y(0.2)=y(0)+0.2*y(0.2) and y(0) is 9.
0.8*y(0.2)=9 and y(0.2)=11.25. Now, we will replace this value into the next iteration o the formula istead of y(0). The equation is like:
y(0.4)=y(0.2)+0.2*y(0.4) and y(0.4) is found to be 14.06.
for h=0.1:
This is also similar to the above solutions but will be longer and have 4 iterations.
first iteration: y(0.1)=9+0.1*y(0.1) --> y(0.1)=10
second iteration: y(0.2)=y(0.1)+0.1*y(0.2) --> y(0.2)=11.11
third iteration: y(0.3)=y(0.2)+0.1*y(0.3) --> y(0.3)=12.34
fourth iteration: y(0.4)=y(0.3)+0.1*y(0.4) --> y(0.4)=13.71
As the step size gets smaller, the answer also gets smaller and more accurate. With even smaller step sizes, there will be a better approximation. However, in case you have more complex equations or smaller step sizes, it is recommended to use a computer software to make an approximation.