# Buy Engineering Applied Exercise Assignment

Many engineering applications of differential calculus involve second derivatives. For example, Newton’s second law
can be written in the form
where “r” represents position and “a” represents acceleration.
But what if the position cannot be expressed as a polynomial, exponential, logarithmic, or trigonometric function? For example, a booster rocket is subject to many unplanned in-flight accelerations, such as high altitude winds and asymmetric thrust; one way to reconstruct those accelerations would be to periodically sample the booster’s position using GPS, then apply numerical analysis.
where is a nodal point on a one-dimensional mesh, and h is the spacing between nodes:
If we estimate using the forward difference formula and using the backward difference formula, we have
which reduces to the central difference formula for the second derivative
Again, this is only a numerical estimate; we are not taking the limit as h goes to zero, so there will be an error in the estimate. We would expect that, as the mesh size h becomes smaller, the error in this estimate would also become smaller; but does the error drop quickly, or only very slowly?
With this background, here is your assignment:
· Assume the function f is defined as
· Use the corresponding derivative rules to find the second derivative and evaluate that second derivative at x = 1.7. Be sure your calculator is in radians mode. Note: To avoid round-off error, retain at least eight decimal places in your calculations.
· Use the “central difference formula for the second derivative” to estimate at x = 1.7 for four different values of the mesh sizes
· h = 0.2
· h = 0.1
· h = 0.05
· h = 0.025
Note: To avoid round-off error, retain at least eight decimal places in your functional evaluations, and retain the maximum possible number of decimal places in calculations of the central difference approximation.
· Use your calculated values to fill in this table:
Calculate second derivatives:
central difference approximation
exact second derivative
error = approximation – exact value
0.2
0.1
0.05
0.025
Note: To avoid round-off error, retain the maximum possible number of decimal places in calculations of the error.
· Answer the following two questions:
· Does the error in the central difference approximation drop as the mesh size h is reduced?
· As the mesh size h is reduced by a factor of 2, by what factor is the error in the central difference approximation reduced?
Be sure to show all of your work in making these calculations.
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