Derivatives / Chain Rule / Taylor Polynomials
If I have the following function f(x,y) = x^3 + x^2y − 16y
There are 6 things I would like to know how to do:
I would like to know how to find the first-order and second-order partial derivatives of f?
If the value of f when x = 3 and y = −2 is 41 but I was to know only that the value of x lies between 2.98 and 3.02, while the value of y lies between −2.03 and −1.97, how would I use the first-order partial derivatives to investigate the accuracy of 41 as an estimate of f(x,y) for values of x and y in these ranges?
How would I use the chain rule to determine, in terms of t, the rate of change of f with time along the path given at time t by x(t) = 4 sec(t), y(t) = 1 − t^2?
If I have a surface where surface z = f(x,y) where f(x,y) is defined above how would I determine the gradient function grad f(x,y) and hence find the slope of the surface at the point (3, −2,41) in the direction of the unit vector j?
How would I determine the second-order Taylor polynomial for f about (−1,2)?
And finally how would I locate and classify all the stationary points of f?
Solution Preview
If I have the following function f(x,y) = x3 + x2y − 16y
There are 6 things I would like to know how to do:
I would like to know how to find the first-order and second-order partial derivatives of f?
If the value of f when x = 3 and y = −2 is 41 but I was to know only that the value of x lies between 2.98 and 3.02, while the value of y lies between −2.03 and −1.97, how would I use the first-order partial derivatives to investigate the accuracy of 41 as an estimate of f(x,y) for values of x and y in these ranges?
Solution. As f(x,y) = x3 + x2y − 16y, we have
and
So, and
Note that dx=0.02 and dy=0.03. Hence,
So, when the value of x lies between 2.98 and 3.02, ...
Solution Summary
The following posting helps with problems involving second-order Taylor polynomials.