Derivatives / Chain Rule / Taylor Polynomials
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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?
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Solution Summary
The following posting helps with problems involving second-order Taylor polynomials.
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, ...
Education
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
Recent Feedback
- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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