Meche 132 Problem Set 8 Spring 2014 UC Berkeley

Question

Meche 132 Problem Set 8version 1spring 2014 Uc Berkeley Below are 4 problems. I will post more problems over the course of the weekend, but this should be good to get started. These are due on Friday, April 11, at 5PM. 1. This purpose of this problem is to remind you how first-order Taylor series (ie., linear approximation) are used to approximate a function of many variables. The temperature in a particular 3-dimensional solid is a function of position, and is known to be T(x,y,z) = 42 + (x− 2)^2 + 3 (y− 4)^2 − 5 (z− 6)^2 + 2 y z. (a) Find the first order approximation (linearization) of the temperature near the location (x̄ = 4, ȳ = 6, z̄ = 0). Use δx, δy and δz as your deviation variables. (b) What is the maximum error between the actual temperature and the first order approximation formula for |δx| ≤ 0.3, |δy| ≤ 0.2, |δz| ≤ 0.1? Solve this numerically, by simply sampling a dense grid over the 3-dimensional cube, and determining the maximum error. (c) More generally, suppose that x̄ ∈ R, ȳ ∈ R, z̄ ∈ R. Find the first order approximation of the temperature near the location (x̄, ȳ, z̄).

Dr. Jack HW Helper · Accepted Answer

The task involves exploring the use of the first-order Taylor series approximation to estimate the temperature at nearby points in a three-dimensional space. This fundamental concept in multivariable calculus facilitates understanding how small deviations from a known point affect a function's value, which has diverse applications ranging from engineering to physical sciences. To begin, consider the temperature function: T(x, y, z) = 42 + (x− 2)^2 + 3(y − 4)^2 − 5(z − 6)^2 + 2 y z Part (a): Linearization Near a Specific Point The goal is to determine a linear approximation (or the tangent plane) of T(x, y, z) near the point (x̄ = 4, ȳ = 6, z̄ = 0). The first-order Taylor expansion is given by: T(x, y, z) ≈ T(x̄, ȳ, z̄) + ∂T/∂x (x̄, ȳ, z̄) (x − x̄) + ∂T/∂y (x̄, ȳ, z̄) (y − ȳ) + ∂T/∂z (x̄, ȳ, z̄) * (z − z̄) Calculating each partial derivative at the point (4, 6, 0) yields: ∂T/∂x = 2(x − 2), so at (4, 6, 0): 2(4 − 2) = 4 ∂T/∂y = 6(y − 4) + 2 z, so at (4, 6, 0): 6(6 − 4) + 2*0 = 12 ∂T/∂z = −10(z − 6) + 2 y, so at (4, 6, 0): −10(0 − 6) + 2*6 = 60 + 12 = 72 The temperature at the base point is: T(4, 6, 0) = 42 + (4− 2)^2 + 3(6− 4)^2 − 5(0− 6)^2 + 260 = 42 + 4 + 34 − 536 + 0 = 42 + 4 + 12 − 180 = −122 Thus, the linear approximation becomes: T_approx(x, y, z) = −122 + 4(x − 4) + 12(y − 6) + 72(z − 0) Part (b): Numerical Estimation of Maximum Error We are asked to assess the maximum deviation between the actual temperature and the linear approximation within bounded deviations: |δx| ≤ 0.3, |δy| ≤ 0.2, |δz| ≤ 0.1. This involves sampling points within the specified cube and evaluating the error: error = |T(x, y, z) − T_approx(x, y, z)| The approach includes creating a dense grid over the ranges: x in [4−0.3, 4+0.3] y in [6−0.2, 6+0.2] z in [0−0.1, 0+0.1] Calculating T(x, y, z) exactly and comparing it with T_approx, then identifying the maximum error. This numerical grid-based method provides an empirical estimate of the maximum approximation error within the local neigh

Meche 132 Problem Set 8 Spring 2014 UC Berkeley

Meche 132 Problem Set 8version 1spring 2014 Uc Berkeley

Paper For Above instruction

Part (a): Linearization Near a Specific Point

Part (b): Numerical Estimation of Maximum Error

Part (c): Generalization of Linearization

Conclusion

References