a. Explanation Given: z = x + y 2 Where, x = u + v , y = u − v Calculation: To find the partial derivatives ∂ z ∂ u , ∂ z ∂ v , where z = x + y 2 Where, x = u + v , y = u − v , first lets express z in terms of u and v , as z = x + y 2 = ( u + v ) + ( u − v ) 2 = u + v + u 2 + v 2 − 2 u v Now, to find ∂ z ∂ u differentiate above with respect to u treating v as constant, as ∂ z ∂ u = ∂ ∂ u ( u + v + u 2 + v 2 − 2 u v ) = b. To determine To find the partial derivatives ∂ z ∂ u , ∂ z ∂ v , where z = x + y 2 Where, x = u + v , y = u − v

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

See similar textbooks

Concept explainers

Linear Algebra

In mathematics, problems can be solved by the arithmetic operators like addition, subtraction, multiplication, and division. In algebra, we represent the numbers by any letter called a variable. If these variables are of degree 1, then it is studied unde…

Videos

Question

Chapter 11.5, Problem 8PS

To determine

To find the partial derivatives ∂z∂u,∂z∂v , where

z=x+y2

Where, x=u+v,y=u−v

To determine

To find the partial derivatives ∂z∂u,∂z∂v , where

z=x+y2

Where, x=u+v,y=u−v

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 11.5, Problem 7PS

Chapter 11.5, Problem 9PS

Students have asked these similar questions

2. Suppose a 4D space exists for variables a, b, c, and d, where a is a function of b, c, and d, then, the correct expression for finding the partial derivative of a with respect to b is (da/ab)..

5. If the gradient of f(r, y) at (1, 2) is 2i – 2j, then the maximum and minimum values for a directional derivative of f at (1, 2) are respectively A. 2/2 and -V2 B. -2/2 and v2 C. 2/2 and 2/2 D. 2/2 and -2v2 6. Suppose the second-order partial derivatives of the function f(r, y) exists at thr critical point (0,0). Which of the following one is true for the critical point to be maximum? A. frr (0,0) = 2 fyy (0,0) = 2 and rry(0,0) = 4 C. frz (0,0) = 4 fyy(0,0) = 4 and Iry(0,0) = 4 B. frz (0,0) = 2 fyy (0,0) = 2 and rry(0,0) = 2 D. frz(0,0) = -4 fyy (0,0) = -4 and rry(0,0) = 4 %3D %3D %3D

Below is a function f. We are interested in how changes in the values of the input variables x, y or parameters a, b, c affect f. To that aim, we writef as a function of both its input variables and its parameters, using the notation f(x, y; a, b, c). The “gradient of f towards v", denoted as Vyf, is the vector of all partial derivatives of f to the variables/parameters listed in v. Fill in the boxes. • f(x, y; a, b, c) = a exp(2x) + by² + cxy • Vxyf(x, y; a, b, c) = Vabef(x, y; a, b, c) = • If ô, ɛ are very small real numbers, what is f(1,2; 3,4+ 8,5 + e) – f(1,2; 3,4,5) approximately? Explain how you obtained the answer. Answer: Explanation:

Chapter 11 Solutions

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

Ch. 11.1 - Prob. 1PS Ch. 11.1 - Prob. 2PS Ch. 11.1 - Prob. 3PS Ch. 11.1 - Prob. 4PS Ch. 11.1 - Prob. 5PS Ch. 11.1 - Prob. 6PS Ch. 11.1 - Prob. 7PS Ch. 11.1 - Prob. 8PS Ch. 11.1 - Prob. 9PS Ch. 11.1 - Prob. 10PS

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

To find the partial derivatives ∂ z ∂ u , ∂ z ∂ v , where z = x + y 2 Where, x = u + v , y = u − v

Solution Summary: The author explains how to find the partial derivatives of partial z, v, and u.

Concept explainers

Videos

To find the partial derivatives ∂z∂u,∂z∂v , where z=x+y2 Where, x=u+v,y=u−v

To find the partial derivatives ∂z∂u,∂z∂v , where z=x+y2 Where, x=u+v,y=u−v

Want to see the full answer?

Chapter 11 Solutions

To find the partial derivatives ∂z∂u,∂z∂v , where

z=x+y2

Where, x=u+v,y=u−v

To find the partial derivatives ∂z∂u,∂z∂v , where

z=x+y2

Where, x=u+v,y=u−v