Explanation Given information: The function is f ( x , y , z ) = 2 x 2 + 3 y 2 + z 2 − 2 x − y − z . Explanation: If f ( x , y , z ) has minimum at point ( x , y , z ) = ( a , b , c ) , then ∂ f ∂ x ( a , b , c ) = 0 , ∂ f ∂ y ( a , b , c ) = 0 and ∂ f ∂ z ( a , b , c ) = 0 . Therefore, find the partial derivatives ∂ f ∂ x , ∂ f ∂ y and ∂ f ∂ z of the given function. By differentiating the given function with respect to x . ∂ f ∂ x = ∂ ∂ x ( 2 x 2 + 3 y 2 + z 2 − 2 x − y − z ) ⇒ ∂ f ∂ x = 4 x − 2 . By differentiating the given function with respect to y . ∂ f ∂ y = ∂ ∂ y ( 2 x 2 + 3 y 2 + z 2 − 2 x − y − z ) ⇒ ∂ f ∂ y = 6 y − 1

Calculus & Its Applications (14th Edition)

14th Edition

ISBN: 9780134437774

Author: Larry J. Goldstein, David C. Lay, David I. Schneider, Nakhle H. Asmar

Publisher: PEARSON

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Question

Chapter 7.3, Problem 47E

To determine

The possible values of x, y, z at which f(x, y, z)=2x2+3y2+z2−2x−y−z assumes its minimum value.

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Chapter 7.3, Problem 46E

Chapter 7.3, Problem 48E

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1. Abel's Theorem. The goal in this problem is to prove Abel's theorem by following a series of steps (each step must be justified). Theorem 0.1 (Abel's Theorem). If y1 and y2 are solutions of the differential equation y" + p(t) y′ + q(t) y = 0, where p and q are continuous on an open interval, then the Wronskian is given by W (¥1, v2)(t) = c exp(− [p(t) dt), where C is a constant that does not depend on t. Moreover, either W (y1, y2)(t) = 0 for every t in I or W (y1, y2)(t) = 0 for every t in I. 1. (a) From the two equations (which follow from the hypotheses), show that y" + p(t) y₁ + q(t) y₁ = 0 and y½ + p(t) y2 + q(t) y2 = 0, 2. (b) Observe that Hence, conclude that (YY2 - Y1 y2) + P(t) (y₁ Y2 - Y1 Y2) = 0. W'(y1, y2)(t) = yY2 - Y1 y2- W' + p(t) W = 0. 3. (c) Use the result from the previous step to complete the proof of the theorem.

2. Observations on the Wronskian. Suppose the functions y₁ and y2 are solutions to the differential equation p(x)y" + q(x)y' + r(x) y = 0 on an open interval I. 1. (a) Prove that if y₁ and y2 both vanish at the same point in I, then y₁ and y2 cannot form a fundamental set of solutions. 2. (b) Prove that if y₁ and y2 both attain a maximum or minimum at the same point in I, then y₁ and Y2 cannot form a fundamental set of solutions. 3. (c) show that the functions & and t² are linearly independent on the interval (−1, 1). Verify that both are solutions to the differential equation t² y″ – 2ty' + 2y = 0. Then justify why this does not contradict Abel's theorem. 4. (d) What can you conclude about the possibility that t and t² are solutions to the differential equation y" + q(x) y′ + r(x)y = 0?

Question 4 Find an equation of (a) The plane through the point (2, 0, 1) and perpendicular to the line x = y=2-t, z=3+4t. 3t, (b) The plane through the point (3, −2, 8) and parallel to the plane z = x+y. (c) The plane that contains the line x = 1+t, y = 2 − t, z = 4 - 3t and is parallel to the plane 5x + 2y + z = 1. (d) The plane that passes through the point (1,2,3) and contains the line x = 3t, y = 1+t, and z = 2-t. (e) The plane that contains the lines L₁: x = 1 + t, y = 1 − t, z = 2t and L2 : x = 2 − s, y = s, z = 2.

Chapter 7 Solutions

Calculus & Its Applications (14th Edition)

Ch. 7.1 - Let f(x,y,z)=x2+y(xz)4. Compute f(3,5,2).Ch. 7.1 - Prob. 2CYU Ch. 7.1 - Let f(x,y)=x23xyy2. Compute f(5,0),f(5,2) and...Ch. 7.1 - Prob. 2E Ch. 7.1 - Let g(x,y,z)=x/(yz). Compute g(1,1),g(0,1) and...Ch. 7.1 - Let f(x,y,z)=x2ey2+z2. Compute f(1,1,1) and...Ch. 7.1 - Let f(x,y)=xy. Show that f(2+h,3)f(2,3)=3h.Ch. 7.1 - Let f(x,y)=xy. Show that f(2,3+k)f(2,3)=2k.Ch. 7.1 - Cost Find a formula C(x,y,z) that gives the cost...Ch. 7.1 - Cost Find a formula C(x,y,z) that gives the cost...

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The possible values of x , y , z at which f ( x , y , z ) = 2 x 2 + 3 y 2 + z 2 − 2 x − y − z assumes its minimum value.

Solution Summary: The author explains that the function f(x,y) = 2x2+3 y 2x-y-z assumes its minimum value. The partial derivatives

The possible values of x, y, z at which f(x, y, z)=2x2+3y2+z2−2x−y−z assumes its minimum value.

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Chapter 7 Solutions