2. Suppose that /(x, v) is a continuous function of two variables such that each of the first and second order partial derivatives are also continuous at the point (2o, Mo). (a) Consider the polynomial Q(x, w) = /(20, wo) + S.(#0, w)(x – 20) + Su(xo, wo)(w – m) 1 + Show that / and Q have the same values, same first derivatives, and same second derivatives at the point (ro, o). (b) Briefly explain - maybe using the Calculus II idea of a Taylor polynomial? - why we might expect that S(a, w) Q(z, v) for (x, v) (#0, 0)- (e) Lat g(a, v) = S(a + #9,V + Mo) – S(2o, to). and suppose that (ro, Mo) is a critical point of J. Show: i. g(0,0) - 0 i. g.(0, 0) -0= 9,(0,0). ii. Show that the second partials of g, evaluated at (0,0) give the same answer as the second partials of I, evaluated at (o, Mo). (d) Explain why the behavior of f at (20, Mo) is the same as the behavior of g at (0,0) - that is, both have a local minimum, both have a local maximum, or both have a saddle point. [Hint: 9 is really just a shift of (e) Build the second-order polynomial approximation for g at (0,0). Use the result of question #1 to explain why we can apply the Second Derivative Test to this polynomial, and hence to g, to classify the critical point at (0,0). (1) Finally, summarize the result of this problem, as it relates to classifying the critical point (2o, o) for f.
2. Suppose that /(x, v) is a continuous function of two variables such that each of the first and second order partial derivatives are also continuous at the point (2o, Mo). (a) Consider the polynomial Q(x, w) = /(20, wo) + S.(#0, w)(x – 20) + Su(xo, wo)(w – m) 1 + Show that / and Q have the same values, same first derivatives, and same second derivatives at the point (ro, o). (b) Briefly explain - maybe using the Calculus II idea of a Taylor polynomial? - why we might expect that S(a, w) Q(z, v) for (x, v) (#0, 0)- (e) Lat g(a, v) = S(a + #9,V + Mo) – S(2o, to). and suppose that (ro, Mo) is a critical point of J. Show: i. g(0,0) - 0 i. g.(0, 0) -0= 9,(0,0). ii. Show that the second partials of g, evaluated at (0,0) give the same answer as the second partials of I, evaluated at (o, Mo). (d) Explain why the behavior of f at (20, Mo) is the same as the behavior of g at (0,0) - that is, both have a local minimum, both have a local maximum, or both have a saddle point. [Hint: 9 is really just a shift of (e) Build the second-order polynomial approximation for g at (0,0). Use the result of question #1 to explain why we can apply the Second Derivative Test to this polynomial, and hence to g, to classify the critical point at (0,0). (1) Finally, summarize the result of this problem, as it relates to classifying the critical point (2o, o) for f.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I just need question 2 d,e, and f
![1. Consider the paraboloid f(r, y) = ar + bry + cy", where we will assume that a, c+ 0. We will investigate the
behavior of f at its critical point.
(a) Show that (0, 0) is the only critical point when - dac # 0.
(b) By completing the square, show that
S(r, y) = a
4ac -
+
(c) Suppose that D = dac – . Without using the second derivative test:
i. Suppose that D > 0 and a > 0. Show that / has a local minimum at (0,0). Hint: Show that
S(0,0) = 0. Use the fact that a and D are both positive to conclude that when r, y +0, /(r, y) > 0.]
ii. Suppose that D >0 and a < 0. Show that f has a local maximum at (0,0).
iii. Finally, suppose that D<0. Show that / has a saddle point. [Hint: Explain why the tangent plane
at (0,0) has equation z = 0. We wish to show that / crosses this tangent plane, by showing there
exist different paths for which / has opposite signs along those paths.]
(d) Point out that the value D as described above is exactly the function D(r, y) involved in the second
derivative test, and that the results match that test.
2. Suppose that f(r, y) is a continuous function of two variables such that each of the first and second order
partial derivatives are also continuous at the point (ro, 0).
(a) Consider the polynomial
Q(r, v) = S(z0, yo) + S=(10, 0)(z – z0) + Solxo, yo)(v – yo)
+lz(20, yo)(x – z0)* + Szs(xo, m)(z – 20)(y - 20) +Suv(u – s0)
Show that / and Q have the same values, same first derivatives, and same second derivatives at the point
(ro, m).
(b) Briefly explain – maybe using the Calculus II idea of a Taylor polynomial? - why we might expect that
S(r, y) = Q(z, v) for (2, v) (ro. Vo)-
(c) Let g(r, y) = S(r+ ro, y + 30) – S(20. B0). and suppose that (ro, Mo) is a critical point of f. Show:
i. g(0,0) = 0
ii. g.(0,0) = 0 = 9,(0,0).
iii. Show that the second partials of g, evaluated at (0,0) give the same answer as the second partials of
1, evaluated at (o, Vo)-
(d) Explain why the behavior of f at (20, 30) is the same as the behavior of g at (0,0) – that is, both have a
local minimum, both have a local maximum, or both have a saddle point. [Hint: g is really just a shift of
(e) Build the second-order polynomial approximation for g at (0,0). Use the result of question #1 to explain
why we can apply the Second Derivative Test to this polynomial, and hence to g, to classify the critical
point at (0,0).
(1) Finally, summarize the result of this problem, as it relates to classifying the critical point (ro, 0) for f.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1863dd2d-66ae-4de0-8154-3d7d6de9d0c9%2F9cc48873-9900-4f30-b1ca-05ab0a2f3dbe%2Ffvngk9u_processed.png&w=3840&q=75)
Transcribed Image Text:1. Consider the paraboloid f(r, y) = ar + bry + cy", where we will assume that a, c+ 0. We will investigate the
behavior of f at its critical point.
(a) Show that (0, 0) is the only critical point when - dac # 0.
(b) By completing the square, show that
S(r, y) = a
4ac -
+
(c) Suppose that D = dac – . Without using the second derivative test:
i. Suppose that D > 0 and a > 0. Show that / has a local minimum at (0,0). Hint: Show that
S(0,0) = 0. Use the fact that a and D are both positive to conclude that when r, y +0, /(r, y) > 0.]
ii. Suppose that D >0 and a < 0. Show that f has a local maximum at (0,0).
iii. Finally, suppose that D<0. Show that / has a saddle point. [Hint: Explain why the tangent plane
at (0,0) has equation z = 0. We wish to show that / crosses this tangent plane, by showing there
exist different paths for which / has opposite signs along those paths.]
(d) Point out that the value D as described above is exactly the function D(r, y) involved in the second
derivative test, and that the results match that test.
2. Suppose that f(r, y) is a continuous function of two variables such that each of the first and second order
partial derivatives are also continuous at the point (ro, 0).
(a) Consider the polynomial
Q(r, v) = S(z0, yo) + S=(10, 0)(z – z0) + Solxo, yo)(v – yo)
+lz(20, yo)(x – z0)* + Szs(xo, m)(z – 20)(y - 20) +Suv(u – s0)
Show that / and Q have the same values, same first derivatives, and same second derivatives at the point
(ro, m).
(b) Briefly explain – maybe using the Calculus II idea of a Taylor polynomial? - why we might expect that
S(r, y) = Q(z, v) for (2, v) (ro. Vo)-
(c) Let g(r, y) = S(r+ ro, y + 30) – S(20. B0). and suppose that (ro, Mo) is a critical point of f. Show:
i. g(0,0) = 0
ii. g.(0,0) = 0 = 9,(0,0).
iii. Show that the second partials of g, evaluated at (0,0) give the same answer as the second partials of
1, evaluated at (o, Vo)-
(d) Explain why the behavior of f at (20, 30) is the same as the behavior of g at (0,0) – that is, both have a
local minimum, both have a local maximum, or both have a saddle point. [Hint: g is really just a shift of
(e) Build the second-order polynomial approximation for g at (0,0). Use the result of question #1 to explain
why we can apply the Second Derivative Test to this polynomial, and hence to g, to classify the critical
point at (0,0).
(1) Finally, summarize the result of this problem, as it relates to classifying the critical point (ro, 0) for f.
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