Chapter 1, Problem 11CC

18. Using the method of variation of parameter, a particular solution to y′′ + 16y = 4 sec(4t) isyp(t) = u1(t) cos(4t) + u2(t) sin(4t). Then u2(t) is equal toA. 1 B. t C. ln | sin 4t| D. ln | cos 4t| E. sec(4t)

Question 4. Suppose you need to know an equation of the tangent plane to a surface S at the point P(2, 1, 3). You don't have an equation for S but you know that the curves r1(t) = (2 + 3t, 1 — t², 3 − 4t + t²) r2(u) = (1 + u², 2u³ − 1, 2u + 1) both lie on S. (a) Check that both r₁ and r2 pass through the point P. 1 (b) Give the expression of the 074 in two ways Ət ⚫ in terms of 32 and 33 using the chain rule მყ ⚫ in terms of t using the expression of z(t) in the curve r1 (c) Similarly, give the expression of the 22 in two ways Əz ди ⚫ in terms of oz and oz using the chain rule Əz მყ • in terms of u using the expression of z(u) in the curve r2 (d) Deduce the partial derivative 32 and 33 at the point P and the equation of მე მყ the tangent plane

Coast Guard Patrol Search Mission The pilot of a Coast Guard patrol aircraft on a search mission had just spotted a disabled fishing trawler and decided to go in for a closer look. Flying in a straight line at a constant altitude of 1000 ft and at a steady speed of 256 ft/s, the aircraft passed directly over the trawler. How fast (in ft/s) was the aircraft receding from the trawler when it was 1400 ft from the trawler? (Round your answer to one decimal places.) 1000 ft 180 × ft/s Need Help? Read It SUBMIT ANSWER