Chapter 14.3, Problem 101E

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

6. The largest interval in which the solution of (cos t)y′′ +t^2y′ − (5/t)y = e^t/(t−3) , y(1) = 2, y′(1) = 0is guaranteed to exist by the Existence and Uniqueness Theorem is:A. (0, ∞) B. (π/2, 3) C. (0,π/2) D. (0, π) E. (0, 3)

12. For the differential equation in the previous question, what is the correct form for a particularsolution?A. yp = Ae^t + Bt^2 B. yp = Ae^t + Bt^2 + Ct + DC. yp = Ate^t + Bt^2 D. yp = Ate^t + Bt^2 + Ct + D  Previous differential equation y′′ − 4y′ + 3y = e^t + t^2

16. The appropriate form for the particular solution yp(x) of y^(3) − y′′ − 2y′ = x^2 + e^2x isA. yp(x) = Ax^2 + Bx + C + De^2x B. yp(x) = Ax^3 + Bx^2 + Cx + Dxe^2xC. yp(x) = Ax^2 +Be^2x D. yp(x) = A+Be^2x +Ce^−x E. yp(x) = Ax^2 +Bx+C +(Dx+E)e^2x

Distance Between Two Ships Two ships leave the same port at noon. Ship A sails north at 17 mph, and ship B sails east at 11 mph. How fast is the distance between them changing at 1 p.m.? (Round your answer to one decimal place.) 20.3 X mph Need Help? Read It Watch It SUBMIT ANSWER

practice problem please help!

Find the first and second derivatives of the function. f(u) = √7 3u − 3 f'(u) 2 (7-34) (½) f"(u) = 9 4(7-3u) 32 X Need Help? Read It Watch It SUBMIT ANSWER