Use your knowledge on Higher-order derivative and Chain Rule to show that z" + 4z' + 8z = 0 if z = e-2x (sin 2x + cos 2x). Z = e-2x (sin 2x + cos 2x) Let u = 2x, then du = 2dx du = 2 dx Z = e-"(sin u + cos u) z'%3Dе"Dx(sin и + cos u) + (sin и + cos u)Dx(е ") z'%3Dе " (сos u — sin u) — е (sinu + cosu) coS и — е "и sin u — e"u sin u — e"и cos u dy = -2e¬u sin u du dy = -2e¬u sin u, du du Substitute = 2 using the Chain Rule, dx dz dy du (-2e¬" sin u)2 = -4e¬u sin u du du dx Substitute u = 2x, z' = -4e-2x sin 2x To get the second derivative, again let u = 2x then du = 2dx du = 2 dx z' = -4e-u sin u -4[e-"Dx(sin u) + sin u Dx(e¬u) 3 -4е И cos и — 4е -И sin u z" = dy — — 4е и соs и — 4e-U sin u du dy = -4e-u du Substitute cOS и — 4e"u sin u, = 2 using the Chain Rule, dx du dz dy du (-4е " cos u — 4е И sin u)2 3D —8е И сosu — 8e u sin u du du dx

Use your knowledge on Higher-order derivative and Chain Rule to show that z" + 4z' + 8z = 0 if z = e-2x (sin 2x + cos 2x). Z = e-2x (sin 2x + cos 2x) Let u = 2x, then du = 2dx du = 2 dx Z = e-"(sin u + cos u) z'%3Dе"Dx(sin и + cos u) + (sin и + cos u)Dx(е ") z'%3Dе " (сos u — sin u) — е (sinu + cosu) coS и — е "и sin u — e"u sin u — e"и cos u dy = -2e¬u sin u du dy = -2e¬u sin u, du du Substitute = 2 using the Chain Rule, dx dz dy du (-2e¬" sin u)2 = -4e¬u sin u du du dx Substitute u = 2x, z' = -4e-2x sin 2x To get the second derivative, again let u = 2x then du = 2dx du = 2 dx z' = -4e-u sin u -4[e-"Dx(sin u) + sin u Dx(e¬u) 3 -4е И cos и — 4е -И sin u z" = dy — — 4е и соs и — 4e-U sin u du dy = -4e-u du Substitute cOS и — 4e"u sin u, = 2 using the Chain Rule, dx du dz dy du (-4е " cos u — 4е И sin u)2 3D —8е И сosu — 8e u sin u du du dx

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

See similar textbooks

Similar questions

Q1: show that1) d/dz [cosz] = -sin z , 2) * 1+tan2 z = sec2 z
find ƒx , ƒy , and ƒz . ƒ(x, y, z) = sin-1 (xyz)
Integrate the trig identity 2 cos j x cos kx = cos(j + k )x + cos(j -k )x to show that cos j x is orthogonal to cos kx, provided j -/- k. What is the result when j = k?
A force of F(x)=x2−cos(3x)+2, x is in meters, acts on an object. What is the work required to move the object from x=3 to x=7?
Differentiate y=excos(x2): y'= A. excos(x2)-2xsin(x2) B. -2xexsin(x2) C. ex(cos(x2)-2xsin(x2)) D. -exsin(x2)
If f(x,y,z)=ln(x2y+sin2(x+y))+227x228y2z229, then ∂4f ∂x2∂y∂z at (1,1,1) is equal to
Differentiate : a. y= x^2 sinx +sin^2x + cos^2x + tan 2x/sin2x/cos2x b. y= In √x^3
Describe the method used to integrate sinm x cosn x, for m even and n odd.
Show that bothy= sinx and y = cos x satisfy y" = -y
Find the firsst derivative of y = arctan [ (3 sin x) / (4 + 5 cos x) ] and the first partial derivative fx, fy, and fz of (x + yw )2 – ez – 2 = 0 and f ( x , y, z ) = 2xyz + tan x + y sin z Make the solution detailed and clear IN A PAPER.
Suppose that the position of one particle at time t is given by x1 = 3 sin(t), y1 = 2 cos(t), 0 ≤ t ≤ 2? and the position of a second particle is given by x2 = −3 + cos(t), y2 = 1 + sin(t), 0 ≤ t ≤ 2?. (a) Graph the paths of both particles. How many points of intersection are there? __ points of intersection (b) Are any of these points of intersection collision points? That is, are the particles ever at the same place at the same time? If so, find the collision points. (Enter your answers as a comma-separated list of ordered pairs of the form (x, y). If there are no collision points, enter DNE.) (x, y) =__ (c) Describe what happens if the path of the second particle is given by x2 = 3 + cos(t), y2 = 1 + sin(t), 0 ≤ t ≤ 2?. The circle is centered at (x, y) = __ There are _ intersection point(s), and there are _ collision point(s).
find div and curl F F(x,y,z)= 7y3z2i- 8x2z5j-3xy4k F(x,y,z) = exyi-cosy j +sin2 z k