Answered: (2 cos t, 2 sin t, t), where 0 | bartleby

Algebra & Trigonometry with Analytic Geometry

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter6: The Trigonometric Functions

Section6.6: Additional Trigonometric Graphs

Problem 78E

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Find the work done by F(x,y, z) = (-y,–x, z) on a particle that moves along the curve R(t)
(2 cos t, 2 sin t, t), where 0 < t < .

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Sketch the curve with the vector equation r(t) = cos(t)i − cos(t)j + sin(t)k. Show the direction of increasing t with an arrow drawn on your curve.
Definition 3.27 Let P(x, y) be a point in R² and 0 be the radian measure of the angle formed by OP (O is the origin) and the positive side of the x-axis. Suppose f is a function of the two variables x and y and U cos 0 î + sin 0 ĵ. The directional derivative of ƒ in the direction of U, denoted by D7#f, is given by f(x + h cos 0, y + h sin 0) – f(x, y) DGf(r, y) = lim h→0 h if this limit exists. Note that the directional derivative is a scalar (a real number). The partial derivatives fr and fy are special cases of the directional derivative. 1 Remark (a) If U = î, then cos 0 = 1 and sin 0 = 0. Thus, f(r + h, y) – f(x,y) D;f(x, y) = lim h-0 h which is the partial derivative of f with respect to x. (6) If U = î, then cos 0 = 0 and sin 0 = 1. Thus, f(x, y + h) – f(r, y) D;f(x, y) = lim h→0 h which is the partial derivative of f with respect to y.
Find r’(t), r(t), and r’(t) for the given value of tå. r' (t) r(to) r'(to) = = = = r(t) = 3 cos(t)i + 3 sin(t)j, to T 2 Sketch the curve represented by the vector-valued function and sketch the vectors r(t) and r'(t). = r(t) starts at (x, y) = (0, 0) and ends at (x, y) = r'(to) starts at the terminal point of r(t) at (x, y) = and ends at (x, y) =
3. Find the directional derivative of f(x, y) = cos(3x+4y) at the point (π/6,π/6) along the direction of the vector (-4,-3).
4. ) Find the directional derivative of the function f(x, y) = sin²(xy) at the point P (2,7) in the direction of the unit vector ū = −²¹+] √5
Find the tangential and normal components. r(t)=(t+1)i+2tj+t^ 2 × k , t = 1 r(t) = (t cos t) i + (t sin t) j + t^2 × k , t = 0 r(t)=t^2 i+(t+(1/3)t^ 3 )j+(t-(1/3)t^ 3 )k , t = 0 r(t)=(e^ t cos t) i +(e^ t sin t)j+ √2e^t k , t=0
If the electric potential at a point (x, y) in the xy-plane is V(x, y), then the electric intensity vector at the point (x, y) is E = – vV(x, y). Suppose that V(x, y) = e¯2x cos 2y. (a) Find the electric intensity vector at (x/4, 0). (b) Show that at each point in the plane, the electric po- tential decreases most rapidly in the direction of the vector E.
Answer and show complete solution.
for the attched r(t): 1) how do I find r'(t) 2) sketch the position vector r(0) 3) sketch the tangent vector r'(0)
Compute the directional derivative of f(r. y) at the point (1, 1) in a direction nornnal to the ellipse r + 2y= at the point (1,0).
c. r(t) = cos (t - /2)i + sin (t d. r(t) = (cos t)i – (sin 1)J• 20 %3D %3D niz mil e. r(t) = cos (r)i + sin (12)j, t0 v 38. Motion along a circle Show that the vector-valued function %3! r(1) = (2i + 2j + k) + cos t il + sin i + j+ k V3 describes the motion of a particle moving in the circle of radius 1 centered at the point (2, 2, 1) and lying in the plane x + y - 2z = 2. 39. Motion along a parabola A particle moves along the top of the s bai 1 smit %3D
: Find the derivative of f(x, y, z) = x³ - xy² - z at the point po (2,1,0) in the direction of vector v = 2i - 3j + 6k.

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