1-2 Find the domain of the vector function. 1. r(t) = (10(1 In(t + 1), t - 2:21
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- A charged particle begins at rest at the origin. Suddenly, a force causes the particleto accelerate according to the vector function a(t) = ⟨ sin(t) , 6t , 2cos(t)⟩Find functions for the velocity, speed and position of the particle at time tThe motion of a point on the circumference of a rolling wheel of radius 2 feet is described by the vector function r(t) = 2(23t sin (23t))i + 2(1 - cos(23t))j - Find the velocity vector of the point. v(t) = Find the acceleration vector of the point. a(t) = Find the speed of the point. s(t) =The position vector r describes the path of an object moving in the xy-plane. Position Vector Point r(t) = 2 cos ti + 2 sin tj (VZ, V2) (a) Find the velocity vector, speed, and acceleration vector of the object. v(t) = s(t) a(t) = (b) Evaluate the velocity vector and acceleration vector of the object at the given point. a(#) =
- The motion of a point on the circumference of a rolling wheel of radius 5 feet is described by the vector function F(t) = 5(24t - sin(24t))i +5(1 - cos(24t))j Find the velocity vector of the point. v(t) Find the acceleration vector of the point. ä(t) 2880 sin (24t)i + 2880 cos (24t)j✔ CABAME 120(1- cos (24t) )i + 120 sin (24t)j✔ Find the speed of the point. s(t) 240 sin (12t) wwwww Submit Question X Q Search EO HThe motion of a point on the circumference of a rolling wheel of radius 4 feet is described by the vector function F(t) = 4(26t – sin(26t))i + 4(1 – cos(26t))3 Find the velocity vector of the point. ü(t) = | 4(26 – 26 cos(26t))i + 4(26 sin( 26t ))j Find the acceleration vector of the point. ä(t) = | 2704 sin(26t )i + 2704 cos( 26t)j v| Find the speed of the point. s(t) = 2704 sin(26t)i+ 2704 cos( 26t)j x syntax error. Check your variables - you might be using an incorrect one.The motion of a point on the circumference of a rolling wheel of radius 5 feet is described by the vector function F(t) = 5(12t - sin(12t))? +5(1 cos(12t))) Find the velocity vector of the point. (t) = 60(1- cos (12t)i + sin(12t)j) × Find the acceleration vector of the point. ä(t) = 720(sin(12t)i + cos (12t)j) ✓ Find the speed of the point. s(t) = 120 sin (6t) (Write i, j, k for 2,5, k.) Submit Question
- (1n |t – 1], e', vî ) 1. Let 7(t) = (a) Express the vector valued function in parametric form. (b) Find the domain of the function. (c) Find the first derivative of the function. (d) Find T(2). (e) Find the vector equation of the tangent line to the curve when t=2. 2. Complete all parts: (a) Find the equation of the curve of intersection of the surfaces y = x? and z = x3 (b) What is the name of the resulting curve of intersection? (c) Find the equation for B the unit binormal vector to the curve when t= 1. Hint: Instead of using the usual formula for B note that the unit binormal vector is orthogonal to 7 '(t) and 7"(t). In fact, an alternate formula for this vector is ア'(t) × ア"(t) ア(t) ×デ"(t)| B(t) =The motion of a point on the circumference of a rolling wheel of radius 4 feet is described by the vector function r(t) = 4(12t - sin(12t))i + 4(1 − cos(12t))j Find the velocity vector of the point. v(t) = Find the acceleration vector of the point. a(t) = Find the speed of the point. s(t) = =6. (a) Find the directional derivative of w = x²y² at the point (1, -3) in the direc- 5T 5T tion of the unit vector u = cos i+sinj. (b) What is the maximum value of the directional derivative of w = ²y at (1,-3) and it what direction is it attained? Q Search A
- Sketch the curve whose vector equation is Solution r(t) = 6 cos(t) i + 6 sin(t) j + 3tk. The parametric equations for this curve are X = I y = 6 sin(t), z = Since x² + y² = + 36. sin²(t) = The point (x, y, z) lies directly above the point (x, y, 0), which moves counterclockwise around the circle x² + y2 = in the xy-plane. (The projection of the curve onto the xy-plane has vector equation r(t) = (6 cos(t), 6 sin(t), 0). See this example.) Since z = 3t, the curve spirals upward around the cylinder as t increases. The curve, shown in the figure below, is called a helix. ZA (6, 0, 0) (0, 6, 37) I the curve must lie on the circular cylinder x² + y² =Find parametric equations for the tangent line to the helix with parametric equations x = 2 cos(t), y = 4 sin(t), and z = t at the point ( 0, 4, 4,7). Solution The vector equation of the helix is r(t) = (2 cos(t), 4 sin(t), t), so r'(t) = (0,4,7) is t (7²) = 2 , so by the equations x = Xo+at, y = Yo + bt, and z = Zo + ct, its parametric equations are the following. The parameter value corresponding to the point 0, 4, to the vector (x(t), y(t), z(t)) = = so the tangent vector there is r' I The tangent line is the line through (0,4, 1) ₁ paralle