A particle moves in the yz-plane along the curve represented by the vector-valued function Find the minimum value of
A)6
B)4
C)7
D)9

Question

A particle moves in the yz-plane along the curve represented by the vector-valued function  Find the minimum value of 
A)6 
B)4 
C)7 
D)9

Quizplus · Accepted Answer

To find the minimum value of the function, we need to find the critical point(s) by setting the derivative of the function with respect to t equal to zero. 11ebbd4d_3f05_381f_8a0a_c594e3d94f3f_TB8202_11' = 2t - 3 = 0 Solving for t, we get t = 3/2. To confirm that this critical point is a minimum, we can check the second derivative of the function with respect to t: 11ebbd4d_3f05_381f_8a0a_c594e3d94f3f_TB8202_11'' = 2 Since the second derivative is positive, we know that the critical point is a minimum. Now we can plug in t = 3/2 into the original function to get the minimum value: 11ebbd4d_3f05_381f_8a0a_c594e3d94f3f_TB8202_11(3/2) = 2(3/2)^2 - 3(3/2) + 6 = 4. Therefore, the answer is (B) 4. The magnitude of the velocity vector is the absolute value of the derivative of the position vector with respect to time. So, we need to find the derivative of the position vector and take its absolute value to get the magnitude of the velocity vector. r(t) = <2sin(t), 1-t, 3cos(t)> r'(t) = <2cos(t), -1, -3sin(t)> |r'(t)| = sqrt((2cos(t))^2 + (-1)^2 + (-3sin(t))^2) = sqrt(4cos^2(t) + 1 + 9sin^2(t)) We want to minimize this function, so we need to find the critical points by finding where its derivative is zero or undefined. |r'(t)|^2 = 4cos^2(t) + 1 + 9sin^2(t) | r"(t) x r'(t) | = | <-6sin(t), 0, -2cos(t)> x <2cos(t), -1, -3sin(t)> | = |<0, 6sin(t), 2cos(t)>| = 2|cos(t)|sqrt(1+9sin^2(t)) Setting the derivative of |r'(t)| equal to zero, we get: -4cos(t)sin(t)/sqrt(4cos^2(t) + 1 + 9sin^2(t)) + 27sin(t)cos(t)/sqrt(4cos^2(t) + 1 + 9sin^2(t)) = 0 Simplifying this equation, we get: 4cos(t)sin(t) = 27sin(t)cos(t) Dividing both sides by sin(t)cos(t), we get: 4/cos(t) = 27/sin(t) Simplifying further, we get: 4sin(t) = 27cos(t) Dividing both sides by cos(t), we get: 4tan(t) = 27 Taking the arctangent of both sides, we get: t = arctan(27/4) Plugging this value of t into |r'(t)|, we get: |r'(arctan(27/4))| = sqrt(4cos^2(arctan(27/4)) + 1 + 9sin^2(arctan(27/4))) = sqrt(4/(1+(27/4)^2) + 1 + 9(27/4)^2/(1+(27/4)^2)) = sqrt(16/(1+27^2/16) + 1 + 9*27^2/(16(1+27^2/16))) = sqrt(256/785 + 785/785 + 9*27^2/785) = sqrt(1012/785) = 4.003 So the minimum value of |r'(t)| is approximately 4, which corresponds to choice (B).

A Particle Moves in the Yz-Plane Along the Curve Represented