Find given the following vector function.
A)cost sint
D)-16 sin t cos t

Question

Find  given the following vector function. 
A)cost sint 
D)-16 sin t cos t

Quizplus · Accepted Answer

The given vector function is: r(t) = (cos t) i + (sin t) j + (t^2) k To find the tangent vector, we need to take the derivative of this function with respect to t: r'(t) = (-sin t) i + (cos t) j + 2t k To find the normal vector, we need to take the derivative of the tangent vector with respect to t: r''(t) = (-cos t) i - (sin t) j + 2k Now, we can use the formula for the curvature: kappa = ||r'(t) x r''(t)|| / ||r'(t)||^3 where ||v|| denotes the magnitude of vector v, and x denotes the cross product. r'(t) x r''(t) = (2sin t cos t) i + (2sin t cos t) j - (sin^2 t + cos^2 t) k = (2sin t cos t) i + (2sin t cos t) j - k ||r'(t)||^3 = sqrt(sin^2 t + cos^2 t + 4t^2)^3 = (sin^2 t + cos^2 t + 4t^2)^(3/2) So, putting it all together: kappa = ||(2sin t cos t) i + (2sin t cos t) j - k|| / (sin^2 t + cos^2 t + 4t^2)^(3/2) = sqrt((2sin t cos t)^2 + (2sin t cos t)^2 + 1) / (sin^2 t + cos^2 t + 4t^2)^(3/2) = sqrt(8sin^2 t cos^2 t + 1) / (sin^2 t + cos^2 t + 4t^2)^(3/2) = sqrt(8sin^2 t cos^2 t + 1) / (4t^2 + 1) Therefore, the correct answer is D) -16 sin t cos t. The given vector function can be written as . We want to find the cross product of this vector function with its derivative. The derivative of the vector function is <-sint, cost, 8>. Taking the cross product, we get: x <-sint, cost, 8> = (8cost^2 + 8sint^2, -8cos(t)sin(t), sint(-8cos(t)) - cost(8sin(t))) = (8, -4sin(2t), -8cos(t)sin(t)) The third component can be simplified using the identity sin(2t) = 2sin(t)cos(t): -8cos(t)sin(t) = -4sin(2t) Therefore, the cross product is <8, -4sin(2t), -4sin(2t)>, and the magnitude of the cross product is: |<8, -4sin(2t), -4sin(2t)>| = sqrt(8^2 + (-4sin(2t))^2 + (-4sin(2t))^2) = sqrt(64 + 16sin^2(2t)). Taking the derivative of this magnitude with respect to t, we get: d/dt [sqrt(64 + 16sin^2(2t))] = 16sin(2t)cos(2t)/(2sqrt(64 + 16sin^2(2t))) = -8sin(2t)cos(2t)/sqrt(64 + 16sin^2(2t)) Plugging in t = pi/4 (which corresponds to 45 degrees), we get: -8sin((2pi)/4)cos((2pi)/4)/sqrt(64 + 16sin^2((2pi)/4)) = -8(1/2)(0)/sqrt(64 + 16) = 0 Therefore, the correct choice is D, -16 sin t cos t.

Find Given the Following Vector Function