1. Let c(t) be the curve c(t) = (cos(2t), sin(3t)) with 0 ≤ t ≤ 2π (a) Sketch the curve (b) Compute the velocity vector of c(t) The velocity vector is the derivative of the path, so v(t) = c0 (t) v(t) = c0 (t) = (− sin(t), 3 cos(3t)) (c) Compute the speed function of c(t) The speed naturally depend on the time t, in particular, speed will be the length of the velocity vector at time t. That is s(t) = |v(t)| = p sin(t)2 + 9 cos(3t)2 (d) Compute the acceleration of c(t) The acceleration is given by the derivative of the velocity a(t) = (− cos(t), −9 sin(3t)) √ 2. Let c(t) be the curve c(t) = (t, t2 / 2, t3 /9) with t ≤ 0 ≤ 2 (a) Compute the velocity vector of c(t) As before, the velocity vector is the derivative of the path with a fancy name. 2 1 v(t) = c0 (t) = (1, √ t, t2 ) 2 3 (b) Compute the speed function of c(t) s(t) = |v(t)| = p p (1 + 2t2 + t4 ) = ((1 + t2 )2 ) = (1 + t2 ) (c) Compute the acceleration of c(t) 2 2 a(t) = v 0 (t) = (0, √ , t) 2 3 (d) Compute the length of the curve c(t) Computing the length of c(t) is simple once we know the speed function (which we computed above. We will integrate s(t) over 0 ≤ t ≤ 2. As it happens this is a very easy integral. Z 0 2 2 t3 14 8 1+t =t+ =2+ = 3 0 3 3 2 3. Sketch the following vector fields, compute the scalar curl and divergence of each field. (a) G(x, y) = (y 2 , y) (b) F (x, y) = (x2 , y) (c) H(x, y) = (−1/x, −1/y) 4. Let F (x, y, z) = (x + y, y + z, x + z) be a vector valued function. To give a better feel for the vector field, here is a plot of the vector field. (a) Compute the divergence of the vector field. ∂F1 ∂F2 ∂F3 + + =1+1+1=3 ∂x ∂y ∂z divF = ∇ · F = (b) Compute the curl of the vector field. We compute the curl as the cross product curlF =∇ × F~ ∂ ∂ ∂ = , , × (F1 , F2 , F3 ) ∂x ∂y ∂z   i j k ∂ ∂ ∂ =det  ∂x ∂y ∂z  F1 F2 F3 In our specific problem we know that F1 (x, y, z) =(x + y) F2 (x, y, z) =(y + z) F3 (x, y, z) =(x + z) Which gives us the specific formula  curl F =  i j k  ∂ ∂x ∂ ∂y ∂ ∂z  x+y y+z x+z Evaluating this determinant gives us the vector (−1, −1, −1) 5. Repeat parts a and b of the previous problem for the vector field G(x, y, z) G(x, y, z) = (yeyx + y, xexy + z + x2 , 2xz + y) To compute the divergence, we use the forumla divG = ∇ · G We will write this out for our particular G as ∂ ∂ ∂ , , · yeyx + y, xexy + z + x2 , 2xz + y = ∂x ∂y ∂z ∂ ∂ ∂ (yeyx + y) + xexy + z + x2 + (2xz + y) = = ∂x ∂y ∂z = y 2 exy + x2 exy + 2x divG = Now to compute the curl we recall that curlG = ∇ × G And for our particular function, this becomes  i ∂ curlG = det  ∂x yeyx + y xy xe j k  ∂ ∂y ∂ ∂z = 2 +z+x 2xz + y = i (1 − 1) − j (2z − 0) + k ((2x + eyx + xyexy ) − (exy + xyex y + 1)) = (0, 2z, 2x − 1)