MATH 234(003) Spring 2017
Worksheet
Name: 1. Compute
df dt ,
where θ and R are constants:
(a) f (x, y) = g(x(t), y(t)) (b) f (x, y) = cos(xy); x(t) = t, y(t) =
1 t
(c) f (x, y) = x2 + y 2 ; x(t) = t; y(t) = t (d) f (x, y) = g(x(t), y(t)); x(t) = cos(t); y(t) = sin(t) (e) f (x, y) = g(x(t), y(t)); x(t) = t; y(t) = t2 (f) f (x, y) = g(x(t), y(t)); x(t) = at; y(t) = bt (g) f (x, y) = x2 − y 2 ; x = g(t); y = −g(t) Solutions: (a) d f (x, y) = gx (x(t), y(t))x0 (t) + gy (x(t), y(t))y 0 (t) dt (b) d 1 f (x, y) = (−y sin(xy))(1) + (−x sin(xy))(− 2 ) = dt t 1 1 = (− sin(1))(1) + (−t sin(1))(− 2 ) = 0 t t (c) d f (x, y) = (2x)(1) + (2y)(1) = 2t + 2t = 4t dt (d) d f (x, y) = gx (x(t), y(t))(− sin(t)) + gy (x(t), y(t))(cos(t)) dt (e) d f (x, y) = gx (x(t), y(t))(1) + gy (x(t), y(t))(2t) dt (f) d f (x, y) = gx (x(t), y(t))a + gy (x(t), y(t))b dt (g) d f (x, y) = (2x)g 0 (t) + (−2y)(−g 0 (t)) = 2g(t)g 0 (t) − 2g(t)g 0 (t) = 0 dt 2. Consider the function f (x, y) = x2 − 3y 2 , and the curve in the plane ~v (t) = (x, y) = (2 + t, −1 − t). (a) Make a big sketch of the curve ~v (t) along with the graph of f . (b) As the point ~v (t) moves in the xy-plane, there’s a point on the graph above or below it. What are the coordinates of this point? (Hint: they will depend on t, and the x-coordinate will equal 2 + t).
(c) Let w(t) ~ be the space curve you computed in the previous part. Compute its tangent vector at time t. df 0 0 (d) Compute df dt at time t. What do you notice about the vector (x (t), y (t), dt (t))? (Look at your answer for part c).
(e) Can you give an explanation for what
df dt
means?
Solution: Here’s a BIG sketch:
The coordinates for w(t) ~ are as follows: w(t) ~ = (2 − t, −t, f (x, y)) = (2 − t, −t, (2 − t)2 − 3(−t)2 ) Its tangent vector is w ~ 0 (t) = (−1, −1, −2(2 − t) + 6t) = (−1, −1, 8t − 4) Now let’s compute df (x(t), y(t)) = fx (2 − t, −t)(−1) + fy (2 − t, −t)(−1) = −2(2 − t) − (−6t) = 8t − 4 dt The vector w ~ 0 equals the vector (x0 (t), y 0 (t), df dt (t))! This maybe won’t surprise us, but it makes for a pretty picture (above): the total derivative is the slope of the yellow curve, which in turn is the lifting of the curve (x, y) to the graph of f .