M408M Second Midterm Exam, November 9, 2006
1. Parametrized curves. A particle follows the curve r(t) =<3+2 cos(2t),4+
2 sin(2t),3t+ 7 >, where trepresents time. Find
a) The speed of the particle at t=π.
b) The unit tangent vector Tat time t=π.
c) The acceleration at time t=π, and
d) The distance traveled (arclength, not straight-line distance) from t=π
to t= 2π.
Please put a BOX around each of your answers.
The velocity is v(t) = r′(t) = h−4 sin(2t),4 cos(2t),3i, so the speed is
v(t) = |v(t)|= 5 and the tangent vector is T=v/v =h−4 sin(2t)/5,4 cos(2t)/5,3/5i.
Evaluated at time t=π, this gives a velocity of h−4,0,3ia tangent vec-
tor of <0,4/5,3/5>and a speed of 5. The acceleration is dv/dt =
h−8 cos(2t),−8 sin(2t),0i=h−8,0,0i. The distance traveled is R2π
πv(t)dt =
5π.
2. Partial derivatives and differentials. Consider the function f(x, y, z) =
ex(1 + ln(y))z2+xy2.
a) Compute fx,fy, and fzas functions of (x, y, z ), and evaluate them at the
point (0,1,3).
fx=ex(1 + ln(y))z2+y2= 10, fy=exz2/y + 2xy = 9, and fz=
2ex(1 + ln(y))z= 6.
b) Use the results of part (a) to estimate the value of f(0.01,0.98,3.02).
f(0,1,3) = 9. df =fxdx+fydy +fzdz = (0.01)(10)−(0.02)(9) +(0.02)6 =
0.04, so f(0.01,0.98,3.02) ≈9.04.
3. Suppose that f(x, y , z) is a function of three variables, and that f(1,4,3) =
13, fx(1,4,3) = 2, fy(1,4,3) = 5, and fz(1,4,3) = −2.
a) Find the directional derivative of fat the point (1,4,3) in the direction
of the point (3,6,4). [Note that (3,6,4) is a point, not a vector.]
The displacement vector from (1,4,3) to (3,6,4) is <2,2,1>, whose
length is √4 + 4 + 1 = 3. The direction of this vector is u=<2/3,2/3,1/3>.
The directional derivative is ∇f·u= (2/3)2 + (2/3)5 −(1/3)2 = 4.
b) Find the equation of the tangent plane to the surface f(x, y , z) = 13 at
the point (1,4,3).
The normal vector is the gradient, <2,5,−2>, so our plane is
2x+ 5y−2z= 2(1) + 5(4) −2(3) = 16.
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