Exam 2 - Solutions
Spring 2012
1. f(u, v) = u3
3+v3
3−v2
2−u+ 2
(a) Determine the location of all critical points of f(u, v).
∇f=hu2−1, v2−vi=h0,0i
u2= 1 →u=±1
v2=v→v= 0,1
Critical points are at: (1,0); (1,1); (−1,0); and (−1,1).
(b) Classify your critical points from part (a).
fuu = 2u, fuv = 0, fvv = 2v−1
D=fuufvv −(fuv )2= 2u(2v−1)
At (1,0), D=−2<0→saddle point
At (1,1), D= 2 >0 and fuu = 2 →local min
At (−1,0), D= 2 >0 and fuu =−2→local max
At (−1,1), D=−2<0→saddle point
(c) Suppose that uis a function of αand β, but vis only a function of β. Write down
the formula for ∂ f
∂β using the chain rule.
∂f
∂β =∂f
∂u
∂u
∂β +∂f
∂v
dv
dβ = (u2−1) ∂u
∂β + (v2−v)dv
dβ
2. We are given: r(t) = ht3, t2i, and C(x, y ) = 20 + x2+xy.
(a)
dC
dt =dC
ds
ds
dt =∇C·r0(t) = h2x+y, xi·h3t2,2ti
At P(8,4), t= 2 and we have:
dC
dt =h20,8i·h12,4i= 272
(b)
dC
ds =∇C·r0(t)
|r0(t)|=h20,8i · h12,4i
|h12,4i| =270
4√10
(c) Begin with C(8,4) = 116
∆C≈dC =dC
dt dt =272
√2∆t=272
5√2
So the new concentration is approximately 116 + 272
5√2.
