MAT 272 Calculus and Analytic Geometry III 10/11/2011 Sample solutions for test 2

MAT 272
Calculus and Analytic Geometry III
Sample solutions for test 2
1. For f (x, y) = x3 − 3x2 + y 3 − 3y, calcufxx (x, y) = 6x − 6
late the 1st partials fx (x, y) = 3x2 − 6x, and
2
fyy (x, y) = 6y
fy (x, y) = 3y − 3. Their common zeros are the
fxy (x, y) = 0
four critical points (2, ±1) and (0, ±1). Calcund
late the 2 partial derivatives, and the discrim- D(x, y) = 36(x − 1)y
inant, and evaluate at all critical points.
(0, −1)
−6
−6
0
+
loc max
10/11/2011
(0, 1)
−6
6
0
−
saddle
(2, −1)
6
−6
0
−
saddle
(2, 1)
6
6
0
+
loc min
2. Write the constraint as the zero-level set of the function g(x, y) = x + y − 5.
Form the Lagrangian L(λ, x, y) = f (x, y) + λg(x, y) and calculate the first partials
Lλ = x + y − 5, Lx = 2(x − 2) + λ,Ly = 2(y − 1) + λ. and equate all to zero.
The resulting system of three linear equations has the unique solution (λc , xc , yc ) =
(−2, 3, 2). Since the gradient of g is never zero, and f is convex, this critical point
must be the unique global minimum.
At any local minimum, the constraint curve must be tangent to the level sets of
f critical, which means that the gradients of f and g must be scalar multiples of
each other.
3.
f (a+∆x,b+∆y)−f (a+∆x,b)
is the slope of the secant lines DL.
∆y
∂f
∂y (a + ∆x, b) is the slope of the tangent line DJ.
∂f
∂f
∂x (a, b) · ∆x + ∂y (a, b) · ∆y is the length of the line segment
(a,b)
√
The slope of the line segment AL is f (a+∆x,b+∆y)−f
.
(∆x)2 +(∆y)2
~ (a, b) · √ h∆x,∆yi .
The slope of the line segment AI is ∇f
(∆x)2 +(∆y)2
HI.
4. For a twice differentiable function z = f (r, θ) and polar coordinates the partial
derivatives simplify to
−y
p
2
∂r
∂
∂θ
∂
θ
θ
x2 + y 2 = √ 2x2 2 = −r cos
= cos θ and ∂x
= ∂x
arctan xy = x y 2 = x2−y
= −rrsin
= sinr θ .
2
∂x = ∂x
r
+y 2
2 x +y
1+( x )
2
∂
∂θ
Using these results the chain rule yields zx = zr · cos θ − zθ sinr θ . and ∂x
(cos θ) = − sin θ · ∂x
= sinr θ .
5. At P the contour of f is parallel to the y-axis, and hence the gradient of f is
parallel to the x-axis. Since it points up, at P it points East.
From above ∂f
∂y (P ) = 0. Using a central difference quotient to estimate the other
f (175,350)−f (100,350)
partial yields ∂f
= 17,000−13,000
≈ 53.33.
∂x (P ) ≈
175−100
75
The average slope between T and B is approximately m = 2,000−27,000
≈ −76.72
725−400
(meters per kilometer), i.e., an downhill slope of about 8%. This occurs fre
quently on freeways, but typically warrants a warning sign (check brakes, etc.).
The cross-section from T to B does not confirm the statement that the “the flanks [exhibit] a concave upward
profile”. as stated, e.g., on http://en.wikipedia.org/wiki/Olympus_Mons.