Exam 2 Study Guide
Math 223 - Vector Calculus
Selected Definitions and Equations
Tangent Plane to the Surface z = f (x, y) at the Point (a, b)
Assuming f is differentiable at (a, b), the equation of the tangent plane is
z = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b)
(1)
Local Linearization of the Surface z = f (x, y) at the Point (a, b)
Provided f is differentiable at (a, b), we can approximatef (x, y):
f (x, y) ≈ f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b)
Remark
(2)
: The above two equations are fundamentally the same, except in the local linearization,
”z =” is replaced with ”f (x, y) ≈”. This is just to illustrate that the tangent plane to the surface can be
used to estimate the values of f (x, y) for points near (a, b).
Local Linearization of the Surface k = f (x, y, z) at the Point (a, b, c)
f (x, y, z) ≈ f (a, b, c) + fx (a, b, c)(x − a) + fy (a, b, c)(y − b) + fz (a, b, c)(z − c)
(3)
The Differential of a Function z = f (x, y)
The differential, df (or dz), at a point (a, b) is the linear function of dx and dy given by the formula
df = fx (a, b)dx + fy (a, b)dy
(4)
The Gradient Vector of a differentiable function f at the Point (a, b)
grad f (a, b) = fx (a, b)~i + fy (a, b)~j
(5)
grad f (a, b, c) = fx (a, b, c)~i + fy (a, b, c)~j + fz (a, b, c)~k
(6)
Note: The gradient in two dimensions is perpendicular to the contours of a function in two variables. If you
are asked to find a vector perpendicular to a level surface, you use the gradient in three dimensions. The
Directional Derivative and the Gradient
If f is differentiable at (a, b) and ~u = u1~i + u2~j is a unit vector, then
f~u (a, b) = fx (a, b)u1 + fy (a, b)u2 = grad f (a, b) · ~u
(7)
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Exam 2 Study Guide
Math 223 - Vector Calculus
Second-Derivative Test for Functions of Two Variables
Suppose (x0 , y0 ) is a point where grad f (x0 , y0 ) = ~0. Let
D = fxx (x0 , y0 )fyy (x0 , y0 ) − (fxy (x0 , y0 ))2
(8)
• If D > 0 and fxx (x0 , y0 ) > 0, then f has a local minimum at (x0 , y0 ).
• If D > 0 and fxx (x0 , y0 ) < 0, then f has a local maximum at (x0 , y0 ).
• If D < 0, then f has a saddle point at (x0 , y0 ).
• if D = 0, this test can’t tell us anything about the classification of the critical point.
Relationship Between Cartesian and Cylindrical Coordinates
Each point in 3-space is represented using 0 ≤ r < ∞, 0 ≤ θ ≤ 2π, −∞ < z < ∞.
p
x = rcosθ
r = x2 + y 2
y = rsinθ
θ = tan−1 ( xy )
z=z
dV = r dr dθ
Relationship Between Cartesian and Spherical Coordinates Each point in 3-space is represented
using 0 ≤ ρ < ∞, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π.
p
x = ρ sinφcosθ
ρ = x2 + y 2 + z 2
y = ρ sinφsinθ
z = ρcosφ
dV = ρ2 sin φ dρ dφ dθ
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Exam 2 Study Guide
Math 223 - Vector Calculus
Additional Review Problems
1.
Let F (u, v) be a function of two variables. Find f 0 (x) if
a) f (x) = F (x, 3)
b) f (x) = F (5x, x2 )
2.
Find the directional derivative using f (x, y, z) = xy + z 2 as you arrive at (0, 1, 1) from the direction of
1, 1, 0).
3.
Check that the point (−1, 1, 2) lies on the surface x2 − xyz = 3. Then, viewing the surface as a level
surface for a function f (x, y, z), find a vector normal to the surface and an equation for the tangent plane
to the surface at (−1, 1, 2).
4.
A student was asked to find the equation of the tangent plane to the surface z = x3 − y 2 at the point
(x, y) = (2, 3). The student’s answer was
z − 3x2 (x − 2) − 2y(y − 3) − 1
(9)
What was the student’s mistake? With that in mind, what is the correct solution?
5.
Find the critical points of f (x, y) = x2 − 2xy + 3y 2 − 8y and classify them as local minima, local maxima,
saddle points, or none of these.
6.
Find the critical points of f (x, y) = x3 + y 3 − 3x2 − 3y + 10 and classify them as local minima, local
maxima, saddle points, or none of these.
For the next three problems, decide whether the integrals are positive, negative, or zero.
Let S be the solid sphere x2 + y 2 + z 2 ≤ 1, and T be the top half of this sphere (i.e. z ≥ 0), and
R be the right half of the sphere withx ≥ 0.
7.
R
8.
R
9.
R
S
sin z dV
T
R
ez dV
sin z dV
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