Test 1 Review Sheet Hints—Math 222
Prof. Frank, Spring 2005
(1) Use the point-direction equation of a line. Your answer won’t be unique.
(2) Think of the span of two vectors. If they are in R3you can take a cross product to get the
normal vector, then plug into the usual formula for a plane.
(3) The length of vis the square root of v·v.
(4) Holding rconstant gives a cylinder of radius r. Holding θconstant gives a half-plane.
Holding zconstant gives a horizontal plane at height z.
(5) Holding ρconstant gives a sphere of radius ρ. Holding θconstant gives a half-plane. Holding
φconstant gives a cone.
(6) If f:Rd→R, the level curve for a constant c∈Ris given by Lc={x∈Rd:f(x) = c}.
If the domain is R2you should think of the level curves as a topographical map telling you
curves of constant elevation.
(7) The graph is a plane passing through the line z=−xand the y-axis. The section in the
xz-plane can certainly be a cubic; every section parallel to the xz-plane will be the same
cubic.
(8) The sphere is given by the equation x2+y2+z2= 25. It is the level surface of the function
f(x, y, z) = x2+y2+z2corresponding to c= 25. Take g(x, y) = p25 −x2−y2to get the
upper hemisphere as a graph. In spherical coordinates the equation is simply ρ= 5.
(9) There will be many such functions. Try and make one up!
(10) The partial derivatives are the slopes of the tangents to the curves z=f(x, y0) and z=
f(x0, y), which are sections of the graph parallel to the xz- and yz-planes, respectively.
(11) The vectors are (1,0, fx(x0, y0)) and (0,1, fy(x0, y0)); the cross product will give you a
normal vector. This normal vector defines the tangent plane since z0=f(x0, y0).
(12) The formula for the linear approximation appears in your notes.
(13) A curve is a set of points in Rm, while a path is a function from Rto Rm. That is, a curve
is the image traced by a path. Many paths can trace out the same curve.
(14) The path can be differentiable at a cusp, in which case the derivative will be zero there.
(15) You’ll have to draw pictures of this. The chain rule is Df (c(t))Dc(t) = ∇f(c(t)) ·c0(t).
(16) Use the fact that g(ρ, θ , φ)=(ρcos θsin φ, ρ sin θsin φ, ρ cos φ) to get Dg. Use the second
special case of the chain rule for part (b).
(17) One one hand, the partial derivatives are the directional derivatives in the coordinate di-
rections. On the other hand, the gradient, which is composed of the partial derivatives, is
used to compute the directional derivative in any direction vvia the formula ∇f·v.
(18) (a) Use your professor’s favorite theorem involving dot products! (b) Parameterize a level
curve for fusing a curve c(t) so that f(c(t)) = kfor some constant k. Differentiate both
sides using the chain rule.
(19) If fis class Ckthen all mixed partials up to the kth order will be equal.
(20) True. You need an example where the partials exist but aren’t continuous. In class, I
referred you to Exercise 24, page 193 for an example of this.
(21) The only thing you need to know about the remainders is that for k= 1 or k= 2:
lim
(h1,h2)→(0,0)
Rk((x0, y0),(h1, h2))
||(h1, h2)||k= 0.
(22) Your linear approximation would look like a tangent plane. Your quadratic approximation
would look like a paraboloid or a hyperboloid.
(23) Consider the sections in the xz- an yz-planes. Those are one-variable functions and as such
can only have local extrema at points where the derivative is 0. This tells you the partials
must be zero.
