Mathematics 241 – Multivariable Calculus Solutions for Midterm Exam 1 September 28, 2007

I.

A) A direction vector for the line is Q − P = (1, 2, −4) − (0, 0, 1) = (1, 2, −5). Parametric equations come from P + t(Q − P) = (0, 0, 1) + t(1, 2, −5), so

x = t
y = 2t
z = 1 − 5t.

The interval of t-values giving the line segment from P to Q for these parametric equations is 0 ≤ t ≤ 1 (note that t = 0 gives (x, y, z) = (0, 0, 1) which is P, and t = 1 gives (1, 2, −4), which is Q). The point equidistant from P and Q along the line is the midpoint of the line segment, which is the point with t = \frac{1}{2}. This can be seen in several different ways. The easiest is just to notice that the parametric equations correspond to motion with constant speed along the line segment, so t = \frac{1}{2} will take us exactly half the distance from P to Q.

B. Let v = Q − P = (1, 2, −5) as above. Also let w = R − P = (1, −1, 0) − (0, 0, 1) = (1, −1, −1). Then a normal vector for the desired plane is

N = (1, 2, −5) × (1, −1, −1) = (−2 − 5, −(−1 + 5), −1 − 2) = (−7, −4, −3).

The Cartesian equation of the plane is

0 = N · (x − 0, y − 0, z − 1) = (−7, −4, −3) · (x, y, z − 1) = −7x − 4y − 3z + 3,

or 7x + 4y + 3z = 3.

C. The desired angle is

θ = cos^{−1} (\frac{v · w}{∥v∥∥w∥})
= cos^{−1} (\frac{4}{\sqrt{30}\sqrt{3}})
≐ 1.13 radians, or
≐ 65.1^{∘}.

II. Note that (x, y, z) · (0, 0, 1) = z, and ∥(x, y, z)∥ = \sqrt{x^{2} + y^{2} + z^{2}}. Hence the equation given can be rewritten as

\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} = \frac{1}{\sqrt{2}}.

Changing to spherical coordinates, we know z = ρ cos(φ), and ρ^{2} = x^{2}+y^{2}+z^{2}. Hence the equation can be rewritten as

\frac{1}{\sqrt{2}} = \frac{ρ cos(φ)}{ρ} = cos(φ).

For 0 ≤ φ ≤ π, this has just one solution, φ = \frac{π}{4} . The set of all points satisfying this equation (equivalent to the original one) is a cone with axis along the z-axis.

Comment: A common error on this problem was to take the equation \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} = \frac{1}{\sqrt{2}} and

equate numerators and denominators to obtain z = 1 and \sqrt{x^{2} + y^{2} + z^{2}} = 2. This is NOT a valid algebraic maneuver. Knowing \frac{a}{b} = \frac{c}{d} does not imply that a = c and b = d; it just implies that ad = bc.

III.

A. Letting (x, y) go to (0, 0) along the line y = mx,

lim_{x→0} f(x, mx) = lim_{x→0} \frac{x · mx − (mx)^{2}}{x^{2} + 4(mx)^{2}}
= lim_{x→0} \frac{x^{2}(m − m^{2})}{x^{2}(1 + 4m^{2})}
= lim_{x→0} \frac{m − m^{2}}{1 + 4m^{2}} (cancel the x^{2}’s)
= \frac{m − m^{2}}{1 + 4m^{2}}

B. From the computations in part A, we see that f(x, mx) is approaching different values as x → 0, depending on the value of m. For example, we get a limiting value of 0 if m = 1, but a limiting value of −\frac{2}{5} if m = −1. This implies that lim_{(x,y)→(0,0)} f(x, y) does not exist. Hence f cannot be continuous at (0, 0), since the definition of continuity is that lim_{(x,y)→(0,0)} f(x, y) must exist and equal f(0, 0).

IV.

A. The Cartesian equation of the graph is z = cos(\sqrt{x^{2} + y^{2}}). In cylindrical coordinates, r^{2} = x^{2} + y^{2} and z is the Cartesian z-coordinate, so the cylindrical equation is just z = cos(r).

B. The equation of the level curve is cos(r) = 1. This implies r = 2nπ some integer n ≥ 0. If n = 0, we get a single point at the origin. If n ≥ 1, then the polar equation r = 2nπ defines a circle of radius 2nπ in the polar coordinate plane. Thus, the level curve consists of an isolated point at the origin together with an infinite collection of circles centered at the origin, with radii 2nπ for n = 1, 2, 3, . . ..

Comment: Of course you can get the same conclusion by going back to the Cartesian coordinate equation of the graph. The set of points in the plane with \sqrt{x^{2} + y^{2}} = 2nπ is the circle described above when n ≥ 1.

C. By the chain rule for 1-variable derivatives:

\frac{∂f}{∂x} = \frac{∂}{∂x} (cos(\sqrt{x^{2} + y^{2}})) = − sin(\sqrt{x^{2} + y^{2}}) · \frac{1}{2}(x^{2} + y^{2})^{−1/2} · 2x.

This can be simplified to give

\frac{∂f}{∂x} = \frac{−x sin(\sqrt{x^{2} + y^{2}})}{\sqrt{x^{2} + y^{2}}}.

Substituting x = \frac{π}{2} , we obtain

\frac{∂f}{∂x} (\frac{π}{2} , 0) = \frac{−\frac{π}{2} sin(\frac{π}{2} )}{\frac{π}{2}} = −1.