Mathematics 241 – Multivariable Calculus Sample Exam Questions for Exam 2 – Solutions October 20, 2007

I. A) By direct computation,

\frac{∂w}{∂s} = −t sin^{2}(st) + t cos^{2}(st) and
\frac{∂w}{∂t} = −s sin^{2}(st) + s cos^{2}(st)

Hence

s\frac{∂w}{∂s} = −st sin^{2}(st) + st cos^{2}(st) = t\frac{∂w}{∂t} .

B) By the Chain Rule,

\frac{∂w}{∂s} = \frac{dw}{dx} \frac{∂x}{∂s} = f′(st)t

and

\frac{∂w}{∂t} = \frac{dw}{dx} \frac{∂x}{∂t} = f′(st)s

Hence

s\frac{∂w}{∂s} = stf′(st) = t\frac{∂w}{∂t} .

II. A) We have f_{x} = \frac{∂f}{∂x} = e^{x−y} and f_{y} = \frac{∂f}{∂x} = −e^{x−y}. These functions are continuous at

all (x, y) ∈ R^{2}. Hence f is differentiable at all (x, y) ∈ R^{2}. (See Theorem 3.5 on page 115 of the text.)

B) By the definition,

D(f) = ( f_{x} f_{y}) = ( e^{x−y} −e^{x−y})

and

D(g) = ( \begin{matrix}(g_{1})_{s} & (g_{1})_{t} \\ (g_{2})_{s} & (g_{2})_{t} \end{matrix} ) = ( \begin{matrix}2s & −2t \\ t & s + 2 \end{matrix} ) .

C) The tangent plane is given by

z = f(1, 1) + f_{x}(1, 1)(x − 1) + f_{y}(1, 1)(y − 1) = 1 + (x − 1) − (y − 1) = 1 + x − y.

This meets the z-axis when (x, y) = (0, 0), and then z = 1. So the intersection point is (0, 0, 1).

D) The directional derivative is given as follows. Let u = v/∥v∥ = (1/\sqrt{10}, 3/\sqrt{10}) be the unit vector in the direction of v. So then

D_{u}f(2, 1) = ∇f(2, 1) · (1/\sqrt{10}, 3/\sqrt{10})
= (f_{x}(2, 1), f_{y}(2, 1)) · (1/\sqrt{10}, 3/\sqrt{10})
= (e, −e) · (1/\sqrt{10}, 3/\sqrt{10})
= −2e/\sqrt{10}.

E) By the Chain Rule, since g(3, 0) = (9, 0),

D(f ∘ g)(3, 0) = D(f)(g(3, 0))D(g)(3, 0)

= ( e^{x−y} −e^{x−y}) |_{(x,y)=(9,0)} ( \begin{matrix}2s & −2t \\ t & s + 2 \end{matrix} ) |_{(s,t)=(3,0)}

= ( e^{9} −e^{9} ) ( \begin{matrix}6 & 0 \\ 0 & 5 \end{matrix} )

= ( 6e^{9} −5e^{9} )

F) Substituting and computing the derivatives directly,

(f ∘ g)(s, t) = e^{s^{2}−t^{2}−st−2t}

So

D(f ∘ g)(s, t) = ( (2s − t)e^{s^{2}−t^{2}−st−2t} (−2t − s − 2)e^{s^{2}−t^{2}−st−2t} )

Hence substituting (s, t) = (3, 0),

D(f ∘ g)(3, 0) = ( 6e^{9} −5e^{9} ) ,

which is the same as above.

III. A) By our general knowledge of directional derivatives, f is increasing the fastest at (1, 1)

in the direction of ∇f(1, 1). We have ∇f(x, y) = (−2ax, −2by) so this means we want a unit vector in the direction of (−2a, −2b):

u = (\frac{−a}{\sqrt{a^{2} + b^{2}}} , \frac{−b}{\sqrt{a^{2} + b^{2}}} ) .

B) The marble would roll in the direction that decreased the altitude the fastest, hence in the direction of −∇f(1, 1) = (2a, 2b), so in direction of

v = (\frac{a}{\sqrt{a^{2} + b^{2}}} , \frac{b}{\sqrt{a^{2} + b^{2}}} ) .

C) To remain at a constant altitude, move in a direction perpendicular to ∇f(1, 1), so in either of the directions specified by

w = ± (\frac{b}{\sqrt{a^{2} + b^{2}}} , \frac{−a}{\sqrt{a^{2} + b^{2}}} )

(These are chosen to make w · u = 0 using the u from part A.)

IV. Apply the Chain Rule, thinking of the polar coordinates map x = r cos θ, y = r sin θ as the “inside function.” We have

\frac{∂z}{∂r} = \frac{∂z}{∂x} \frac{∂x}{∂r} + \frac{∂z}{∂y} \frac{∂y}{∂r}

= \frac{∂z}{∂x} cos θ + \frac{∂z}{∂y} sin θ and

\frac{∂z}{∂θ} = \frac{∂z}{∂x} \frac{∂x}{∂θ} + \frac{∂z}{∂y} \frac{∂y}{∂θ}

= −\frac{∂z}{∂x}r sin θ + \frac{∂z}{∂y} r cos θ

Hence starting from the right side of the desired equation,

(\frac{∂z}{∂r} )^{2} + \frac{1}{r^{2}} (\frac{∂z}{∂θ} )^{2} = (\frac{∂z}{∂x} cos θ + \frac{∂z}{∂y} sin θ)^{2} + \frac{1}{r^{2}} (−\frac{∂z}{∂x}r sin θ + \frac{∂z}{∂y} r cos θ)^{2}

= (\frac{∂z}{∂x})^{2} (cos^{2} θ + sin^{2} θ) + 2\frac{∂z}{∂x} \frac{∂z}{∂y} (cos θ sin θ − cos θ sin θ)

+ (\frac{∂z}{∂y} )^{2} (cos^{2} θ + sin^{2} θ)

= (\frac{∂z}{∂x})^{2} + (\frac{∂z}{∂y} )^{2} ,

which is what we wanted to show(!)

V. A) We have x′(t) = (3 cos(3t), −3 sin(3t), 3t^{1/2}), so

∥x′(t)∥ = 3\sqrt{cos^{2}(3t) + sin^{2}(3t) + t} = 3\sqrt{1 + t}.

Hence the desired arclength is

L = ∫_{0}^{1} 3\sqrt{1 + t} dt = 2(1 + t)^{3/2}|_{0}^{1} = 2(2\sqrt{2} − 1) = 4\sqrt{2} − 2.

B) The unit tangent vector is

T = \frac{1}{∥x′(t)∥}x′(t) = \frac{1}{3\sqrt{1 + t}}(3 cos(3t), −3 sin(3t), 3t^{1/2})

= (\frac{cos(3t)}{\sqrt{1 + t}}, \frac{− sin(3t)}{\sqrt{1 + t}}, \sqrt{\frac{t}{1 + t}}) .

C) x′(t) is given above in part A. Then

x′′(t) = (−9 sin(3t), −9 cos(3t), \frac{3}{2}t^{−1/2}).

Using the alternate curvature formula, we compute

x′(t) × x′′(t) = (−\frac{9 sin(3t)}{2\sqrt{t}} + 27\sqrt{t} cos(3t), −27\sqrt{t} sin(3t) − \frac{9 cos(3t)}{2\sqrt{t}},
−27(cos^{2}(3t) − sin^{2}(3t)))

Hence κ(t) = ∥x′(t) × x′′(t)∥/∥x′(t)∥^{3} =

\frac{81}{4} \frac{36t^{2} + 36t + 1}{t} · \frac{1}{(3\sqrt{1 + t})^{3}} = \frac{3}{4} \frac{36t^{2} + 36t + 1}{(1 + t)\sqrt{1 + t}}

VI. A) This is clear – if any two of x, y, z are zero, then all the component functions of

F(x, y, z) = (yz, xz, xy) are zero. B) This will not be covered on the exam (as announced on Friday, 10/19). C) This will not be covered on the exam (as announced on Friday, 10/19). D) f = xyz + C for any constant. E) The curve given by x = y = z = \frac{1}{1−t} is a flow line of F since

x′(t) = y′(t) = z′(t) = (−1)(1 − t)^{−2}(−1) = \frac{1}{(1 − t)^{2}} = y(t)z(t) = x(t)z(t) = x(t)y(t).