OHSx XM521 Multivariable Differential Calculus:Homework Solutions §15.1

(18) Sketch (or describe) the domain of f(x, y) = lnxy.

Solution: The domain of ln is the set of positive reals, so lnxy is defined if and only ifxy > 0. This region consists of the first and third quadrants of the xy-plane, not includingpoints on the coordinate axes.

(20) (a) Describe the domain of f(x, y) =√4−x2

y2+3 .

Solution: Since the denominator of f is always defined, and the numerator is definedwhenever 4− x2 ≥ 0, the domain of f is

{(x, y)∣∣ 4− x2 ≥ 0} = {(x, y)

∣∣ x2 ≤ 4} = {(x, y)∣∣ − 2 ≤ x ≤ 2}.

Geometrically, this set is the vertical strip in the xy-plane with left boundary equal tothe line x = −2, and right boundary equal to the line x = 2. These boundary lines areincluded in the domain of f .

(34) Sketch (or describe) the level curve z = k for z = x2 + 9y2, k = 0, 1, 2, 3, 4.

Solution:

z = k = 0: The equation becomes 0 = x2 + 9y2. The only point that satisfies this equationis (0, 0). So the level curve is just the point (0, 0).

z = k = 1: The equation becomes 1 = x2 + 9y2. This is an ellipse centered at the originwith major axis of length 1 (in the x-direction) and minor axis of length

√1/9 = 1/3 (in

the y-direction).

z = k = 2: The equation becomes 2 = x2 + 9y2. This is an ellipse centered at the originwith major axis of length

√2 (in the x-direction) and minor axis of length

√2/9 =

√2/3

(in the y-direction).

z = k = 3: The equation becomes 3 = x2 + 9y2. This is an ellipse centered at the originwith major axis of length

√3 (in the x-direction) and minor axis of length

√1/3 =

√3/3

(in the y-direction).

z = k = 4: The equation becomes 4 = x2 + 9y2. This is an ellipse centered at the originwith major axis of length 2 (in the x-direction) and minor axis of length

√4/9 = 2/3 (in

the y-direction).

1

(38) Sketch (or describe) the level surface f(x, y, z) = k where f(x, y, z) = x2 + y2 − z2, k = 0.

Solution: Setting f(x, y, z) = k = 0, we obtain 0 = x2 +y2−z2, or z2 = x2 +y2. This isa double cone with common vertex at the origin, and opening in the positive and negativez-directions.

(42) Describe the level surfaces for f(x, y, z) = 3x− y + 2z.

Solution: Setting f(x, y, z) = k, we obtain k = 3x − y + 2z. For each value of k, thisequation defines a plane with normal vector 〈3,−1, 2〉. For a given finite set of values fork, we obtain a corresponding finite set of parallel planes, all of which are perpendicular tothe vector 〈3,−1, 2〉.

(46) Let f(x, y) = yex. Find an equation of the level curve that passes through the given point.

(a) (ln 2, 1).

Solution: When x = ln 2 and y = 1, f(x, y) = 1 · eln 2 = 2. Therefore, (ln 2, 1) lies onthe level curve f(x, y) = 2, which can be written 2 = yex, or equivalently y = 2e−x.

(b) (0, 3).

Solution: When x = 0 and y = 3, f(x, y) = 3 · e0 = 3. Therefore, (0, 3) lies on the levelcurve f(x, y) = 3, which can be written 3 = yex, or equivalently y = 3e−x.

(c) (1,−2).

Solution: When x = 1 and y = −2, f(x, y) = −2 · e1 = −2e. Therefore, (1,−2)lies on the level curve f(x, y) = −2e, which can be written −2e = yex, or equivalentlyy = −2e−x+1.

2

OHSx XM521 Multivariable Differential Calculus:Homework Solutions §15.2

(20) Show that the limit does not exist by considering the limits as (x, y) → (0, 0) along thecoordinate axes.

(a) lim(x,y)→(0,0)

x− yx2 + y2

.

Solution: Consider the curve C (the nonnegative y-axis) defined by x = 0, y = t, t ≥ 0.Substituting in for x and y, and taking the limit as t→ 0+ gives

limt→0+

0− t0 + t2

= limt→0+

−1

t= −∞.

This alone suffices to prove that the (general) limit does not exist as a finite number. (If itdid exist, then the limit along every curve must be that finite number as well.) A similarargument shows that the limit along the nonnegative x-axis is +∞.

(b) lim(x,y)→(0,0)

cosxy

x+ y.

Solution: Consider the curve C (the nonnegative y-axis) defined by x = 0, y = t, t ≥ 0.Substituting in for x and y, and taking the limit as t→ 0+ gives

limt→0+

cos 0 · t0 + t

= limt→0+

1

t= +∞.

Since the limit along C does not exist (as a finite number), the (general) limit also doesnot exist.

(26) Determine whether lim(x,y)→(0,0)

x4 − 16y4

x2 + 4y2exists. If so, find its value.

Solution: Observe that x4 − 16y4 = (x2 − 4y2)(x2 + 4y2), so

x4 − 16y4

x2 + 4y2=

(x2 − 4y2)(x2 + 4y2)

(x2 + 4y2)= x2 − 4y2,

whenever x2 + 4y2 6= 0, which happens only when (x, y) = (0, 0). Therefore,

lim(x,y)→(0,0)

x4 − 16y4

x2 + 4y2= lim

(x,y)→(0,0)(x2 − 4y2) = 0.

The last equality follows because lim(x,y)→(0,0) x2 = 0, lim(x,y)→(0,0) 4y2 = 0, and the limit

of a difference is the difference of the limits. (A good exercise would be to prove directlyfrom the definition of general limit that lim(x,y)→(0,0) x

2 = 0 and lim(x,y)→(0,0) 4y2 = 0.)

(31) Determine whether lim(x,y,z)→(0,0)

sin(x2 + y2 + z2)√x2 + y2 + z2

exists. If so, find its value.

1

Solution: We will prove that lim(x,y,z)→(0,0,0)

sin(x2 + y2 + z2)√x2 + y2 + z2

= 0. To this end, let ε > 0.

Choose δ = ε. Observe that for all (x, y, z) 6= (0, 0, 0),∣∣∣∣ sin(x2 + y2 + z2)√x2 + y2 + z2

∣∣∣∣ =

∣∣∣∣ sin(x2 + y2 + z2)

x2 + y2 + z2

∣∣∣∣ ·√x2 + y2 + z2 ≤√x2 + y2 + z2.

The last inequality follows because | sin t|/|t| ≤ 1 for all t. Therefore, if

0 ≤√x2 + y2 + z2 < δ,

then ∣∣∣∣ sin(x2 + y2 + z2)√x2 + y2 + z2

− 0

∣∣∣∣= ∣∣∣∣ sin(x2 + y2 + z2)√x2 + y2 + z2

∣∣∣∣ ≤√x2 + y2 + z2 < δ = ε.

We’ve shown that for all ε > 0 there exists δ > 0 such that

∣∣∣∣ sin(x2 + y2 + z2)√x2 + y2 + z2

− 0

∣∣∣∣< ε

whenever 0 <√

(x− 0)2 + (y − 0)2 + (z − 0)2 < δ. Therefore,

lim(x,y,z)→(0,0,0)

sin(x2 + y2 + z2)√x2 + y2 + z2

= 0.

(38) (a) Show that the value ofx3y

2x6 + y2approaches 0 as (x, y)→ (0, 0) along any straight line

y = mx, or along any parabola y = kx2.

• Solution: Let C1 be the curve x = t, y = mt, −∞ < t < ∞, which corresponds to the

line y = mx. Substituting these values for x and y into x3y2x6+y2 and taking the limit as

t→ 0 gives

limt→0

t3 ·mt2t6 +m2t2

= limt→0

mt2

2t4 +m2.

If m = 0, then

limt→0

mt2

2t4 +m2= lim

t→0

0

2t4= 0.

If m 6= 0, then limt→0(2t4 +m2) = m2 6= 0, which implies that

limt→0

mt2

2t4 +m2=

limt→0mt2

limt→0(2t4 +m2)=

0

m2= 0.

So, in both cases the limit along C1 is zero.

Now let C2 be the curve x = t, y = kt2, −∞ < t <∞, which corresponds to the parabola

y = kx2. Substituting these values for x and y into x3y2x6+y2 and taking the limit as t → 0

gives

limt→0

t3 · kt2

2t6 + k2t4= lim

t→0

kt

2t2 + k2.

A similar analysis to the one given for C1 shows that this last limit is zero. Therefore, thelimit along C2 is also zero.

(b) Show that the value of lim(x,y)→(0,0)

x3y

2x6 + y2does not exist by letting (x, y) → (0, 0)

along the curve y = x3.

2

• Solution: Let C3 be the curve x = t, y = t3, −∞ < t < ∞, which corresponds to the

curve y = x3. Substituting these values for x and y into x3y2x6+y2 and taking the limit as

t→ 0 gives

limt→0

t3 · t3

2t6 + t6= lim

t→0

t6

3t6= lim

t→0

1

3=

1

3.

Since the limit along C3 is different from the limit along C1 (or C2), the general limit

lim(x,y)→(0,0)x3y

2x6+y2 does not exist.

(43) Let f(x, y) =x2

x2 + y2. Is it possible to define f(0, 0) so that f will be continuous at (0, 0)?

Solution: The answer is no. f is continuous at (0, 0) if and only if lim(x,y)→(0,0) f(x, y)is defined and equals f(0, 0). So to prove that f cannot be continuous at (0, 0) no matterhow f(0, 0) is chosen, it suffices to show that lim(x,y)→(0,0) f(x, y) does not exist. To thisend, let C1 be the curve x = t, y = 0, −∞ < t <∞. Then

limt→0

f(x(t), y(t)) = limt→0

t2

t2 + 0= 1.

So the limit along C1 is 1. However, if C2 is the curve x = 0, y = t, −∞ < t < ∞, thenthe limit along C2 is

limt→0

f(x(t), y(t)) = limt→0

0

0 + t2= 0.

Since the limits along these two curves are different, lim(x,y)→(0,0) f(x, y) does not exist,and therefore f cannot be continuous at (0, 0) no matter how f(0, 0) is chosen.

(44) Let f(x, y) = xy ln(x2 + y2). Is it possible to define f(0, 0) so that f will be continuous at(0, 0)?

Solution: Define f(0, 0) = 0. We will prove that given this value, f is continuous at(0, 0). First, note that limt→0+ t ln t = 0. (Just use L’Hospital’s rule.) Now suppose ε > 0.Since limt→0+ t ln t = 0, there exists δ1 > 0 such that if 0 < t < δ1, then |t ln t| < ε.Let δ =

√δ1. Observe that 0 ≤ (|x| − |y|)2 = x2 − 2|x||y| + y2, which implies that

2|x||y| ≤ x2 +y2. In particular |xy| ≤ x2 +y2, and |xy ln(x2 +y2)| ≤ |(x2 +y2) ln(x2 +y2)|.Now suppose that 0 <

√(x− 0)2 + (y − 0)2 < δ. Then 0 < x2 + y2 < δ2 = δ1. Therefore,

by choice of δ1,

|xy ln(x2 + y2)− 0| = |xy ln(x2 + y2)| ≤ |(x2 + y2) ln(x2 + y2)| < ε.

(The last inequality follows because the quantity t = x2 + y2 is assumed to be less thanδ1, and by definition of δ1 this forces |t ln t| < ε.)

We’ve shown that for any ε > 0, there exists δ > 0 such that |xy ln(x2 + y2) − 0| < εwhenever 0 <

√(x− 0)2 + (y − 0)2 < δ. Consequently, lim(x,y)→(0,0) f(x, y) = 0, and thus

assigning this limit value to f(0, 0) makes f continuous at (0, 0).

3

OHSx XM521 Multivariable Differential Calculus:Homework Solutions §15.3

(30) Confirm that the mixed second-order partial derivatives of f(x, y) = ex−y2

are the same.

Solution: First we compute fxy:

fx(x, y) = ∂∂xe

x−y2

= ex−y2

.(Differentiate f w.r.t. xtreating y as a constant.)

fxy(x, y) = ∂∂yfx(x, y) = ∂

∂y ex−y2

= −2yex−y2

.(Differentiate fx w.r.t. ytreating x as a constant.)

Now we compute fyx:

fy(x, y) = ∂∂y e

x−y2

= −2yex−y2

.(Differentiate f w.r.t. ytreating x as a constant.)

fyx(x, y) = ∂∂xfy(x, y) = ∂

∂x (−2yex−y2

) = −2yex−y2

.(Differentiate fy w.r.t. xtreating y as a constant.)

So fxy(x, y) = −2yex−y2

= fyx(x, y).

(39) Using implicit differentiation, compute ∂z/∂x and ∂z/∂y for x2 + z sinxyz = 0.

Solution: To find ∂z/∂x we differentiate each term with respect to x:

2x+ ∂z/∂x · sinxyz + z · ∂∂x

(sinxyz) = 0. (∗)

Note that we’ve used the product rule to differentiate z sinxyz. To find ∂∂x (sinxyz) we

differentiate with respect to x, treating y as a constant and z as a function of x and y.Thus,

∂

∂x(sinxyz) = (cosxyz) · ∂

∂x(xyz)

= (cosxyz) · (yz + xy(∂z/∂x))

= yz cosxyz + xy(∂z/∂x) cosxyz.

Plugging this back into equation (∗) gives

2x+ ∂z/∂x · sinxyz + z · (yz cosxyz + xy(∂z/∂x) cosxyz) = 0.

Finally, solving for ∂z/∂x gives

∂z/∂x =−2x− yz2 cosxyz

sinxyz + xyz cosxyz.

To find ∂z/∂y, we could compute as we did above. Alternatively, we could exploit symme-try. Note that when differentiating with respect to y, the x2 term in the original equationmakes no contribution; and with the exception of that term the equation is unchangedif x and y are interchanged. Therefore, to find ∂z/∂y, we take ∂z/∂x, ignore the −2xcontribution from the x2 term, then interchange x and y. This gives

∂z/∂y =−xz2 cosxyz

sinxyz + xyz cosxyz.

1

(42) The accompanying figure (in the textbook) shows a contour plot for an unspecified functionf(x, y). Make a conjecture about the sign of the partial derivatives fx(x0, y0) and fy(x0, y0),and explain your reasoning.

Solution: From an examination of the figure and assuming nothing unexpected happensbetween the contour lines, it appears that as we move along the line y = y0, near x0, and inthe direction of increasing x, we move across contour lines of successively decreasing value.In other words, it appears that at least on some line segment x0 − ε < x < x0 + ε, y = y0,the values of f are decreasing. So we expect that the rate of change of f in the x-direction,at the point (x0, y0), is negative. Equivalently, we expect fx(x0, y0) to be negative.

On the other hand, it appears that as we move along the line x = x0, near y0, and in thedirection of increasing y, we move across contour lines of successively increasing value. Inother words, it appears that at least on some line segment y0 − ε < y < y0 + ε, x = x0, thevalues of f are increasing. So we expect that the rate of change of f in the y-direction, atthe point (x0, y0), is positive. Equivalently, we expect fy(x0, y0) to be positive.

(48) Express the derivatives in “subscript” notation.

(a)∂3f

∂y2∂x.

Solution:∂3f

∂y2∂x=

∂

∂y

(∂

∂y

(∂f

∂x

))= fxyy.

(b)∂4f

∂x4.

Solution:∂4f

∂x4=

∂

∂x

(∂

∂x

(∂

∂x

(∂f

∂x

)))= fxxxx.

(c)∂4f

∂y2∂x2.

Solution:∂4f

∂y2∂x2=

∂

∂y

(∂

∂y

(∂

∂x

(∂f

∂x

)))= fxxyy.

(d)∂5f

∂x2∂y3.

Solution:∂5f

∂x2∂y3=

∂

∂x

(∂

∂x

(∂

∂y

(∂

∂y

(∂f

∂y

))))= fyyyxx.

(70) Find fx and fy for f(x, y) =∫ xy

1et

2

dt.

Solution: According to the fundamental theorem of calculus,

d

dx

(∫ x

1

et2

dt

)= ex

2

.

Moreover, if g(x) is any differentiable function of x, then according to the chain rule,

d

dx

(∫ g(x)

1

et2

dt

)= e(g(x))

2

· g′(x).

Therefore, letting g(x) = xy and holding y constant and differentiating f with respect tox, we obtain

fx(x, y) = e(xy)2

· y = yex2y2

.

2

Similarly, holding x constant and differentiating f with respect to y, we obtain

fy(x, y) = xex2y2

.

(74) Show that the function satisfies the heat equation:∂z

∂t= c2

∂2z

∂x2, where c is a positive

constant.

(a) z = e−t sin(x/c).

Solution: ∂z/∂t = −e−t sin(x/c). ∂z/∂x = e−tc−1 cos(x/c), and

∂2z

∂x2=

∂

∂x

(e−tc−1 cos(x/c)

)= −e−tc−2 sin(x/c).

Therefore,

c2∂2z

∂x2= c2 · −e−tc−2 sin(x/c)

= −e−t sin(x/c)

=∂z

∂t,

as desired. Part (b) works the same way.

(78) A point moves along the intersection of the elliptic paraboloid z = x2 + 3y2 and the planex = 2. At what rate is z changing with y when the point is at (2, 1, 7)?

Solution: Let f(x, y) = x2 + 3y2 and let the curve C be the intersection of the ellipticparaboloid z = x2 + 3y2 and the plane x = 2. (C is simply the graph of f when restrictedto the line x = 2 in the xy-plane.) From our discussion of partial derivatives, we know thatfy(2, 1) is the rate of change of f with respect to y along the curve C at the point (2, 1).So, along the intersection of the elliptic paraboloid z = x2 + 3y2 and the plane x = 2, therate z is changing with y at the point (2, 1, 7) is

fy(x, y) = 6y∣∣∣(x,y)=(2,1)

= 6.

3

OHSx XM521 Multivariable Differential Calculus:Homework Solutions §15.4

(16) Let w = rs/(r2 + s2); r = uv, s = u− 2v. Use a chain rule to find ∂w/∂u and ∂w/∂v.

Solution: w is a function of r and s, each of which is a function of u and v. So,

∂w

∂u=

∂w

∂r

∂r

∂u+

∂w

∂s

∂s

∂u

=s3 − r2s

(r2 + s2)2· v +

r3 − rs2

(r2 + s2)2· 1.

Similarly,

∂w

∂v=

∂w

∂r

∂r

∂v+

∂w

∂s

∂s

∂v

=s3 − r2s

(r2 + s2)2· u +

r3 − rs2

(r2 + s2)2· (−2).

Substituting r = uv and s = u − 2v into these expressions would give ∂w/∂u and ∂w/∂vsolely in terms of u and v, if desired.

(18) Use a chain rule to find the value of

∂f

∂u

∣∣∣∣u=1,v=−2

and∂f

∂v

∣∣∣∣u=1,v=−2

if f(x, y) = x2y2 − x + 2y; x =√u, y = uv3.

Solution: f is a function of x and y, each of which can be viewed as a function of uand v. (Note that despite v not appearing in the formula for x, x may still be considereda function of u and v: on input (u, v), x outputs

√u.) Using the chain rule

∂f

∂u=

∂f

∂x

∂x

∂u+

∂f

∂y

∂y

∂u

= (2xy2 − 1) · 1/2√u + (2x2y + 2) · v3,

and

∂f

∂v=

∂f

∂x

∂x

∂v+

∂f

∂y

∂y

∂v

= (2xy2 − 1) · 0 + (2x2y + 2) · 3uv2.

When (u, v) = (1,−2), (x, y) = (√u, uv3) = (1,−8). Plugging these value for u, v, x, and

y into the formulas for ∂f/∂u and ∂f/∂v gives

∂f

∂u= (2 · 1 · (−8)2 − 1) · 1/2

√1 + (2 · 12 · (−8) + 2) · (−2)3 = 351/2,

and∂f

∂v= (2 · 1 · (−8)2 − 1) · 0 + (2 · 12 · (−8) + 2) · 3 · 1 · (−2)2 = −168.

1

(30) Suppose that a particle moving along a metal plate in the xy-plane has velocity v = i− 4j(cm/s) at the point (3, 2). Given that the temperature of the plate at points in the xy-planeis T (x, y) = y2 lnx, x ≥ 1, in degrees Celsius, find dT/dt at the point (3, 2).

Solution: Let r(t) = x(t)i + y(t)j be the path traveled by the particle. Then thevelocity of the particle is given by v(t) = r′(t) = x′(t)i + y′(t)j, and the temperatureT (x, y) = T (x(t), y(t)) is a function of t. By the chain rule

dT

dt=

∂T

∂xx′(t) +

∂T

∂yy′(t) = (y2/x) · x′(t) + 2y lnx · y′(t).

When (x, y) = (3, 2), i − 4j = v(t) = r′(t) = x′(t)i + y′(t)j, so x′(t) = 1 and y′(t) = −4.Plugging these values for x, y, x′(t), and y′(t) into the above equation for dT/dt gives

dT

dt

∣∣∣∣(x,y)=(3,2)

= (22/3) · 1 + 2 · 2(ln 3) · (−4) = 4/3− 16 ln 3.

(43) Recall from Formula (6) of Section 15.3 that under appropriate conditions a plucked stringsatisfies the wave equation

∂2u

∂t2= c2

∂2u

∂x2,

where c is a positive constant.

(a) Show that a function of the form u(x, t) = f(x + ct) satisfies the wave equation.

Solution: Here f is assumed to be a twice differentiable function of one variable. Tofind ∂u/∂t, we hold x constant and differentiate f(x+ ct) with respect to t using the chainrule for functions of one variable:

∂u

∂t= f ′(x + ct) · c = cf ′(x + ct),

and in turn∂2u

∂t2= cf ′′(x + ct) · c = c2f ′′(x + ct).

Similarly, differentiating with respect to x gives

∂u

∂x= f ′(x + ct) · 1 = f ′(x + ct),

and∂2u

∂x2= f ′′(x + ct).

So∂2u

∂t2= c2f ′′(x + ct) = c2

∂2u

∂x2.

(b) Show that a function of the form u(x, t) = g(x− ct) satisfies the wave equation.

Solution: Computing as we did above, we obtain

∂u

∂t= g′(x− ct) · (−c) = −cg′(x− ct),

∂2u

∂t2= −cg′′(x− ct) · (−c) = c2g′′(x− ct);

2

and∂u

∂x= g′(x− ct) · 1 = g′(x− ct),

∂2u

∂x2= g′′(x− ct).

So∂2u

∂t2= c2g′′(x− ct) = c2

∂2u

∂x2.

(c) Show that a function of the form u(x, t) = f(x + ct) + g(x − ct) satisfies the waveequation.

Solution: From parts (a) and (b) it suffices to show that if u1 and u2 satisfy the waveequation, then so does u1 + u2. To this end, suppose u1 and u2 both satisfy the waveequation. Then

∂2

∂t2(u1 + u2) =

∂2u1

∂t2+

∂2u2

∂t2

= c2∂2u1

∂x2+ c2

∂2u2

∂x2

= c2(∂2u1

∂x2+

∂2u2

∂x2

)= c2

∂2

∂x2(u1 + u2),

which shows that u1 + u2 satisfies the wave equation as well.

(d) It can be proved that every solution of the wave equation is expressible in the formstated in part (c). Confirm that u(x, y) = sin t sinx satisfies the wave equation in whichc = 1, and then use appropriate trigonometric identities to express this function in theform f(x + t) + g(x− t).

Solution: If u(x, y) = sin t sinx, then

∂u/∂t = cos t sinx,

∂2u/∂t2 = − sin t sinx,

∂u/∂x = sin t cosx,

∂2u/∂x2 = − sin t sinx.

Thus,∂2u/∂t2 = − sin t sinx = 1 · ∂2u/∂x2,

as desired.

Now let f(s) = −(1/2) cos s and g(s) = (1/2) cos s. Using the trig identity cos(a ± b) =cos a cos b∓ sin a sin b, we have

f(x + t) + g(x− t) = −(1/2) cos(x + t) + (1/2) cos(x− t)

=cos(x− t)− cos(x + t)

2

=cosx cos t + sinx sin t− (cosx cos t− sinx sin t)

2

3

= sinx sin t.

(52) Prove: If f , fx, and fy are continuous on a circular region containing A(x0, y0) andB(x1, y1), then there is a point (x∗, y∗) on the line segment joining A and B such that

f(x1, y1)− f(x0, y0) = fx(x∗, y∗)(x1 − x0) + fy(x∗, y∗)(y1 − y0).

This result is the two-dimensional version of the Mean-Value Theorem. [Hint: Express theline segment joining A and B in parametric form and use the Mean-Value Theorem forfunctions of one variable.]

Solution: The straight line segment from A to B is given parametrically by

x(t) = x0 + (x1 − x0)t, y(t) = y0 + (y1 − y0)t, 0 ≤ t ≤ 1.

Let g(t) = f(x(t), y(t)). Since fx and fy are continuous throughout the circular region, fis differentiable throughout this region and g is differentiable on 0 ≤ t ≤ 1. Applying thechain rule to compute g′(t),

g′(t) =d

dtf(x(t), y(t)) = fx(x(t), y(t))x′(t) + fy(x(t), y(t))y′(t)

= fx(x(t), y(t))(x1 − x0) + fy(x(t), y(t))(y1 − y0).

By the Mean-Value Theorem applied to g on the interval 0 ≤ t ≤ 1, there exists t∗ in thisinterval such that

g(1)− g(0)

1− 0= g′(t∗).

Let (x(t∗), y(t∗)) = (x∗, y∗). Observe that g(1) − g(0) = f(x(1), y(1)) − f(x(0), y(0)) =f(x1, y1)− f(x0, y0); and by our formula for g′ above,

g′(t∗) = fx(x(t∗), y(t∗))(x1 − x0) + fy(x(t∗), y(t∗))(y1 − y0)

= fx(x∗, y∗)(x1 − x0) + fy(x∗, y∗)(y1 − y0).

Therefore, equating our formulas for g(1)− g(0) and g′(t∗) gives

f(x1, y1)− f(x0, y0) = fx(x∗, y∗)(x1 − x0) + fy(x∗, y∗)(y1 − y0),

as desired.

(53) Prove: If fx(x, y) = 0 and fy(x, y) = 0 throughout a circular region, then f(x, y) is constanton that region.

Solution: Let (x0, y0) be any point in the region. Let k = f(x0, y0), and let (x1, y1)be any other point in the region. Note that f is differentiable, and therefore continuous,on the circular region because its first partials are constant (and therefore continuous!) onthat region. By the previous exercise there exists a point (x∗, y∗) in the region such that

f(x1, y1)− f(x0, y0) = fx(x∗, y∗)(x1 − x0) + fy(x∗, y∗)(y1 − y0).

Since the first partial derivatives of f are zero, the right side is zero. Therefore, f(x1, y1)−f(x0, y0) = 0, implying that f(x1, y1) = f(x0, y0) = k. Since (x1, y1) was arbitrary,f(x, y) = k for all (x, y) in the region.

4

OHSx XM521 Multivariable Differential Calculus:Homework Solutions §15.5

(12) Show that the surfaces

z =√x2 + y2 and z =

1

10(x2 + y2) +

5

2

intersect at (3, 4, 5) and have a common tangent plane at that point.

Solution: First, it is easy to check that both surfaces pass through the point (3, 4, 5). As

for tangent planes, if z =√x2 + y2, then

zx(x, y) =x√

x2 + y2, zy(x, y) =

y√x2 + y2

,

and the equation of the tangent plane at (3, 4, 5) is

z − 5 = zx(3, 4)(x− 3) + zy(3, 4)(y − 4)

=3

5(x− 3) +

4

5(y − 4).

If z = 110 (x2 + y2) + 5

2 , then

zx(x, y) =x

5, zy(x, y) =

y

5,

and the equation of the tangent plane at (3, 4, 5) is

z − 5 = zx(3, 4)(x− 3) + zy(3, 4)(y − 4)

=3

5(x− 3) +

4

5(y − 4),

which is the same equation obtained for the tangent plane at (3, 4, 5) for the other surface.

(24) Show that if α 6= 1 and β 6= 1, then the local linear approximation of the function f(x, y) =xαyβ at (1, 1) is

xαyβ ≈ 1 + α(x− 1) + β(y − 1).

Solution: In general, the local linear approximation of f at (x0, y0) is

f(x, y) ≈ f(x0, y0) + fx(x0, y0)(x− x0) + fy(x0, y0)(y − y0).

If f(x, y) = xαyβ , then

fx(x0, y0) = αxα−10 yβ0 , fy(x0, y0) = βxα0 yβ−10 .

So the local linear approximation at (1, 1) is

xαyβ ≈(xα0 y

β0 + αxα−10 yβ0 (x− x0) + βxα0 y

β−10 (y − y0)

)∣∣∣∣(x0,y0)=(1,1)

≈ 1 + α(x− 1) + β(y − 1).

There is no need for the assumptions α, β 6= 1.

1

(48) Show that if f is differentiable and z = xf(x/y), then all tangent planes to the graph ofthis equation pass through the origin.

Solution: To avoid notational confustion, let g(x, y) = xf(x/y). Let (x0, y0, g(x0, y0))be a point on the graph of g. Computing the first partials gx and gy gives

gx(x, y) = f(x/y) + xf ′(x/y) · 1/y = f(x/y) + (x/y)f ′(x/y),

gy(x, y) = xf ′(x/y) · (−x/y2) = (−x2/y2)f ′(x/y).

So the tangent plane at (x0, y0, g(x0, y0)) is given by

z = g(x0, y0) + gx(x0, y0)(x− x0) + gy(x0, y0)(y − y0)

= x0f(x0/y0) + (f(x0/y0) + (x0/y0)f ′(x0/y0))(x− x0) + (−x20/y20)f ′(x0/y0)(y − y0).

Substituting x = y = 0, we obtain

z = x0f(x0/y0)− x0f(x0/y0)− x20y0f ′(x0/y0) +

x20y0f ′(x0/y0).

On the right-hand side, the first two terms cancel, as do the last two. Therefore, for thetangent plane at the point (x0, y0, g(x0, y0)), x = y = 0 implies z = 0; so this tangent planepasses throught the origin. Since the point (x0, y0, g(x0, y0)) was arbitrary, all tangentplanes to the graph of g pass through the origin.

(50) Show that the equation of the plane that is tangent to the paraboloid z = x2

a2 + y2

b2 at(x0, y0, z0) can be written in the form

z + z0 =2x0x

a2+

2y0y

b2.

Solution: To avoid notational confustion, let g(x, y) = x2

a2 + y2

b2 . Computing first partialsgives gx(x, y) = 2x/a2 and gy(x, y) = 2y/b2. So the tangent plane at (x0, y0, z0 = g(x0, y0))has equation

z − z0 = gx(x0, y0)(x− x0) + gy(x0, y0)(y − y0)

=2x0a2

(x− x0) +2y0b2

(y − y0)

=2x0x

a2− 2x20

a2+

2y0y

b2− 2y20

b2

=2x0x

a2+

2y0y

b2− 2

(x20a2

+y20b2

).

But z0 = g(x0, y0) =x20

a2 +y20b2 , so the equation for the tangent plane becomes

z − z0 =2x0x

a2+

2y0y

b2− 2z0,

which simplifies to

z + z0 =2x0x

a2+

2y0y

b2

as desired.

2

(51) Prove: If the surfaces z = f(x, y) and z = g(x, y) intersect at P (x0, y0, z0), and if f and gare differentiable at (x0, y0), then the normal lines at P are perpendicular if and only if

fx(x0, y0)gx(x0, y0) + fy(x0, y0)gy(x0, y0) = −1.

Solution: The tangent plane to the graph of f at P has normal vector 〈fx(x0, y0), fy(x0, y0),−1〉.Similarly, the tangent plane to the graph of g at P has normal vector 〈gx(x0, y0), gy(x0, y0),−1〉.Each normal line at P is parallel to the corresponding normal vector. Therefore, the nor-mal lines are perpendicular if and only if the two normal vectors are perpendicular, whichhappens if and only if 〈fx(x0, y0), fy(x0, y0),−1〉 • 〈gx(x0, y0), gy(x0, y0),−1〉 = 0. Butnotice that

〈fx(x0, y0), fy(x0, y0),−1〉 • 〈gx(x0, y0), gy(x0, y0),−1〉 = 0

if and only if

fx(x0, y0)gx(x0, y0) + fy(x0, y0)gy(x0, y0) + 1 = 0

if and only if

fx(x0, y0)gx(x0, y0) + fy(x0, y0)gy(x0, y0) = −1.

3

OHSx XM521 Multivariable Differential Calculus:Homework Solutions §15.6

(22) Find the directional derivative of f at P in the direction of a vector making the counter-clockwise angle θ with the positive x-axis.

f(x, y) =x− yx+ y

; P (−1,−2); θ = π/2.

Solution: A quick calculation shows that ∇f(x, y) = 2(x+y)2 〈y,−x〉, so ∇f(−1,−2) =

29 〈−2, 1〉. The vector j is a unit vector making a counterclockwise angle of π/2 with thepositive x-axis. Therefore, the directional derivative of f at P in the direction of j is

Djf(−1,−2) = ∇f(−1,−2) • j =2

9〈−2, 1〉 • 〈0, 1〉 =

2

9.

(32) Find a unit vector in the direction in which f increases most rapidly at P ; and find therate of change of f at P in that direction.

f(x, y) =x

x+ y; P (0, 2).

Solution: At (x, y), ∇f(x, y) points in the direction of most rapid increase, and that rateof increase is given by ||∇f(x, y)||. In this case, ∇f(x, y) = 1

(x+y)2 〈y,−x〉; so ∇f(0, 2) =14 〈2, 0〉 = 〈1/2, 0〉. The vector i is a unit vector in this direction, and the rate of change off at P in this direction (of maximal increase) is ||∇f(0, 2)|| = ||〈1/2, 0〉|| = 1/2.

(42) Find a unit vector u that is normal at P (2,−3) to the level curve of f(x, y) = 3x2y − xythrough P .

Solution: At (x, y), ∇f(x, y) is normal to the level curve of f through (x, y). In this case∇f(x, y) = 〈6xy− y, 3x2 − x〉; so ∇f(2,−3) = 〈−33, 10〉. A unit vector in this direction is(1/√

1189)〈−33, 10〉.

(52) The temperature at a point (x, y) on a metal plate in the xy-plane is T (x, y) = xy1+x2+y2

degrees Celsius.

(a) Find the rate of change of temperature at (1, 1) in the direction of a = 2i− j.

Solution: It is easy to check that

∇T (x, y) =1

(1 + x2 + y2)2〈y − x2y + y3, x− xy2 + x3〉,

∇T (1, 1) =1

9〈1, 1〉,

and a unit vector u in the direction of a = 2i− j is

u = a/||a|| = (2i− j)/√

5 =2√5i− 1√

5j.

1

So the rate of change of temperature at (1, 1) in the direction of a is

DuT (1, 1) = ∇T (1, 1) • (2√5i− 1√

5j)

=1

9〈1, 1〉 • 〈 2√

5,−1√

5〉

=1

9√

5.

(b) An ant at (1, 1) wants to walk in the direction in which the temperature drops mostrapidly. Find a unit vector in that direction.

Solution: −∇T (x, y) gives the direction of most rapid decrease of T from the point (x, y).

In this case −∇T (1, 1) = 〈−19 ,−19 〉. A unit vector in this direction is 〈−

√2

2 , −√2

2 〉.

(58) (c) Prove that if f and g are differentiable, then ∇(fg) = f∇g + g∇f .

Solution: Expanding the left-hand side using the definition of ∇ and the product rule,then re-grouping gives

∇(fg) = 〈(fg)x, (fg)y〉

= 〈fgx + gfx, fgy + gfy〉

= 〈fgx, fgy〉+ 〈gfx, gfy〉

= f〈gx, gy〉+ g〈fx, fy〉

= f∇g + g∇f,

as desired.

(59) A heat-seeking particle is located at the point P on a flat metal plate whose temperatureat a point (x, y) is T (x, y). Find parametric equations for the trajectory of the particle ifit moves continuously in the direction of maximum temperature increase.

T (x, y) = 5− 4x2 − y2; P (1, 4).

Solution: Let x = x(t), y = y(t) be parametric equations for the trajectory of the particle.We want the particle to move continuously in the direction of maximum temperatureincrease. Equivalently, we want the velocity vector 〈x′(t), y′(t)〉 to point in the directionof ∇f(x(t), y(t)). Since ∇f(x, y) = 〈−8x,−2y〉, the previous condition will be satisfied if〈x′(t), y′(t)〉 = 〈−8x(t),−2y(t)〉 for all t. To find such functions x(t) and y(t), we mustsolve the differential equations

x′(t) = −8x(t), y′(t) = −2y(t).

Because the heat-seeking particle begins at the point P (1, 4), we also require that x(0) = 1and y(0) = 4. The solutions to the differential equations are x(t) = c1e

−8t and y(t) =c2e−2t, where c1 and c2 are any constants. However, to satisfy the initial conditions

x(0) = 1 and y(0) = 4, we must have c1 = 1 and c2 = 4. Therefore,

x(t) = e−8t, y = 4e−2t, t ≥ 0,

2

gives one parametrization of the trajectory of the heat-seeking particle.

Note: There are many other parametrizations. If f(t) is any positive scalar function, thenfunctions x(t), y(t) satisfying

x′(t) = −8f(t)x(t), y′(t) = −2f(t)y(t)

also give a trajectory in which the velocity points in the direction of the gradient. Solvingthese equations (with the initial conditions) gives x(t) = e−8F (t), y(t) = 4e−2F (t), where

F (t) =∫ t

0f(u) du is an antiderivative of f . Since f is nonnegative, F is monotone increasing

and F (0) = 0. So x(t) = e−8F (t), y(t) = 4e−2F (t), t ≥ 0, traces out the same curve asx(t) = e−8t, y = 4e−2t, t ≥ 0, though perhaps at a different rate.

(65) Prove: If f , fx, and fy are continuous on a circular region, and if ∇f(x, y) = 0 throughoutthe region, then f(x, y) is constant on the region. [Hint: See Exercise 52, Section 15.4.]

Solution: Using Exercise 52, Section 15.4, we proved the assertion of Exercise 53, Section15.4, which states: “If fx(x, y) = 0 and fy(x, y) = 0 throughout a circular region, thenf(x, y) is constant on that region.” But ∇f(x, y) = 0 throughout the circular region ifand only if fx(x, y) = fy(x, y) = 0 throughout that region. So the desired result followsimmediately from Exercise 53, Section 15.4.

(66) Prove: If the function f is differentiable at the point (x, y) and if Duf(x, y) = 0 in twononparallel directions, then Duf(x, y) = 0 in all directions.

Solution: Suppose that u and v are nonparallel unit vectors, andDuf(x, y) = Dvf(x, y) =0. Since u and v are not parallel, every vector in 2-space is a “linear combination” of uand v. In other words, every vector w (in 2-space) can be written as αu + βv for some αand β. (Do you see why?) In particular, if w is a unit vector, then there exists α and βsuch that w = αu + βv. So,

Dwf(x, y) = ∇f(x, y) •w

= ∇f(x, y) • (αu + βv)

= ∇f(x, y) • αu +∇f(x, y) • βv

= α∇f(x, y) • u + β∇f(x, y) • v

= αDuf(x, y) + βDvf(x, y).

But by assumption Duf(x, y) = Dvf(x, y) = 0. So Dwf(x, y) = 0. Since w was arbitrary,the directional derivative of f is zero in all directions.

3

OHSx XM521 Multivariable Differential Calculus:Homework Solutions §15.7

(23) Given that the directional derivative of f(x, y, z) at the point (3,−2, 1) in the direction ofa = 2i− j− 2k is −5 and that ||∇f(3,−2, 1)|| = 5, find ∇f(3,−2, 1).

Solution: Let u = a/||a|| = 2/3i− 1/3j− 2/3k be the unit vector in the direction of a.According to the hypotheses,

Duf(3,−2, 1) = ∇f(3,−2, 1)•u = −5,

which implies that

||∇f(3,−2, 1)|| cos θ = ||∇f(3,−2, 1)|| ||u|| cos θ = ∇f(3,−2, 1)•u = −5,

where θ is the angle between ∇f(3,−2, 1) and u. By assumption ||∇f(3,−2, 1)|| = 5,implying that cos θ = −1, θ = π, and ∇f(3,−2, 1) points in the opposite direction of u. So∇f(3,−2, 1) is a vector of magnitude 5 pointing in the direction of −u = −2/3i + 1/3j +2/3k. Consequently, ∇f(3,−2, 1) = −5u = −10/3i + 5/3j + 10/3k.

(28) Find two unit vectors that are normal to the surface sinxz − 4 cos yz = 4 at the pointP (π, π, 1).

Solution: Let F (x, y, z) = sinxz − 4 cos yz. Then sinxz − 4 cos yz = 4 is a level surfaceof F containing the point P . So the vectors ±∇F (P ) are normal to this level surface. Aquick computation shows that

∇F (x, y, z) = Fx(x, y, z)i + Fy(x, y, z)j + Fz(x, y, z)k

= z cosxzi + 4z sin yzj + (x cosxz + 4y sin yz)k,

and ∇F (P ) = −i− πk. Therefore, the two desired unit vectors normal to the surface are

± ∇F (P )

||∇F (P )||= ± 1√

1 + π2(−i− πk).

(56) Let u = rs2 ln t, r = x2, s = 4y + 1, t = xy3. Find ∂u/∂x and ∂u/∂y.

Solution: We view r, s, and t as functions of x and y. According to the chain rule

∂u

∂x=

∂u

∂r

∂r

∂x+∂u

∂s

∂s

∂x+∂u

∂t

∂t

∂x

= (s2 ln t)(2x) + (2rs ln t)(0) + (rs2/t)(y3)

= 2xs2 ln t+ rs2y3/t.

Similarly,

∂u

∂y=

∂u

∂r

∂r

∂y+∂u

∂s

∂s

∂y+∂u

∂t

∂t

∂y

= (s2 ln t)(0) + (2rs ln t)(4) + (rs2/t)(3xy2)

= 8rs ln t+ 3rs2xy2/t.

To express these first partials in terms of x and y only, simply substitute for r, s, and tusing their defining equations r = x2, s = 4y + 1, t = xy3.

1

(61) Let f be a differentiable function of one variable, and let z = f(x+ 2y). Show that

2∂z

∂x− ∂z

∂y= 0.

Solution: Let u = x+ 2y. By the chain rule

∂z

∂x=df

du

∂u

∂x=df

du· 1 =

df

du,

∂z

∂y=df

du

∂u

∂y=df

du· 2 = 2

df

du.

Therefore,

2∂z

∂x− ∂z

∂y= 2

df

du− 2

df

du= 0,

as desired.

(71) Given that the equations u = u(x, y, z), v = v(x, y, z), and w = w(x, y, z) are all differen-tiable, show that

∇f(u, v, w) =∂f

∂u∇u+

∂f

∂v∇v +

∂f

∂w∇w.

Solution: By the chain rule,

fx(u(x, y, z), v(x, y, z), w(x, y, z)) = fuux + fvvx + fwwx,

fy(u(x, y, z), v(x, y, z), w(x, y, z)) = fuuy + fvvy + fwwy,

fz(u(x, y, z), v(x, y, z), w(x, y, z)) = fuuz + fvvz + fwwz.

So

∇f(u(x, y, z), v(x, y, z), w(x, y, z)) = 〈fuux + fvvx + fwwx, fuuy + fvvy + fwwy, fuuz + fvvz + fwwz〉

= fu〈ux, uy, uz〉+ fv〈vx, vy, vz〉+ fw〈wx, wy, wz〉

= fu∇u+ fv∇v + fw∇w

=∂f

∂u∇u+

∂f

∂v∇v +

∂f

∂w∇w.

(77) We showed in Exercise 24 of Section 7.9 that

d

dx

∫ g(x)

h(x)

f(t) dt = f(g(x))g′(x)− f(h(x))h′(x).

Derive this same result by letting u = g(x) and v = h(x) and then differentiating thefunction

F (u, v) =

∫ u

v

f(t) dt

with respect to x.

2

Solution: First, observe that F (u, v) = −∫ v

uf(t) dt. So by the Fundamental Theorem of

Calculus, ∂F/∂u = f(u) and ∂F/∂v = −f(v). Applying the chain rule to compute dF/dxgives

dF

dx=

∂F

∂u

du

dx+∂F

∂v

dv

dx

= f(u)g′(x) + (−f(v)h′(x))

= f(g(x))g′(x)− f(h(x))h′(x).

Since F (u, v) =∫ u

vf(t) dt =

∫ g(x)

h(x)f(t) dt, we’ve shown

d

dx

∫ g(x)

h(x)

f(t) dt = f(g(x))g′(x)− f(h(x))h′(x),

as desired.

(78) Two surfaces f(x, y, z) = 0 and g(x, y, z) = 0 are said to be orthogonal at a point Pof intersection if ∇f and ∇g are nonzero at P and the normal lines to the surfaces areperpendicular at P . Show that if ∇f(x0, y0, z0) 6= 0 and ∇g(x0, y0, z0) 6= 0, then thesurfaces f(x, y, z) = 0 and g(x, y, z) = 0 are orthogonal at the point (x0, y0, z0) if and onlyif

fxgx + fygy + fzgz = 0

at this point. [Note: This is a more general version of the result in Exercise 51 of Section15.5.]

Solution: Let P denote (x0, y0, z0). Suppose ∇f(P ) 6= 0 and ∇g(P ) 6= 0 (and the levelsurfaces f(x, y, z) = 0 and g(x, y, z) = 0 intersect at P ). We know that ∇f(P ) is normalto the level surface of f at P , and therefore ∇f(P ) is parallel to the normal line to thissurface at P . Similarly, ∇g(P ) is parallel to the normal line to the level surface of g at P .So

the surfaces are orthogonal at P

if and only if

their normal lines are perpendicular at P

if and only if

∇f(P ) is perpendicular to ∇g(P )

if and only if

∇f(P )•∇g(P ) = 0

if and only if

〈fx, fy, fz〉•〈gx, gy, gz〉 = 0 at P

if and only if

fxgx + fygy + fzgz = 0 at P.

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OHSx XM521 Multivariable Differential Calculus:Homework Solutions §15.8

(10) Locate all relative maxima, relative minima, and saddle points of f(x, y) = x2 +xy− 2y−2x+ 1.

Solution: A quick calculation shows that fx(x, y) = 2x + y − 2 and fy(x, y) = x − 2.So fx and fy exist everywhere. (In fact, f is differentiable everywhere.) So any relativeextrema, or saddle points, must occur where fx = fy = 0. Setting fy(x, y) = 0, gives x = 2.Plugging this into the equation fy(x, y) = 0 and solving for y gives y = −2. Thereforefx(x, y) = fy(x, y) = 0 only at the point (2,−2). At this point fxx = 2, fyy = 0, andfxy = 1. So at this point D = fxxfyy− (fxy)2 = −1. Since D < 0, (2,−2) is a saddle pointand there are no relative extrema.

Graph of 15.8(10)

(25) Recall from Theorem 6.1.5 that if a continuous function of one variable has exactly onerelative extremum on an interval, then that relative extremum is an absolute extremumon the interval. This exercise shows that this result does not extend to functions of twovariables.

(a) Show that f(x, y) = 3xey − x3 − e3y has only one critical point and that a relativemaximum occurs there.

Solution: f is everywhere differentiable, so any extrema occur at points where fx = fy =0. Computing the necessary partials gives

fx(x, y) = 3ey − 3x2,

fy(x, y) = 3xey − 3e3y,

fxx(x, y) = −6x,

fyy(x, y) = 3xey − 9e3y,

fxy(x, y) = 3ey.

Setting fx = 0 and fy = 0 gives the two equations ey = x2 and x = e2y respectively.Therefore, fx(x, y) = fy(x, y) = 0 implies ey = (e2y)2 = e4y, or 1 = e3y. Since theexponential function is one-to-one, the only y satisfying this equation is y = 0; so x =e2y = e0 = 1. Therefore, f has only one possible relative extremum. At (1, 0), fxx = −6and

D = fxxfyy − (fxy)2 = (−6)(−6)− 32 = 27.

Since fxx < 0 and D > 0, (1, 0) is a relative maximum, and it’s the only extrema of anysort.

(b) Show that f does not have an absolute maximum.

Solution: There are many different ways to prove this. Here’s one: When y = 0,f(x, y) = f(x, 0) = −x3 + 3x− 1. So

limx→−∞

f(x, 0) = limx→−∞

−x3 + 3x− 1 = +∞.

Since the function f(x, 0) (defined for x ∈ R) has no absolute maximum, neither can thefunction f(x, y) (defined for all (x, y) ∈ R2).

Graph of 15.8(25)

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(30) Find the absolute extrema of the function f(x, y) = xey−x2−ey on the rectangular regionR with vertices (0, 0), (0, 1), (2, 1), and (2, 0).

Solution: Since f is differentiable everywhere, the absolute extrema occur either wherefx = fy = 0 or on the boundary of R. Computing the first partials gives fx(x, y) = ey−2xand fy(x, y) = xey− ey. So fx = 0 implies 2x = ey, and fy = 0 implies xey = ey, or x = 1.Therefore, fx = fy = 0 only when x = 1 and y = ln 2, and this is the only possible extremain the interior of R. Now we check the four boundary lines of R.

When x = 0, f(x, y) = f(0, y) = −ey. On the interval 0 ≤ y ≤ 1, this has a minimum aty = 1 and a maximum at y = 0.

When x = 2, f(x, y) = f(2, y) = ey − 4. On the interval 0 ≤ y ≤ 1, this has a minimum aty = 0 and a maximum at y = 1.

When y = 0, f(x, y) = f(x, 0) = x−x2−1. On the interval 0 ≤ x ≤ 2, this has a minimumat x = 2 and a maximum at x = 1/2.

When y = 1, f(x, y) = f(x, 1) = ex − x2 − e. On the interval 0 ≤ x ≤ 2, this has aminimum at x = 0 and a maximum at x = e/2.

So the points to check for extrema are: (1, ln 2), (0, 1), (0, 0), (2, 0), (2, 1), (1/2, 0), (e/2, 1).Plugging each of these into f , we find that f(2, 0) = −3 is the absolute minimum, whilef(1/2, 0) = −3/4 is the absolute maximum.

Graph of 15.8(30)

(36) Find the points on the surface x2 − yz = 5 that are closest to the origin.

Solution: We must find the points (x, y, z) that lie on the surface x2 − yz = 5 and

minimize the function√x2 + y2 + z2. Note that for any nonnegative function g defined on

the surface, the points at which g is minimal are exactly the points at which the function g2

is minimal. Thus, it suffices to minimize the function x2 +y2 +z2 subject to the constraintthat (x, y, z) must lie on the surface x2 − yz = 5. Since on this surface x2 = 5 + yz, weare reduced to finding the minimum of the function f(y, z) = 5 + yz + y2 + z2. Sincex2 can be any nonnegative number, the domain of f is R = {(y, z)

∣∣ 5 + yz ≥ 0}. Notethat this domain is closed but not bounded. It’s easy to compute fy and fz and to seethat fy = fz = 0 only if y = z = 0. It is also easy to check that at (y, z) = (0, 0),D = fyyfzz − (fyz)

2 > 0. Since D is positive and fyy > 0 at (y, z) = (0, 0), this point is arelative minimum of f .

Although this is a relative minimum and the only relative minimum, this does not implythat it must be the absolute minimum. The absolute minimum might occur on the bound-ary curves yz = −5, or the function might not have any absolute minimum. Thus, more isrequired to establish that (0, 0) is indeed the absolute minimum of f . To this end we couldrestrict the domain of f to the closed and bounded region R′ consisting of points (y, z) in R(i.e., (y, z) satisfying yz ≥ −5) such that (y, z) lies in the bounded region y2 +z2 ≤ 7. (Wewant to choose a bounded region such that on its boundary f is larger than the allegedminimum of 5.)

Note that if (y, z) is on the boundary circle of R′, then y2 +z2 = 7, and yz+5 ≥ 0 becausewe’re considering only points within R. (Actually, it’s not hard to check that the disky2 + z2 ≤ 7 lies completely within the region R.) Therefore, for any point (y, z) on theboundary of R′, we have

f(y, z) = (5 + yz) + (y2 + z2) ≥ 0 + 7 = 7 > 5.

Since f must assume an absolute minimum on this closed and bounded domain R′, thatabsolute minimum is 5, and it occurs at (0, 0). Finally, for any (y, z) in R that lies outsideof R′, it should be clear that f(y, z) ≥ y2 + z2 > 7. Therefore, 5 must be the absoluteminimum of f on the entire domain R.

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Alternatively, to see that 5 is the absolute minimum of f , note that y2 + z2 + yz ≥ 0 if yand z have the same sign or if one of them is zero. Now suppose y and z are nonzero andhave different signs. Then y2+z2+yz = y2+z2−|y||z| ≥ y2+z2−2|y||z| = (|y|−|z|)2 ≥ 0.So for all y and z, y2 +z2 +yz ≥ 0, implying that f(y, z) = 5+y2 +z2 +yz ≥ 5. Therefore,5 must be the absolute minimum of f .

Consequently, (±√

5, 0, 0) are the points on the surface x2 − yz = 5 that are closest to theorigin.

Graph of 15.8(36)

(44) A length of sheet metal 27 inches wide is to be made into a water trough by bending uptwo sides as shown in the accompanying figure. Find x and φ so that the trapezoid-shapedcross section has a maximum area.

Solution: The height of the trough is h = x sinφ and the bases have lengths B1 = 27−2xand B2 = 27− 2x+ 2x cosφ. So the area is

A(x, φ) = (1/2)(B1 +B2)h = 27x sinφ− 2x2 sinφ+ x2 sinφ cosφ.

We want to find the maximum ofA on the regionR = {(x, φ)∣∣ 0 ≤ x ≤ 27/2, 0 ≤ φ ≤ π/2}.

(It should be clear that for any given x, the maximal area occurs for some φ ≤ π/2.) Thepartial derivatives of A exist everywhere, so to find the maximum we find the interiorpoints of R at which ∇A = 0, and then examine possible extrema on the boundary of R.To this end,

Ax(x, φ) = (27− 4x) sinφ+ 2x sinφ cosφ,

Aφ(x, φ) = 27x cosφ− 2x2 cosφ+ x2 cos 2φ.

Solving Ax = 0 for x gives,

x =27

4− 2 cosφ.

Plugging this value into Aφ and setting Aφ equal to zero gives

Aφ =(27)2

4− 2 cosφcosφ− 2(27)2

(4− 2 cosφ)2cosφ+

(27)2

(4− 2 cosφ)2cos 2φ = 0.

Multiplying through by 4− 2 cosφ, simplifying, and solving gives cosφ = 1/2, or φ = π/3.Therefore,

x =27

4− 2 cosπ/3= 9.

Direct computation shows that at (x, φ) = (9, π/3), Axx < 0 and D = AxxAφφ −A2xφ > 0.

Therefore, (x, φ) = (9, π/3) is a relative maximum, and A(9/π/3) = 35√

3/4 = 243√

3/5.

Now we check the boundary of the region R. When φ = 0, or x = 0, the area is zero.When x = 27/2, the area is

A(27/2, φ) =(27)2 cosφ sinφ

4=

(27)2 sin 2φ

8.

This has a maximum of (27)2/8 when φ = π/4. So A(27/2, π/4) = (27)2/8. Checkingthe last boundary line, φ = π/2, we obtain A(x, π/2) = 27x− 2x2, which has a maximumof (27)2/8 at x = 27/4. Finally, you can verify that 243

√3/4 > (27)2/8; so the absolute

maximum of A on the region R is A(9, π/3) = 243√

3/4.

(54) Find:

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(a) a continuous function f(x, y) that is defined on the entire xy-plane and has no absoluteextrema on the xy-plane.

Solution: There are many possible solutions. Any function whose graph is a plane notparallel to the xy-plane will have arbitrarily large positive outputs as well as arbitrarilylarge negative outputs. For example, f(x, y) = x.

(b) a function f(x, y) that is defined everywhere on the rectangle 0 ≤ x ≤ 1, 0 ≤ y ≤ 1and has no absolute extrema on the rectangle.

Solution: Again there are many possible solutions. Here’s one: define f(x, y) by

f(x, y) =

{x, if 0 < x < 1,1/2, if x = 0 or x = 1.

Observe that the range of f is the open interval 0 < z < 1. Also, outputs of f get arbitrarilyclose to 0, but 0 is not an output of f ; so f has no absolute minimum. Similarly, outputsof f get arbitrarily close to 1, but 1 is not an output of f ; so f has no absolute maximum.Note that the rectangular region {(x, y)

∣∣ 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} is closed and bounded; soany continuous function must have an absolute maximum and absolute minimum on thisdomain. Therefore, any function that satisfies the desired conditions cannot be continuousthroughout this closed and bounded region.

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OHSx XM521 Multivariable Differential Calculus:Homework Solutions §15.9

(8) Use Lagrange multipliers to find the maximum and minimum values of f(x, y, z) = 3x +6y + 2z subject to the constraint 2x2 + 4y2 + z2 = 70.

Solution: Let g(x, y, z) = 2x2 + 4y2 + z2 − 70. We want to find the maximum andminimal values of f subject to the constraint g(x, y, z) = 0. To use the Method of Lagrangemultipliers, we look for the (constrained) extrema at points (x, y, z) where ∇f(x, y, z) =λ∇g(x, y, z). Computing ∇f and ∇g, this vector equation reduces to the three scalarequations

3 = λ4x,6 = λ8y,2 = λ2z.

Solving these equations for x, y, and z respectively in terms of λ, plugging these val-ues into the constraint equation g(x, y, z) = 0, and solving for λ gives λ = ±1/4. Us-ing the 3 equations above, λ = 1/4 implies (x, y, z) = (3, 3, 4); and λ = −1/4 implies(x, y, z) = (−3,−3,−4). From the formula for f it should be clear that f(3, 3, 4) = 35 isthe (constrained) maximum and f(−3,−3,−4) = −35 is the (constrained) minimum.

(Note: The surface defined by 2x2 + 4y2 + z2 = 70 is a closed and bounded set—an el-lipsoid. Since f is continuous, the Extreme Value Theorem implies that f must have anabsolute minimum and absolute maximum on the ellipsoid; and the theory underlying La-grange Multipliers guarantees that any such max and min will be solutions to the Lagrangeequations.)

(14) Find the point on the plane 4x+ 3y + z = 2 that is closest to (1,−1, 1).

Solution: We want to minimize the distance from a point (x, y, z) to (1,−1, 1) subject tothe constraint 4x+ 3y+ z = 2. This can be accomplished by minimizing the square of thedistance. In other words, we must minimize the function f(x, y, z) = (x−1)2+(y+1)2+(z−1)2 subject to the constraint g(x, y, z) = 0, where g(x, y, z) = 4x+ 3y + z − 2. Computing∇f and ∇g, and setting ∇f(x, y, z) = λ∇g(x, y, z) gives the three scalar equations

2(x− 1) = λ4,2(y + 1) = λ3,2(z − 1) = λ.

Solving for x, y, and z gives x = 1 + 2λ, y = −1 + 3λ/2, and z = 1 + λ/2. Plugging theseinto the constraint equation g(x, y, z) = 0 and solving yields λ = 0. From the three scalarequations above, this implies that x = 1, y = −1, and z = 1 — which happens to be thepoint that distances are being measured from! Since the distance from this point to itselfis zero, this point must give the (constrained) minimum.

(18) Suppose that the temperature at a point (x, y) on a metal plate is T (x, y) = 4x2−4xy+y2.An ant, walking on the plate, traverses a circle of radius 5 centered at the origin. Whatare the highest and lowest temperatures encountered by the ant?

Solution: The circle of radius 5 centered at the origin has equation x2 + y2 = 25. So wewant to find the maximum and minimum of the function T (x, y) subject to the constraintg(x, y) = 0, where g(x, y) = x2 + y2 − 25. Using Lagrange multipliers, we look for points

1

(x, y) such that ∇T (x, y) = λ∇g(x, y). Computing ∇T and ∇g, this vector equation isequivalent to the pair of scalar equations

8x− 4y = λ2x,−4x+ 2y = λ2y.

Multiplying the second equation by 2, then adding the two equations gives 0 = λx+ 2λy.So either x = −2y or λ = 0. Plugging x = −2y into the constraint equation g(x, y) = 0and solving for y gives y = ±

√5, which implies x = ∓2

√5. On the other hand, if λ = 0,

then either of the pair of scalar equations implies that y = 2x. Plugging this into theconstraint equation g(x, y) = 0 and solving for x gives x = ±

√5, which implies y = ±2

√5.

Therefore the only points where possible (constrained) maxima or minima can occur are(2√

5,−√

5), (−2√

5,√

5), (√

5, 2√

5), and (−√

5,−2√

5). Checking the values of T at thesepoints, we find that T (

√5, 2√

5) = T (−√

5,−2√

5) = 0 is the (constrained) minimum, whileT (2√

5,−√

5) = 125 is the (constrained) maximum.

(30) The accompanying figure shows the intersection of the elliptic paraboloid z = x2 +4y2 andthe right circular cylinder x2 + y2 = 1. Use Lagrange multipliers to find the highest andlowest points on the curve of intersection.

Solution: The highest and lowest points on the curve of intersection correspond to thosepoints with the maximal and minimal z values respectively. Thus, we want to find themaximum and minimum of the function z(x, y) = x2 + 4y2 subject to the constraintx2+y2 = 1. As usual let g(x, y) = x2+y2−1 so that the constraint equation is g(x, y) = 0.Using Lagrange multipliers, we look for points (x, y) such that ∇z(x, y) = λ∇g(x, y).Computing ∇z and ∇g, this vector equation is equivalent to the pair of scalar equations

2x = λ2x,8y = λ2y,

which simplify to x = λx and 4y = λy. Solving the first of these two equations givesx = 0 or λ = 1. Plugging x = 0 into the constraint equation g(x, y) = 0 gives y = ±1.On the other hand, if λ = 0, then the equation 4y = λy implies y = 0, in which casex = ±1. Therefore, the four possible (constrained) extrema are (0,±1) and (±1, 0), justas the picture suggests. Plugging these values into z(x, y) we find that z(0,±1) = 4 is the(constrained) maximum, while z(±1, 0) = 1 is the (constrained) minimum.

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