Mathematics 205
HWK 11 Solutions
Section 14.7 p687
p
Problem 7, §14.7, p687.
For the function f (x, y) =
x2 + y 2 , calculate all four partial
derivatives and verify that fxy = fyx (where defined and continuous).
Solution.
fx (x, y) =
fy (x, y) =
fxy (x, y) =
fxx (x, y) =
fyx (x, y) =
fyy =
∂ p 2
2x
x
( x + y2 ) = p
=p
∂x
2 x2 + y 2
x2 + y 2
∂ p 2
y
( x + y2 ) = p
∂y
x2 + y 2
h
i
∂
1
1
2y
−xy
x(x2 + y 2 )− 2 ) = x(− )
3 =
3
∂y
2 (x2 + y 2 ) 2
(x2 + y 2 ) 2
p
Ã
!
x2 + y 2 − x √ 2x 2
∂
x
y2
x +y
p
=
=
3
∂x
x2 + y 2
(x2 + y 2 ) 2
x2 + y 2
i
1
∂ h 2
−xy
y(x + y 2 )− 2 =
3 = fxy (x, y)
2
∂x
(x + y 2 ) 2
Ã
!
∂
y
x2
p
=
3
∂y
(x2 + y 2 ) 2
x2 + y 2
Problem 9, §14.7, p687. For the function f (x, y) = 3 sin 2x cos 5y, calculate all four partial
derivatives and verify that fxy = fyx (where defined and continuous).
Solution.
fx (x, y) =
fy (x, y) =
fxx(x, y) =
fxy (x, y) =
fyx (x, y) =
fyy (x, y) =
∂
(3 sin 2x cos 5y) = 3(cos 5y)(cos 2x)(2) = 6 cos 2x cos 5y
∂x
∂
(3 sin 2x cos 5y) = 3(sin 2x)(− sin 5y)(5) = −15 sin 2x sin 5y
∂y
∂
(6 cos 2x cos 5y) = −12 sin 2x cos 5y
∂x
∂
(6 cos 2x cos 5y) = −30 cos 2x sin 5y
∂y
∂
(−15 sin 2x sin 5y) = −30 cos 2x sin 5y = fxy (x, y)
∂x
∂
(−15 sin 2x sin 5y) = −75 sin 2x cos 5y
∂y
Page 1 of 8
A. Sontag
October 14, 2003
Math 205 HWK 11 Solns continued
§14.7 p687
Problem 13, §14.7, p687.
cos(x + 3y).
Find the quadratic Taylor polynomial about (0, 0) for the function
Solution. Using the notation f (x, y) = cos(x + 3y), we have
fx (x, y) = − sin(x + 3y)
fy (x, y) = −3 sin(x + 3y)
fxx (x, y) = − cos(x + 3y)
fxy (x, y) = −3 cos(x + 3y) = fyx (x, y)
fyy (x, y) = −9 cos(x + 3y)
Evaluating at the point (0, 0), we find
f (0, 0) = cos 0 = 1
fx (0, 0) = − sin 0 = 0,
fxx (0, 0) = − cos 0 = −1,
fy (0, 0) = −3 sin 0 = 0
fxy (0, 0) = fyx (0, 0) = −3 cos 0 = −3,
fyy (0, 0) = −9 cos 0 = −9
Using these values in the formula
f (0, 0) + fx (0, 0)x + f(0, 0)y +
¢
1 ¡
fxx (0, 0)x2 + 2fxy (0, 0)xy + fyy (0, 0)y 2
2!
we find the desired Taylor polynomial to be
Q(x, y) = 1 + 0x + 0y +
¢
1¡
x2
9y 2
−1x2 − 6xy − 9y 2 = 1 −
− 3xy −
2
2
2
u2
Recall that the Maclaurin series for the function cos u begins with the expression 1 − . If we use
2
u = x + 3y, the quadratic polynomial that we get is
1
1 − (x + 3y)2
2
which can be expanded to the form
1
1 − (x2 + 6xy + 9y 2 )
2
which agrees, as we would want it to, with the Taylor polynomial we just found.
Page 2 of 8
A. Sontag
October 14, 2003
Math 205 HWK 11 Solns continued
§14.7 p687
Problem 15, §14.7, p687.
1
1+2x−y .
Find the quadratic Taylor polynomial about (0, 0) for the function
Solution. Writing f (x, y) =
1
= (1 + 2x − y)−1 and differentiating, we have
1 + 2x − y
−2
(1 + 2x − y)2
fx (x, y) = −(1 + 2x − y)−2 (2) =
fy (x, y) = −(1 + 2x − y)−2 (−1) =
1
(1 + 2x − y)2
fxx(x, y) = (−2)(−2)(2)(1 + 2x − y)−3 =
fxy (x, y) = (−2)(−2)(−1)(1 + 2x − y)−3 =
−4
= fyx (x, y)
(1 + 2x − y)3
fyy (x, y) = (−2)(1 + 2x − y)−3 (−1) =
Evaluating at x = 0, y = 0, we find
f (0, 0) = 1,
fxx (0, 0) = 8,
fx (0, 0) = −2,
8
(1 + 2x − y)3
2
(1 + 2x − y)3
fy (0, 0) = 1
fx,y (0, 0) = fyx (0, 0) = −4,
fyy (0, 0) = 2.
Using these numbers in the formula
f (0, 0) + fx (0, 0)x + f(0, 0)y +
we obtain the desired Taylor polynomial:
¢
1 ¡
fxx (0, 0)x2 + 2fxy (0, 0)xy + fyy (0, 0)y 2
2!
Q(x, y) = 1 − 2x + y +
or, more simply,
¢
1 ¡ 2
8x − 8xy + 2y 2
2!
Q(x, y) = 1 − 2x + y + 4x2 − 4xy + y 2 .
1
begins with the quadratic expression 1 + u + u2 . If we
1−u
use u = y − 2x in this expression, we obtain
Recall that the Maclaurin series for
1 + (y − 2x) + (y − 2x)2
which can be expanded and rearranged to read
1 − 2x + y + 4x2 − 4xy + y 2
the same as our 2-variable Taylor polynomial.
Page 3 of 8
A. Sontag
October 14, 2003
Math 205 HWK 11 Solns continued
§14.7 p687
Problem 19, §14.7, p687. Use the given level curves of the function z = f (x, y) to decide the
sign (positive, negative, or zero) of each of the following partial derivatives at the point P indicated
on the diagram. (The level curves in this case are vertical lines. The indicated z-values, from left
to right, are 1, 2, 3, 4, 5. The spacing between the lines decreases a bit from left to right, so that
the contour lines for z = 4 and z = 5 are closer together than those for z = 1 and z = 2. The point
P is more or less in the middle of the diagram, on the contour line where z = 3.)
(a) fx (P )
(b) fy (P )
(c) fxx (P )
(d) fyy (P )
(e) fxy (P )
Solution. (a) Since the z-values increase as x increases (i.e. as we move to the right), fx (P ) is
positive.
(b) Since vertical lines are contour lines, the z-values remain constant when x is held constant.
Therefore fy (P ) is zero. In fact fy (x, y) = 0 everywhere. [In other words, the values of f (x, y)
do not change when y changes, so f (x, y) can be viewed as a function of x alone, which makes
fy (x, y) = 0 everywhere.]
(c) We already know that fx is everywhere positive, or in other words that f (x, y) increases as x
increases with y held constant. Moreover, the spacing of the contours tells us that as x increases,
f increases at an increasing rate. [The contours on the right are closer together than those on the
left, for equal increments of contour level.] Therefore fxx (P ) is positive.
(d) Since fy (x, y) = 0 everywhere, and the derivative of the constant function 0 is also the constant
function 0, fyy (x, y) = 0 everywhere. In particular, fyy (P ) = 0.
(e) To evaluate fxy (P ), we would find the value at P of the partial with respect to y of the derivative
fx (x, y). As was already noted in part (a), f (x, y) is essentially a function of x alone. This makes
fx (x, y) a function of x alone, which makes fxy (x, y) = 0 everywhere. Therefore fxy (P ) = 0.
Problem 25, §14.7, p687. Use the given level curves of the function z = f (x, y) to decide the
sign (positive, negative, or zero) of each of the following partial derivatives at the point P indicated
on the diagram. (The level curves in this case are straight lines running more or less parallel to
→ −
−
→
the vector − i + j . The indicated z-values, from lower left to upper right, are 5, 4, 3, 2, 1. The
spaces between the lines appear to be equal. The point P is more or less in the middle of the
triangular diagram, on the contour line where z = 3.)
(a) fx (P )
(b) fy (P )
(c) fxx (P )
(d) fyy (P )
(e) fxy (P )
Solution. (a) As (x, y) moves straight to the right through P , the z-values decrease. Therefore
fx (P ) is negative.
(b) As (x, y) moves straight upward through P , the z-values also decrease. Therefore fy (P ) is
Page 4 of 8
A. Sontag
October 14, 2003
Math 205 HWK 11 Solns continued
§14.7 p687
negative.
(c) Since the contours (for equal increments of contour level) are equally spaced, this function must
be linear. Therefore each of the functions fx and fy is a constant function. Hence fxx (x, y) = 0
everywhere. In particular, fxx (P ) = 0.
(d),(e). The same reasoning as for part (c) shows that fyy (P ) = 0, as does fxy (P ).
√
Problem 29, §14.7, p687. For the function f (x, y) = x + 2y, find the linear, L(x, y), and
quadratic Q(x, y), Taylor polynomials valid near (1, 0). Compare the values of the approximations
L(0.9, 0.2) and Q(0.9, 0.2) with the exact value of the function f (0.9, 0.2).
Solution. The linear approximation will be: for (x, y) near (1, 0),
f (x, y) ≈ L(x, y) = f (1, 0) + fx (1, 0)(x − 1) + fy (1, 0)(y − 0).
Similarly, the quadratic approximation will be: for (x, y) near (1, 0),
1
1
f (x, y) ≈ Q(x, y) = L(x, y) + fxx (1, 0)(x − 1)2 + fxy (1, 0)(x − 1)(y − 0) + fyy (1, 0)(y − 0)2 .
2
2
We have
f (x, y) =
p
1
x + 2y = (x + 2y) 2
fx (x, y) =
1
1
(x + 2y)− 2
2
1
1
1
(x + 2y)− 2 (2) = (x + 2y)− 2
2
µ ¶2
1
3
fxx (x, y) = −
(x + 2y)− 2
2
µ ¶2
3
3
1
1
fxy (x, y) = fyx (x, y) = −
(x + 2y)− 2 (2) = − (x + 2y)− 2
2
2
fy (x, y) =
1
3
3
fyy (x, y) = − (x + 2y)− 2 (2) = −(x + 2y)− 2
2
Therefore
f (1, 0) =
fx (1, 0) =
1
fxx (1, 0) = − ,
4
1
,
2
√
1=1
fy (1, 0) = 1
1
fxy (1, 0) = − ,
2
fyy (1, 0) = −1.
For the linear approximation, we have, for (x, y) near (1, 0),
1
f (x, y) ≈ 1 + (x − 1) + y
2
Page 5 of 8
A. Sontag
October 14, 2003
Math 205 HWK 11 Solns continued
§14.7 p687
which gives
1
f (0.9, 0.2) ≈ L(0.9, 0.2) = 1 + (−0.1) + 0.2 = 1 − 0.05 + 0.2 = 1.15.
2
Similarly, for the quadratic approximation we have, for (x, y) near (1, 0)
1
1 1
1
1
f (x, y) ≈ 1 + (x − 1) + y − ( )( )(x − 1)2 − (x − 1)y − y 2
2
2 4
2
2
which gives
f (0.9, 0.2) ≈ Q(0.9, 0.2)
1
1
1
1
= 1 + (−0.1) + 0.2 − (−0.1)2 − (−0.1)(0.2) − (0.2)2
2
8
2
2
= 1.15 − 0.00125 + 0.01 − 0.02
= 1.16 − 0.02125
= 1.13875.
For the exact value, my calculator gives
f (0.9, 0.1) =
√
0.9 + 0.4 =
√
1.3 = 1.14018
so the linear approximation is a little high, the quadratic approximation is a little low, and the
quadratic approximation is closer to the exact value than the linear one is.
Problem 32, §14.7, p687.
answer.
If z = f (x) + yg(x), what can you say about zyy ? Explain your
Solution. Since f (x) and g(x) depend on x alone, each of them can be viewed as a constant
when we take partial derivatives with respect to y. Therefore zy = 0 + g(x) and zyy = 0. In other
words, zyy is the constant function 0.
Page 6 of 8
A. Sontag
October 14, 2003
Math 205 HWK 11 Solns continued
§14.7 p687
Problem 34, §14.7, p687. The figure below shows a graph of z = f (x, y).
Is fxx (0, 0) positive or negative? Is fyy (0, 0) positive or negative? Give reasons for your answers.
Solution. The section of x = f (x, y) corresponding to fixed y = 0 appears to be a curve that
is concave upward, and has z decreasing (as x increases) until x = 0 and then has z increasing
as x increases. The fact that this curve is concave upward except possibly at x = 0 tells us that
fxx (0, 0) is either positive or perhaps zero.
Similarly, the section corresponding to fixed x = 0 appears to be a curve that is concave downward,
has z increasing as y increases until y = 0 and then has z decreasing. The concavity tells us that
fyy (0, 0) must be negative or perhaps zero.
Page 7 of 8
A. Sontag
October 14, 2003
Math 205 HWK 11 Solns continued
§14.7 p687
Problem 35, §14.7, p687. A contour diagram for the smooth function z = f (x, y) is shown
below. (This sketch also includes the required arrows for part (d), not shown in the text.)
(a) Is z an increasing or decreasing function of x? Of y? Explain.
(b) Is fx positive or negative? How about fy ? Explain.
(c) Is fxx positive or negative? How about fyy ? Explain.
(d) Sketch the direction of grad f (or ∇f ) at points P and Q.
(e) Is grad f longer at P or at Q? How do you know?
Solution.
(a) With y fixed (so with (x, y) restricted to a horizontal line), z increases as x
increases. In other words, z is an increasing function of x. On vertical lines, however, z decreases
as y increases, so z is a decreasing function of y.
(b) From part (a), we know that fx must be positive (or perhaps occasionally zero, but that doesn’t
seem to happen here). Part (a) also tells us fy is negative (perhaps occasionally zero).
(c) To study fyy examine what happens on vertical lines. Take x = 3, for instance, and/or x = 2, 4
and plot the z values for various values of y. The z values decrease, as we already noted, but the
higher up we are, the faster the z values seem to decrease, since the contour lines are more densely
spaced as we move up. (A one-unit change in y near the bottom gives about a one-unit change in
z near the bottom, while a one-unit change in y near the top gives more than a one-unit change
in z. So the higher y is, the faster z is dropping. Altogether, this would give us a graph that is
decreasing and concave downward. The downward concavity means fyy must be negative. (Near
the top, this denser spacing looks like it might not hold. Probably we’re dealing with a poorly
executed sketch here. If not, then the sign of the second derivative will also change near the top.)
Similarly, we can study fxx by examining what happens on horizontal lines. As we move to the
right, the z values increase, as we already noted, but they seem to be increasing at a faster rate as
we move to the right. This would give us an increasing graph that is concave upward. So fyy is
positive.
Page 8 of 8
A. Sontag
October 14, 2003
Math 205 HWK 11 Solns continued
§14.7 p687
(d) The vector ∇f (Q) should be perpendicular to the contour through Q, and it should point
toward values of f that are higher than the value at Q. Similarly, ∇f (P ) should be perpendicular
to the contour through P , and it should point toward higher f values than the value at P . The
(red) arrows drawn in on the graph above show the directions.
(e) Near P the contours are more densely spaced than they are at Q. This means that f is changing
faster at P than it is at Q. Therefore grad f should be longer (have greater magnitude) at P than
it does at Q. In other words, ∇f (P ) is longer than ∇f (Q).
Page 9 of 8
A. Sontag
October 14, 2003