MA 102 (Mathematics II)
Department of Mathematics, IIT Guwahati
Answers: Tutorial Sheet No. 5
February 15, 2017
(1) Let f : R2 → R be differentiable at (0, 0). Suppose that for U := (3/5, 4/5) and V :=
√
√
√
(1/ 2, 1/ 2), we have DU f (0, 0) = 12 and DV f (0, 0) = −4 2. Then determine fx (0, 0)
and fy (0, 0).
Answer. fx (0, 0) = −92 and fy (0, 0) = 84.
(2) Find the direction where the directional derivative is greatest for the function f (x, y) =
3x2 y 2 − x4 − y 4 at the point (1, 2).
Answer.
√1 (î
2
− ĵ).
y
1
ln(x2 + y 2 ) + tan−1 ( ), P = (1, 3). Find the direction in which f (x, y) is
2
x
increasing the fastest at P . Find the derivative of f (x, y) in this direction.
(3) Let f (x, y) =
√
√
Answer. The direction is U = −1/ 5î + 2/ 5ĵ. The derivative in the direction of U is
√
fU (1, 3) = fx (1, 3)u1 + fy (1, 3)u2 = 1/ 5
(4) A heat-seeking bug is a bug that always moves in the direction of the greatest increase
in heat. Find the direction along which the heat-seeking bug will move when it is placed
at the point (2, 1) on a metal plate heated so that the temperature at (x, y) is given by
T (x, y) = 50y 2 e
−1
(x2 +y 2 )
5
.
√
√
Answer. The bug will move in the direction −1/ 5 î + 2/ 5 ĵ.
(5) Let f (x, y, z) = x2 + 2xy − y 2 + z 2 . Find the gradient of f at (1, −1, 3) and the equations
of the tangent plane and the normal line to the surface f (x, y, z) = 7 at (1, −1, 3).
∂f
∂f
Answer. We have ∇f (1, −1, 3) = ∂f
(1,
−2,
3),
(1,
−1,
3),
(1,
−1,
3)
= (0, 4, 6). The
∂x
∂y
∂z
tangent plane to the surface f (x, y, z) = 7 at the point (1, −1, 3) is given by
2y + 3z = 7.
The Normal Line to the surface f (x, y, z) = 7 at the point (1, −1, 3) is given by x = 1, y =
−1 + 4t, z = 3 + 6t, t ∈ R.
(6) Find DU f (2, 2, 1), where f (x, y, z) = 3x − 5y + 2z and U is the unit vector in the direction
of outward normal to the sphere x2 + y 2 + z 2 = 9 at (2, 2, 1).
Answer. We have U =
DU f (2, 2, 1) = ∇f (2, 2, 1)
(2,2,1)
=
22 +22 +12
• U = 63 − 10
3
√
( 23 , 23 , 31 ) and ∇f (2, 2, 1) = (3, −5, 2). Therefore,
+
2
3
= − 23 .
2
(7) At what rate is the area of a rectangle changing if its length is 8 ft and increasing at 3 ft/s
while its width is 6 ft and increasing at 2 ft/s?
Answer. 34 square feet/sec.
(8) Find equations for the tangent plane and the normal line to the level surface x2 +y 2 +z 2 = 4
√
at the point P0 = (−1, 1, 2)
√
Answer. Equation of the tangent plane is x − y − 2z + 4 = 0. Equation of the normal
√
√
line is (x, y, z) = (−1, 1, 2) + t(−2, 2, 2 2), t ∈ R.
(9) Find equations for the tangent plane and normal line to the surface z = 6 − 3x2 − y 2 at
the point P0 = (1, 2, −1).
Answer. Equation of the tangent plane is 6x + 4y + z − 13 = 0. Equation of the normal
line is (x, y, z) = (1, 2, −1) + t(6, 4, 1), t ∈ R.
(10) Find the equation of the tangent plane to the graphs of the following functions at the given
point:
(a) f (x, y) = x2 − y 4 + exy at the point (1, 0, 2)
√ π
y
(b) f (x, y) = tan−1 at the point (1, 3, ).
x
3
Answer.
(a) z = 2x + y.
√
(b) 3 3x − 3y + 12z − 4π = 0.
(11) Check the following functions for differentiability, and then find the Jacobian Matrix.
(a) f (x, y) = (ex+y + y, xy 2 ) (b) f (x, y) = (x2 + cos y, ex y)
(c) f (x, y, z) = (zex , −yez ).
Answer. All the three functions are differentiable. The Jacobian matrices are
"
(a) Jf (x, y) =
y2
"
(b) Jf (x, y) =
ex+y ex+y + 1
2x
#
.
2xy
#
− sin y
.
yex
ex
"
#
zex 0
ex
(c) Jf (x, y, z) =
.
0 −ez −yez
(12) Let z = x2 + y 2 , and x = 1/t, y = t2 . Compute
dz
by (a) expressing z explicitly in terms
dt
of t and (b) chain rule.
Answer. −2t−3 + 4t3 .
2
(13) Let w = 4x + y 2 + z 3 and x = ers , y = log
r+s
∂w
, z = rst2 . Find
.
t
∂s
3
2
Answer. 8rsers +
(14) If w =
√
2
r+s
log r+s
+ 3r3 s2 t6 .
t
x + yz 3 , x(r, s) = 1 + r2 + s2 , y(r, s) = rs, z(r, s) = 3r, then find ∂w/∂r and
∂w/∂s using the chain rule.
Answer.
r
∂w
=√
+ 108r3 s.
2
2
∂r
1+r +s
s
∂w
=√
+ 27r4 .
2
2
∂s
1+r +s
(15) Let f (x, y) = x(1 + xy)2 for (x, y) ∈ R2 . Calculate all the first order and second order
partial derivatives of f in R2 .
Answer. Easy.
(16) For the following functions, compute the mixed partial derivatives at all points in R2 .
Further find out at each point, whether the mixed derivatives are equal or not?
(a) f (x, y) = x sin y + y sin x + xy
xy(x2 − y 2 )
for (x, y) 6= (0, 0) and f (0, 0) = 0
(b) f (x, y) =
x2 + y 2
Answer. (a)
fx (x, y) = sin y + y cos x + y
fy (x, y) = x cos y + sin x + x
fxy (x, y) = 1 + cos y + cos x
fyx = 1 + cos y + cos x.
Mixed derivatives are equal everywhere.
(b) We have fx (0, 0) = 0 = fy (0, 0).
For (x, y) 6= (0, 0), we have
fx (x, y) =
x4 y + 4x2 y 3 − y 5
(x2 + y 2 )2
fy (x, y) =
x5 − 4x3 y 2 − xy 4
(x2 + y 2 )2
x8 + 10x6 y 2 − 10x2 y 6 − y 8
fxy (x, y) =
(x2 + y 2 )4
fyx (x, y) =
x8 + 10x6 y 2 − 10x2 y 6 − y 8
.
(x2 + y 2 )4
4
Again,
fx (0, t) − fx (0, 0)
= −1
t→0
t
fy (t, 0) − fy (0, 0)
fyx (0, 0) = lim
= 1.
t→0
t
fxy (0, 0) = lim
Hence, mixed partial derivatives are equal at every (x, y) 6= (0, 0) only.
(17) Let F : R2 → R3 be defined by F (x, y) = (sin x cos y, sin x sin y, cos x cos y). Show that
F is differentiable in R2 and find its Jacobian matrix.
Answer. The Jacobian matrix is given by


cos x cos y − sin x sin y


.
JF (x, y) = 
cos
x
sin
y
sin
x
cos
y


− sin x cos y − cos x sin y
(18) Using Taylor’s formula find the quadratic and cubic approximations of the function f (x, y) =
ex cos(y) near the origin.
(19) Find the first three terms in the Taylor’s formula for the function f (x, y) = cos x cos y
at origin. Find a quadratic approximation of f near the origin. How accurate is the
approximation if |x| ≤ 0.1 and |y| ≤ 0.1?
(20) Find all the critical points of f (x, y) = sin x sin y in the domain −2 ≤ x ≤ 2, −2 ≤ y ≤ 2.
Answer. The critical points are (0, 0), ( π2 , π2 ), (− π2 , π2 ), ( π2 , − π2 ) and (− π2 , − π2 ).
(21) Find all the local maxima, local minima and saddle points of the following functions:
(a) f (x, y) = x2 + xy + y 2 + 3x − 3y + 4
(b) f (x, y) = 6x2 − 2x3 + 3y 2 + 6xy
Answer. (a) (−3, 3) is the only critical point, which is a local minimum.
(b) Critical points are (0, 0) and (1, −1). We have (0, 0) is a local minimum and (1, −1)
is a saddle point.
(22) Let f (x, y) = xy − x2 , and let R be the square region given by R = {(x, y) ∈ R2 : 0 ≤
x ≤ 1 and 0 ≤ y ≤ 1}. Find the extreme values of f on R.
Answer. The maximum value 1/4 is attained at (1/2, 1) and the minimum value −1 is
attained at (1, 0).
(23) Verify that f (x, y, z) = x4 + y 4 + z 4 − 4xyz has a critical point (1, 1, 1), and determine
the nature of this critical point by computing the eigenvalues of its Hessian matrix.
5
Answer. Easy to check that (1, 1, 1) is a

12

H=
−4
−4
crtical point. The Hessian at (1, 1, 1) is

−4 −4

12 −4
.
−4 12
The eigenvalues of H are 4, 16, 16. Hence, H is a positive definite matrix and f has a local
minimum at (1, 1, 1).
(24) Using the method of Lagrange multipliers, find the extremum values of f (x, y) = xy
subject to the constraint g(x, y) = x2 + y 2 − 10 = 0.
√
√
√
√
Answer. We have (x, y, λ) = (± 5, ± 5, 1/2) and (x, y, λ) = (± 5, ∓ 5, −1/2). Hence,
maximum value of f is 5 and the minimum value is −5.
(25) Using the method of Lagrange multipliers, find the points on the curve xy 2 = 54 nearest
to the origin.
√
Answer. (3, ±3 2) are the points on the curve nearest to the origin.
(26) The Fermat Principle in optics states that the path AP B taken by a ray of light in passing
across the plane separating two different optical media is such that the travel time t is
minimized. Use this principle to deduce the Law of Refraction (by method of Lagrange
multipliers):
v1
sin θ1
=
sin θ2
v2
where θ1 is the angle of incidence, θ2 is the angle of refraction, and v1 , v2 are the speeds of
light in the two media.
Solution. We must minimize the travel time
t = f (θ1 , θ2 ) =
a
b
+
v1 cos θ1 v2 cos θ2
subject to the condition
g(θ1 , θ2 ) = a tan θ1 + b tan θ2 = d.
6
Using Lagrange multipliers, we set
L(θ1 , θ2 , λ) = f (θ1 , θ2 ) − λ(g(θ1 , θ2 ) − d),
and simultaneously solve
∂L
∂L
= 0,
= 0, g(θ1 , θ2 ) = d.
∂θ1
∂θ2
The first two of these give
a
b
sec θ1 tan θ1 − λa sec2 θ1 = 0 =
sec θ2 tan θ2 − λb sec2 θ2 .
v1
v2
Since a, b, and sec θ are all non-zero,
sin θ1
sin θ2
=λ=
.
v1
v2