Math 2433 Homework #5
Solutions
Section 15.6.
Find the gradient of f , evaluate the gradient at point P and find the rate of change of f
at P in the direction of the vector u.
√
10. f (x, y, z) = x + yz, P (1, 3, 1), u = 72 , 37 , 67 .
Solution: We have
1
z
y
√
∇f (x, y, z) = hfx , fy , fz i =
, √
, √
.
2 x + yz 2 x + yz 2 x + yz
Thus
∇f (1, 3, 1) =
1 1 3
, ,
4 4 4
.
It can be seen that u is a unit vector so
2 3 6
2
3
18
23
Du f (1, 3, 1) = ∇f (1, 3, 1) ·
=
, ,
+
+
= .
7 7 7
28 28 28
28
Find the directional derivative of the function at the given point in the direction of the
vector v.
14. g(r, s) = tan−1 (rs), (1, 2), v = i + 3j..
Solution: We have gr = 1+rs2 s2 , gs = 1+rr2 s2 . Hence, ∇g(1, 2) = √25 i + √15 j. Also, a unit
vector u in the direction of v is u =
v
|v|
=
√1 i
10
+
√3 j.
10
Thus,
1
5
Du g(1, 2) = ∇g(1, 2) · u = √ = √ .
50
2
√
16. f (x, y, z) = xyz, (3, 2, 6), v = h−1, −2, 2i.
Solution: This is straightforward so I will leave it to you to finish.
19. Find the directional derivative of f (x, y) =
−→
√
xy at P (2, 8) in the direction of Q(5, 4).
−→
Solution: We have P Q= Dh3, −4i. Thus
E a unit vector u in the direction of P Q is u =
y
4
x
3
√
√
5 , − 5 . Also, ∇f (x, y) = 2 xy , 2 xy so
2
3 4
1
·
= .
,−
Du f (2, 8) = ∇f (2, 8) · u = 1,
4
5 5
5
1
2
Solutions
Find the maximum rate of change of f at the given point and the direction in which it
occurs.
23. f (x, y) = sin(xy), (1, 0).
Solution: We have
∇f (x, y) = hy cos(xy), x cos(xy)i
and thus |∇f (1, 0)| = |h0, 1i| = 1 is the maximum rate of change and it occurs in the
direction of the vector h0, 1i.
24.f (x, y, z) = (x + y)/z,
Solution: We have
(1, 1, −1).
1 1 (x + y)
∇f (x, y, z) =
, ,−
z z
z2
√
and thus |∇f (1, 1, −1)| = |h−1, −1, −2i| = 6 is the maximum rate of change and it
occurs in the direction of the vector h−1, , −1, −2i.
28. Find the directions in which the directional derivative of f (x, y) = ye−xy at the point
(0, 2) has value 1.
D
E
Solution: Let v =< a, b > be a vector and u = √a2a+b2 , √a2b+b2 a unit vector in the
same direction such that Du f (0, 2) = 1. We calculate
∇f (x, y) = −y 2 e−xy , (1 − xy)e−xy
and thus
Du f (0, 2) = h−4, 1i ·
√
a
b
,√
2
2
2
a +b
a + b2
b − 4a
=√
=1
a2 + b2
which gives us
(b − 4a)2 = a2 + b2
or a(15a − 8b) = 0, so either a = 0, in which case b = 1 so h0, 1i is one such direction, or b = 15
8 a, so all the directions in which the directional derivatives are given by
a
,
a
=
6
0 and h0, 1i together.
a, 15
8
32. The temperature T in a metal ball is inversely proportional to the distance from the
center of the ball, which we take to be the origin. The temperature at the point (1, 2, 2)
is 120.
3
Math 2433 Homework #5
(i) Find the rate of change of T at (1, 2, 2) in the direction of the point (2, 1, 3).
(ii) Show that at any point in the ball the direction of greatest increase in temperature
is given by a vector that points toward the origin.
Solution:
(i) The vector joining
is given by h1, −1, 1i and a unit vector in this
D the two points
E
1
1
1
direction is u = √3 , − √3 , √3 . Also, we have T (x, y, z) = √ 2 k 2 2 , where k is
x +y +z
a constant. Since T (1, 2, 2) = 120, we find that k = 360, so
∇T (x, y, z) = −
(x2
360
hx, y, zi
+ y 2 + z 2 )3/2
and the rate of change of T in the specified direction is just Du T (1, 2, 2). I will leave
the calculation for you to do.
(ii) Since
∇T (x, y, z) = −
(x2
360
hx, y, zi ,
+ y 2 + z 2 )3/2
a unit vector in the direction of ∇T and hence the direction of greatest increase in
T is
1
hx, y, zi
−p
x2 + y 2 + z 2
which is a vector pointing towards the origin.
Find equations of (a) the tangent place and (b) the normal line to the given surface at
the specified point.
40. y = x2 − z 2 , (4, 7, 3).
43. z + 1 = xey cos z, (1, 0, 0).
Solution: In each of the problems above, you just have to use the equation
z − c = fx (a, b)(x − a) + fy (a, b)(y − b)
for the tangent plane, where z = f (x, y) is defined implicitly as a function of x and y, and
similarly the symmetric equations for the normal line. I won’t do the calculations as they
are straightforward. If you are having trouble with this, please talk to me.
49. Show that the equation of the tangent plane to the ellipsoid x2 /a2 + y 2 /b2 + z 2 /c2 = 1
at the point (x0 , y0 , z0 ) can be written as
xx0 yy0 zz0
+ 2 + 2 = 1.
a2
b
c
4
Solutions
Solution: The equation of the tangent plane to the ellipsoid at (x0 , y0 , z0 ) is
2 2 c x0
c y0
z − z0 = − 2
(x − x0 ) + − 2
(y − y0 ).
a z0
b z0
Since (x0 , y0 , z0 ) also lies on the ellipse, we have
x20 y02 z02
+ 2 + 2 = 1.
a2
b
c
This last equation together with the one above gives the desired result.