MATH 209, Lab 3
Richard M. Slevinsky∗
Problems
1. Find equations for the tangent plane and the normal line to the surface x y + sin z = 2 at (1, 2, 0);
To form the tangent plane and the normal line, we require the vector n which is normal to plane (and
therefore parallel to the line). If our function can be written as f (x, y, z) = x y + sin z − 2 = 0, as
above, then this vector is in the direction of the gradient:
∂f ∂f ∂f
,
,
,
∇f (x, y, z) =
∂x ∂y ∂z
= hy, x, cos zi ,
n = ∇f (x, y, z)|(x,y,z)=(1,2,0) = h2, 1, 1i .
Then the tangent plane is written as n · (r − r0 ) = 0, where r is a variable vector and r0 is the point
in question:
h2, 1, 1i · (hx, y, zi − h1, 2, 0i) = 0.
As well, the normal line is written as r(t) = r0 + nt, where t is a parameter:
hx(t), y(t), z(t)i = h1, 2, 0i + h2, 1, 1it,
= h1 + 2 t, 2 + t, ti.
2. Find the greatest rate of change of the function f (x, y) = ex−y at (x, y) = (1, 1). In which direction
does this occur?
The gradient of f (x, y) is:
∇f (x, y) =
∂f ∂f
,
∂x ∂y
,
= ex−y , −1ex−y = ex−y h1, −1i ,
∇f (x, y)|(x,y)=(1,1) = h1, −1i .
Therefore, the greatest rate of change of the function f (x, y) is |∇f | =
in the direction of the gradient, that is: u = √12 h1, −1i.
Exercises
∗
Contact: [email protected]
1
√
12 + 1 2 =
√
2, and this occurs
1. Find the directional derivative of f (x, y, z) = x2 − y z + z 2 x at the point P (1, −4, 3) in the direction
of the point Q(2, −1, 8) [note: you must first find the vector from P to Q]. What is the maximum rate
of increase of f in any direction at P ?
The vector from P to Q is QP = (2 − 1, −1 + 4, 8 − 3) = (1, 3, 5) =⇒ u =
of f at P is:
∂f ∂f ∂f
∇f (x, y, z) =
,
,
,
∂x ∂y ∂z
√1 h1, 3, 5i.
35
The gradient
= h2 x + z 2 , −z, 2 x z − yi,
∇f (x, y, z)|(x,y,z)=(1,−4,3) = h11, −3, 10i.
Therefore, the directional derivative in the direction of the point Q is:
1
Du f = u · ∇f = √ h1, 3, 5i · h11, −3, 10i,
35
11 − 9 + 50
52
√
=
=√ .
35
35
The maximum rate of increase of f in any direction at P is:
|∇f | = |h11, −3, 10i| ,
√
√
= 121 + 9 + 100 = 230.
2. Find equations for the tangent plane and the normal line to the ellipsoid 4 x2 + 9 y 2 + z 2 − 49 = 0 at
the point P (1, −2, 3);
To form the tangent plane and the normal line, we require the vector n which is normal to plane (and
therefore parallel to the line). If our function can be written as f (x, y, z) = 4 x2 + 9 y 2 + z 2 − 49 = 0,
as above, then this vector is in the direction of the gradient:
∂f ∂f ∂f
∇f (x, y, z) =
,
,
,
∂x ∂y ∂z
= h8 x, 18 y, 2 zi ,
n = ∇f (x, y, z)|(x,y,z)=(1,−2,3) = h8, −36, 6i .
2
Then the tangent plane is written as n · (r − r0 ) = 0, where r is a variable vector and r0 is the point
in question:
h8, −36, 6i · (hx, y, zi − h1, −2, 3i) = 0.
As well, the normal line is written as r(t) = r0 + nt, where t is a parameter:
hx(t), y(t), z(t)i = h1, −2, 3i + h8, −36, 6it,
= h1 + 8 t, −2 − 36 t, 3 + 6 ti.
3. The directional derivative of f (x, y, z) at a given point P is greatest in the direction of the vector
h2, 2, −1i, and takes the value 2 in this direction. Find the directional derivative of f (x, y, z) at P in
the direction of the vector h1, −1, 2i [hint: for u a unit vector, then Du f = u · ∇f = |∇f | cos θ];
3
The maximal value of the directional derivative is given as 2. Therefore we know that |∇f | = 2. We
must then find the angle between the direction h2, 2, −1i and h1, −1, 2i, or more precisely the cosine of
this angle. From linear algebra, we recall that:
cos θ =
=
=
=
=
u·v
,
||u||||v||
h1, −1, 2i · h2, 2, −1i
,
||h1, −1, 2i||||h2, 2, −1i||
2−2−2
√
√
,
1+1+4 4+4+1
−2
√ √ ,
6 9
−2
√ .
3 6
Therefore, the directional derivative in the direction u = h1, −1, 2i is Du f =
−4
√ .
3 6
4. Prove that the sphere x2 + y 2 + z 2 = a2 and the cone x2 + y 2 − z 2 = 0 are orthogonal at every point
of intersection.
Orthogonality requires their gradients to be orthogonal:
ns = ∇fs = h2 x, 2 y, 2 zi,
nc = ∇fc = h2 x, 2 y, −2 zi.
Therefore, we require the dot product to be equal to 0:
ns · nc = h2 x, 2 y, 2 zi · h2 x, 2 y, −2 zi,
= 4 x2 + 4 y 2 − 4 z 2 ,
but the values of z 2 must be at once on the sphere, and as well on the cone. Since they are on the
cone, z 2 = x2 + y 2 , and therefore:
ns · nc = 4 x2 + 4 y 2 − 4(x2 + y 2 ) = 0.
4
5