Calculus
Homework # 12
Homework # 12
11.3 # 32: (3 pts, p. 767) Find all of the first partial derivatives of the function:
w = zexyz
Note that w = w(x, y, z) is the function so the partial derivatives needed are the derivatives
with respect to z, y and z. These are:
∂w
= yz 2 exyz
∂x
∂w
= xz 2 exyz
∂y
∂w
= exyz + xyzexyz = (1 + xyz)exyz
∂z
11.4 # 3: (3 pts, p. 778) Find an equation of the tangent plane to the given surface at the
specified point:
√
(1, 1, 1)
z = xy
See Example 1 on page 771 for this:
√
y
fx (x, y) = √ = 12
Find the partial derivatives of x and y at this point
2 x
√
y
fy (x, y) = √ = 12
2 x
z − 1 = 21 (x − 1) + 21 (y − 1) The equation of the tangent plane
z = 12 x + 12 y
x + y − 2z = 0
The plane
Another equation for the plane
11.4 # 39: (4 pts, p. 779) Find an equation of the tangent plane to the parametric surface at
the given point. If you have software that graphs parametric surfaces, use a computer to graph
the surface and the tangent plane.
x=u+v
z =u−v
y = 3u2
(2, 3, 0)
See Example 7 on page 777 for this. First, find the values of u and v which correspond to
the point (x, y, z) = (2, 3, 0):
y = 3u2
√
3y
u=
3
√
3y
+v
x=
3 √
3y
v =x−
3
Solve for u
Solve for v
1
Calculus
Homework # 12
From this, (x, y, z) = (2, 3, 0) corresponds to (u, v) = (1, 1). Continuing with Example 7:
∂x
∂u
∂y
∂u
∂z
∂u
∂x
∂v
∂y
∂v
∂z
∂v
=1
Partial derivatives at the point
= 6u = 6
=1
=1
=0
= −1
∂y
∂z
∂x
i+
y+
k
∂u
∂u
∂u
= i + 6j + k
∂x
∂y
∂z
rv =
i+
y+
k
∂v
∂v
∂v
=i−k
i j k ru × rv = 1 6 1 1 0 −1
(
)
(
)
= (−6 − 0)i + 1 − (−1) j + 0 − 6 k
= −6i + 2j − 6j
ru =
Find the tangent vectors
Find the normal vector
Since this normal vector is −2 · ⟨3, −1, 3⟩, the vector ⟨3, −1, 3⟩ will be used as the normal vector.
So the equation for the tangent plane is:
3(x − 2) − 1(y − 3) + 3(z − 0) = 0
3x − y + 3z = 3
An equation of the tangent plane
Simplified
2