Name: Solutions
Math 53: Quiz 6
October 14, 2014
1. (1 point) Let f : R2 → R be differentiable and let û =
(
)
√1 , √5
. Suppose that Dû f (1, 2) = − √1110 and Dv̂ f (1, 2) =
26
26
We have
(
√3 , − √1
10
10
√7 .
26
)
and v̂ =
Find ∇f (1, 2).
−11
3fx − fy
√ = ∇f (1, 2).û = √
⇒ 3fx − fy = −11
10
10
7
f + 5f
√ = ∇f (1, 2).v̂ = x√ y ⇒ fx + 5fy = 7
26
26
so that 16fx = −48 ⇒ fx = −3 and fy = 2. Thus, ∇f (1, 2) = (−3, 2).
2. (1 point) Consider the surface defined implicitly by
zex+2y+3z = 2.
Find the equation of the plane tangent to the surface at (−4, −1, 2).
Let F (x, y, z) = zex+2y+3z . Then, the vector normal to the surface at the given point
is
⃗n = ∇F (−4, −1, 2) = (zex+2y+3z , 2zex+2y+3z , (3z + 1)ex+2y+3z )(−4,−1,2) = (2, 4, 7).
The equation of the tangent plane then is
2x + 4y + 7z = (−4, −1, 2).(2, 4, 7) = 2.
1
3. (1 point) Let u = f (x, y, z) where x = r sin(ϕ) cos(θ), y = r sin(ϕ) sin(θ) and z =
r cos(ϕ). Find urϕ .
We first compute ur . Since u depends on r via x, y and z, we have
∂u ∂x ∂u ∂y ∂u ∂z
+
+
∂x ∂r ∂y ∂r ∂z ∂r
= fx sin(ϕ) cos(θ) + fy sin(ϕ) sin(θ) + fz cos(ϕ).
ur =
Note that ur depends on ϕ explicitly as well as via x, y and z. We therefore have
urϕ = [fx cos(ϕ) cos(θ) + fy cos(ϕ) sin(θ) − fz sin(ϕ)] +
[fxx sin(ϕ) cos(θ) + fyx sin(ϕ) sin(θ) + fzx cos(ϕ)](r cos(ϕ) cos(θ)) +
[fxy sin(ϕ) cos(θ) + fyy sin(ϕ) sin(θ) + fzy cos(ϕ)](r cos(ϕ) sin(θ)) +
[fxz sin(ϕ) cos(θ) + fyz sin(ϕ) sin(θ) + fzz cos(ϕ)](−r sin(ϕ)).
2