HW3 Due 7/16
1. Find the tangent plane to x3 + z sin(zy) + y 3 log x2 = 1 at (1, 0, 0).
The normal vector is
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∇(x3 +z sin(zy)+y 3 log x2 )1,0,0 = (3x2 + 2yx , z 2 cos(zy)+3y 2 log x2 , sin(zy)+zy cos(zy))1,0,0 =
(3, 0, 0) so the tangent plane is 3(x − 1) = 0 or x = 1.
2. Find the rate of change of f (x, y, z) = xyz in the direction normal to the surface yx2 +
xy 2 + yz 2 = 3 at (1, 1, 1).
The direction normal to the surface is ∇(yx2 + xy 2 + yz 2 )|1,1,1 = (3, 4, 2). And the change
is the directional derivative of f in this direction which is:
∇f(1,1,1) ·
(3,4,2)
√
29
= (yz, xz, xy)|(1,1,1) ·
(3,4,2)
√
29
=
√9
29
3. Let p = (1, 1, 0) and S be the surface x2 + y 2 − z 2 = 1. Find the distance from p to S.
Suggestion: You may use the fact that the shortest line connecting p to S is perpendicular
to S.
The normal to S at (x, y, z) is (2x, 2y, −2z). The vector connecting (x, y, z) to p is
(1 − x, 1 − y, −z). So at the closest point we have (2x, 2y, −2z) = c(1 − x, 1 − y, −z) for
some c ∈ R.
So x = y since the satisfy the same relation with c. And −2z = −cz ⇒ z = 0 or c = 2.
If c = 2 then 2x = 2 − 2x ⇒ x = 21 = y and plugging into x2 + y 2 − z 2 = 1 we get
1
− z 2 = 1 ⇒ z 2 = − 12 impossible with real numbers. So we must have z = 0.
2
√
Now plugging x2 + y 2 = 1 and x = y impies x = y = ±1/ 2.
q
√
√
√
√
Distance is the smaller of 2(1 ± 1/ 2)2 = 2|1 ± 1/ 2| = | 2 ± 1| which is smallest
√
when we take 0 −0 so the distance is 2 − 1.
4. Compute the directional derivative of f (x, y, z) = z 2 x + x sin(xy) at x = 1, y = 0, z = 1:
(a) In the direction of ~v = (1, 2, 1)
∇f(1,0,1) = (z 2 + sin(xy) + xy cos(xy), x2 cos(xy), 2zx)|(1,0,1) = (1, 1, 2), so this is
(1, 1, 2) ·
(1,2,1)
√
6
=
√5
6
(b) Along w
~ = (2, 4, 2).
This is (1, 1, 2) · (2, 4, 2) = 10.
5. Find the points on x2 + 4y 2 − z 2 = 4 where the tangent plane is parallel to the plane
2x + 2y = 5.
We need to find a point on x2 + 4y 2 − z 2 = 4 where (2x, 8y, −2z) = c(2, 2, 0) for some
c ∈ R.
Setting equal components we get z = 0, x = c, 4y = c, so x = 4y.
1
√
Plugging into x2 + 4y 2 − z 2 = 4 gives (4y)2 + 4y 2 = 4 ⇒ y = ±1/ 5.
√
√
Now the points are (±4/ 5, ±1/ 5, 0).
6. Consider the graph z = f (x, y) = 100 − x2 − 3y 2 .
(a) What is the highest point of the graph (maximum z-value)?
Since x2 + 3y 2 ≥ 0 the largest z-value is z = 100 when x = y = 0.
(b) What is the gradient of f at this point?
Since this is a maximum ∇f(0,0) = 0.
√
7. Suppose a particle is ejected from the surface x2 + y 2 − z 2 = −1 at (1, 1, 3) along the
normal to the surface pointing towards the xy-plane with a speed of 10 units/sec. When
and where does it cross the xy-plane?
√
The normal direction is (2x, 2y, −2z)|(1,1,√3) = (2, 2, −2 3) a unit vector in this direction
√
is √15 (1, 1, − 3).
√
√
So the particle’s path is parametrized by `(t) = (1, 1, 3) + t √105 (1, 1, − 3).
√
√ √
So we hit the xy-plane when 3 − t2 5 3 = 0 ⇒ t = 2√1 5 and our location at this time
is: `( 2√1 5 ) = (2, 2, 0).
8. Consider the spiral c(t) = (et cos t, et sin t). Show the angle between c(t) and c0 (t) is
constant.
c0 (t)
= (et cos t, et sin t) + (−et sin t, et√
cos t) so that e2t = c(t) · c0 (t) = |c(t)|c0 (t)| cos θ(t) =
√
et 2et cos θ(t) and now cos θ(t) = 1/ 2 ⇒ θ(t) = π/4 is constant.
9. The position of a certain particle on a helix is given by c(t) = (cos t, sin t, t2 ).
(a) Sketch the particles motion
It is a helix lying on the top half (z > 0) of a cylinder of radius 1. Also the helix is getting
steeper.
(b) Find the speed of the particle at time t.
p
√
speed = |c0 (t)| = (− sin t)2 + (cos t)2 + (2t)2 = 1 + 4t2
(c) Is c(t) ever orthogonal to c0 (t)?
c(t) · c0 (t) = 2t3 so yes only at t = 0.
(d) Find an equation for the tangent line to c(t) at t = 4π
`(t) = c(4π) + tc0 (4π) = (1, 0, 16π 2 ) + t(0, 1, 8π) works.
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(e) Where does this line (from part d) intersect the xy-plane?
Where the z-component is 0, that is 16π 2 + 8πt = 0 ⇒ t = −2π and the point of
intersection is `(−2π) = (1, −2π, 0)
(f) What is the angle of intersection of this line and the xy-plane?
Let α be the angle we want and θ = ∠(c0 (4π), k̂) (c0 (4π) is the direction of the line and k̂
0
k̂
√ 8π
.
is normal to the xy-plane). Then θ + α = π/2 so that sin α = cos θ = c|c(4π)·
0 (4π)| =
1+64π 2
8π
◦
Then α = arcsin( √1+64π2 ) ≈ .47π = 86 .
10. Suppose a planet P moves counterclockwise along a circular orbit with radius 10 around
a star S and completes one revolution in 2π units if time. Suppose thata moon M moves
clockwise around planet P in a circle of radius 3 units completing one revolution in π/4
time and a satelite s moves counterclockwise around the moon in a circle of radius 1 unit
completing one revolution in π/2 units time.
Let the star S be at the origin and at t = 0 the planet be at (10, 0), the moon at (13, 0)
and the satelite at (14, 0). Find a parametric curve that gives the position of the satelite
relative to the sun.
Suggestion: vector addition should be useful.
The planet’s position is P (t) = 10(cos t, sin t).
the Moon’s position relative to the planet is M (t) = 3(cos(8t), − sin(8t))
the satelite’s position relative to the Moon is s(t) = (cos(4t), sin(4t)).
Now the satelite relative to the sun is
c(t) = P (t) + M (t) + s(t) = (10 cos t + 3 cos 8t + cos 4t, 10 sin t − 3 sin 8t + sin 4t).
Pluggin into wolfram looks like:
11. Let c be the curve where the plane 3x+y−7z = 0 intersects the ellipsoid x2 +2y 2 +3z 2 = 39.
Parametrize the tangent line to c at the point (5, −1, 2).
The tangent line to this curve lies in the tangent plane to the ellipsoid at (5, −1, 2) and
the plane 3x + y − 7z = 0. The 2 normals are ∇(x2 + 2y 2 + 3z 2 )(5,−1,2) = (10, −4, 12) and
(3, 1, −7).
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So the tangent line to c has direction (10, −4, 12) × (3, 1, −7) = (16, 106, 22) ∼ (8, 53, 11).
So one possible parametrization is `(t) = (5, −1, 2) + t(8, 53, 11)
12. Find the maximum value of f (x, y) = xy along the path c(t) = (cos t, sin t).
Along this path we have f (cos t, sin t) = cos t sin t =
1
.
2
1
2
sin 2t, which has maximum value
13. Suppose f : Rn → R is differentiable and f (−~x) = f (~x) for any ~x ∈ Rn . Find df~0 .
Let ~v ∈ Rn be any vector and set g(t) = f (−t~v ). Then differentiating g(t) = g(−t) gives
g 0 (t) = −g 0 (−t) by chain rule. So at t = 0 we have g 0 (0) = −g 0 (0) ⇒ g 0 (0) = 0.
Now 0 = g 0 (0) = df~0 (~v ) so df~0 (~v ) = 0 for all ~v that is df~0 is the 0 matrix.
14. Captain Ralph is in trouble near Mercury. The temperature of his vessel’s hull is given
2
2
2
by T (x, y, z) = e−x −2y −3z when his position is at (x, y, z) (x, y, z in meters). His current
coordinates are (1, 1, 1).
(a) In what direction should he proceed to decrease temperature the fastest?
He should proceed in the direction of −∇T(1,1,1) = e−6 (2, 4, 6).
(b) If his ship travels at a speed e8 m/s how fast will his temperature decrease if he heads
in the direction from part a?
If he traveled at unit speed the temperature would decrease at a rate of |∇T(1,1,1) |. Changing speed scales the rate of decrease proportionally
(∇T ·cv̂ = c∇T · v̂), so his temperature
√
8
2
will decrease at a rate of e |∇T(1,1,1) | = e 56
√
(c) Unfortunately the hull of his ship will crack if cooled at a rate greater than 14e2
degrees/sec. Describe a set of directions in which he may head to bring the temperature
down at no more than this rate.
√
√
We seek directions (unit vectors) v̂ s.t. − 14e2 ≤ ∇T(1,1,1) · e8 v̂ = e2 56 cos θ where
θ = ∠(v̂, (−2, −4, −6)).
So these are unit vectors where cos θ ≥ −1/2 i.e. −2π/3 ≤ θ ≤ 2π/3.
A more careful answer (since we don’t want to head right into the sun) is θ ∈ (−2π/3, −π/3)∪
(π/3, 2π/3).
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