# Parameterize the line tangent to the path at timet=0

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```					1. On the moon Pandora, the composition of the atmosphere at the point (x, y, z) is given
by the function                                          
(x2 +y)
z
F (x, y, z) =  xez−1 + y 
                
x + y 3 − 3z 2
In which the ﬁrst coordinate denotes the amount of Ar, the second coordinate denotes
the amount of CO2 , and the third coordinate denotes the amount of N (all measured in
ppm). A Na’vi warrior riding a ikran travels along the path
         
cos(πt)
p(t) =  sin(πt) 
t
(a) Find the rate of change of Ar experienced by the Na’vi warrior at time t = 1.
(b) Find the rate of change of CO2 experienced by the Na’vi warrior at time t = 1.
(c) Find the rate of change of N experienced by the Na’vi warrior at time t = 1.

Solution: By relabeling F (x, y, z) in the following manner
                 
(x2 +y)
z
F (x, y, z) = Ar(x, y, z)CO2 (x, y, z)N (x, y, z) =  xez−1 + y 
               
x + y 3 − 3z 2

we can see that the solution we want is ∂Ar , ∂CO2 , ∂N . These are all computed using
∂t      ∂t ∂t
the chain rule. We begin by computing the matrix of partial derivatives of compute
F (x, y, z),                                                  
2x/z 1/z −(x2 + y)/z 2
F (x, y, z) =  ez−1     1       xez−1    
2
1     3y        −2z
In addition we will need to compute Dp(t),
            
−π sin(πt)
 π cos(πt) 
1
Now the chain rule allows us to compute D(F ◦ p)(1) = DF (p(1)) · Dp(1). When
t = 1 we have p(t) = (0, 1, 1) and Dp(t) = (−1, 0, 1), which we can substitute into
the chain rule to make our computations easier
            
2 1 −2       −1         −4
DF (0, 1, 1) = 1 1 1  ·  0  =  0 
1 0 −2        1          1

So, we have computed that    ∂Ar
∂t
= −4, ∂CO2 = 0, ∂N = 1
∂t       ∂t
2. The mighty nations of Notwen and Zinbiel have been at war for many years over whose
scientists ﬁrst discovered calculus. Through years of painstaking research, Notwen-
ian scientists have determined that the Zinbiel workers require both wood (denoted by
W )and metal (denoted by M ) to build a cannon, and that the rate of cannon production
(in cannons per month) is given by the equation
P = 5M 1/2 W 1/2
Historically the Zinbielians have used 16 units of wood and 4 units metal.
(a) Historically, how many cannons are produced per month?
(b) One month, Notwen spies report that the Zinbiels are producing 2 additional units of
wood per month and 10 additional cannons per month. Deduce how the production
of metal has changed.
(c) Years later, wood and metal production levels have returned to W = 16, M = 4
when a treaty is signed calling for Zinbiel to produce half the number of cannons. If
spies report that both metal and wood production have been decreased by 2 units
per month, can Notwen be sure Zinbiel is honoring the treaty (i.e. could they be
producing more cannons than the treaty allows?)

Solution:
(a) The production depends on only M and W . With W = 16 and M = 4, the
production P = 5(4)1/2 161/2 = 40 cannons per month.

(b) Using the chain rule, we can express the rate of change of production in terms
of the rate of change of wood and metal. Speciﬁcally
∂P   ∂P ∂M   ∂P ∂W
=       +
∂t   ∂M ∂t   ∂W ∂t
So if ∂P = 10 and
∂t
∂W
∂t
= 2 we can solve for    ∂M
∂t
in the previous equation and
ﬁnd ∂M = 3/2.
∂t

(c) We will check that production is at most 20 cannons, or that production has
decreased by 20 or more, that is ∂P ≤ −20. To do this we will use the values
∂t
∂M
∂t
= −2 and ∂W = −2 in the expression
∂t

∂P   ∂P ∂M   ∂P ∂W
=       +       .
∂t   ∂M ∂t   ∂W ∂t
With M = 4 and W = 16 we ﬁnd
∂P
= −5/2 − 10 = −12.5.
∂t
So, the Zinebiel’s could secretly be producing 8 more cannons per month than
allowed by the treaty.
3. An itsy bitsy spider is trying to get out of the parabolic water spout. Suppose the spider
is currently at the point (2, −1, 7) and the depth of the water spout is given by the
equation z = x2 + 3y 2 . In which direction should the ant set out in order to climb out
of the bowl fastest? Should it follow a straight line path from then on?

Solution: The gradient f (x, y) = (2x, 6y) points in the direction of steepest in-
crease at teach point (x, y). So for x = 2, y = −1 the ant should head in the direction
√
of the vector f (2, −1) = (4, −6). That is given by the unit vector n = (2, −3)/ 13.
The ant should not continue to walk in a straight line since the direction of the gra-
√            √
dient changes. Indeed at the point (x, y) = (2 + 2t/ 13, −1 − 3t/ 13), the direction
√
of of√steepest increase is the unit vector parallel to f (x, y) = (4 + 4t/ 13, −6 −
18t/ 13), which is not the same as the direction n of the straight line.

4. Stranded on a makeshift raft, Odysseus is swept into the whirlpool Charybdis! Charybdis
creates her whirlpool by applying a centripetal force with magnitude

f (x, y) = x−2 + y −2 ,

toward the center of the whirlpool. Odysseus’ raft follows the elliptical path

p(t) = (2 cos(πt), 3 sin(πt)).

(a) Sketch a few level curves of f (x, y). Sketch the path of Odysseus’ raft on the same
set of axes. Is the inward force constant along the entire path of Odysseus’ raft?
(b) When t = 1, in what direction is the raft traveling? If Odysseus manages to escape
Charybdis’ maw and continues to travel in this direction, what line will he follow?
(c) Show that the rate of change of f (x, y) in the direction of the path from part (b)
is equal to 0. (You will need to compute a directional derivative.) Why is this?
Explain in terms of the level curves of f (x, y).
(d) At t = 7/4, Odysseus wants to move as quickly as possible to a place where Charyb-
dis’ force is lower. In what direction should he paddle?

Solution:
(a) The level curves are circles centered at the origin, and Odysseus’ path is an
ellipse centered at the origin with major axis 3 and and minor axis 2. The force
is not constant on the path of the raft since it doesn’t travel on a level curve.

(b)
p (t) = (−2π sin(πt), 3π cos(πt))
p (1) = (0, −3π)
When t = 1, Odysseus is traveling in the direction (0, −3π) (or (0, −1)). If
Odysseus continues in this direction, his new path will start at the point p(1) =
(−2, 0) at move in direction (0, −1), so it will have equation

l(t) = (−2, 0) + t(0, −1) = (−2, −t).

(c) We will compute a directional derivative at the point (x0 , y0 ) = (−2, 0) in di-
rection v = (0, −1) (note that this is a unit vector).
¯
f (x0 , y0 ) · v = (2x, 2y)|(−2,0) · (0, −1) = (−4, 0) · (0, −1) = 0
¯
The directional derivative is 0, meaning the force is constant in this direction.
This makes sense because the level curve passing through the point (−2, 0) is
the circle x2 + y2 = 4, and so v is tangent to the level curve.
¯
(d) We need to ﬁnd the direction v in which the directional derivative at p(7/4) =
√ 3                               ¯
( 2, √2 ) is smallest (or negative with the largest absolute value). The direc-
tional derivative f (x0 , y0 ) · v will be smallest when f (x0 , y0 ) and v are point-
¯                                        ¯
ing in opposite directions, so we want v in the direction opposite to the gradient.
¯
√ 3                              √ √
f ( 2, √ ) = (2x, 2y)|(√2, √ ) = ( 2, 3 2)
3
2                    2

√      √
Odysseus should paddle in the direction (− 2, −3 2) to experience the fastest
decrease in force.

5. Find the tangent plane to the surface p(x, y, z) = 6 at the point (1, 1, 1) if p(x, y, z) =
x2 + 2y 2 + 3z 2 . Write the equation of the plane as a graph over the xy plane (i.e. as a
function z = h(x, y)).

Solution: To ﬁnd the tangent plane we will remember that the gradient is normal
to level sets. We begin by ﬁnding the gradient
p(x, y, z) = (2x, 4y, 6z)
At the point (1, 1, 1) we have
p(1, 1, 1) = (2, 4, 6)
So we can write the tangent plane using a normal vector and a point.
p(1, 1, 1) · (x − 1, y − 1, z − 1) = 0
(2, 4, 6) · (x − 1, y − 1, z − 1) = 2x − 2 + 4y − 4 + 6z − 6 = 0
−2x − 4y + 12        −x − 2y + 6
z=                     =
6                  3
6. Find a direction for which the directional derivative of the function w(x, y, z) = y(x2 +
z 2 ) − z 3 at the point (1, 1/2, 1) is zero.

Solution: In general we compute the directional derivative based on the gradient.
We can compute the partial derivatives

∂w
= 2xy
∂x
∂w
= x2 + z 2
∂y
∂w
= 2yz − 3z 2
∂z

Which gives the gradient, evaluated at the point (1, 1/2, 1)

f (1, 1/2, 1) = (2xy, x2 + z 2 , 2yz − 3z 2 ) = (1, 2, −2)

The directional derivative is given as f · u so we must ﬁnd a vector u = (x, y, z)
such that f · u = 0. Expanding this we have

0=      f · u = x + 2y − 2z = 0

Now we just need to ﬁnd a solution to this equation. Setting x = 1 y = 1 we can
solve z = 3/2. This gives the vector (1, 1, 3/2) and we can check that it is in fact a
direction gives 0 change in w, since

(1, 2, −2) · (1, 1, 3/2) = 1 + 2 − 3 = 0

7. The temperature T (x, y) at a point (x, y) in the plane is a diﬀerentiable function such
that ∂T (3, −1) = 2 and ∂T (3, −1) = 5. Find the directional derivative of T at the point
∂x                   ∂y                             √      √
(3, −1) in the direction given by the unit vector u = 1/ 2, 1/ 2 .

Solution: The directional derivative in direction u is given by

1 1          7
T · u = (2, 5) ·   √ ,√        =√ .
2 2          2
8. A bug crawls along a level curve of the temperature function T at speed one. What are
the possible velocity vectors v for the bug at the moment it crawls through the point
(3, −1)?

Solution: The directional derivative for T in the direction of a level curve must be
zero. This gives the equation T · v = 0. The speed of the bug is one, so we also
have v = 1.
Then the equation T · v = 0, can be written as 2v1 + 5v2 = 0. This gives v2 =
− (2/5) v1 . One can also note by inspection that (−5, 2) √ a solution and that any
is              √
multiple of (−5, 2) is a solution. Since v = 1, (−5, 2) / 29 or (5, −2) / 29, take
note that we have carefully normalized the solution to unit length.

9. Compute the value of x2 ∂w + y 2 ∂w at the point (1, −1) if w = f ( x+y ) and f (0) = −2.
∂x       ∂y                                  xy

Solution: Deﬁne
x+y
u(x, y) =
xy

Then f ( x+y ) = f (u(x, y)) is deﬁned in terms of function composition, so we should
xy
ﬁrst think of the chain rule. The chain rule says

Dw = Dw(x, y) =
∂w    ∂w
=    ∂x    ∂y    =
= D(f ◦ u) =
= Df (u(x, y)) · Du(x, y)

We can compute

∂u    ∂u       −1   −1
Du(x, y) =        ∂x    ∂y   =   x2   y2

∂w         ∂w
This allows us to compute    ∂x
and   ∂y
at the point (1, −1)

Dw(1, −1) = Df (0) · −1, −1 = 2, 2

Which ﬁnally allows us to compute

1·2+1·2=4

```
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