Math 53 Worksheet Solutions - Partial Derivatives
p
1. If f (x, y) = 3 x3 + y 3 , find fx (0, 0).
Solution. Note that
fx (x, y) =
3x2
,
3(x3 + y 2 )2/3
so if we try to plug in x = 0 and y = 0, we get division by zero. This doesn’t mean
that fx (0, 0) doesn’t exist, it just means we need to find another way to calculate it.
Instead, we return to the definition.
f (0 + h, 0) − f (0, 0)
(h3 )1/3
fx (0, 0) = lim
= lim
= 1.
h→0
h→0
h
h
2. The ellipsoid 4x2 + 2y 2 + z 2 = 16 intersects the plane y = 2 in an ellipse. Find
parametric equations for the tangent line to this ellipse at the point (1, 2, 2).
Solution. The key is that since we’re looking at the plane y = 2, the slope is actually
∂z
. So, we implicitly differentiate.
just ∂x
∂
∂
4x2 + 2x2 + z 2 =
16 = 0.
∂x
∂x
So we get
8x + 2z
and solving for
Then
∂z
,
∂x
∂z
= 0,
∂x
we have
∂z
4x
=− .
∂x
z
∂z = −2.
∂x x=1, z=2
So the tangent line passes through the point (1, 2, 2) and has direction vector
h1, 0, −2i. We can write the equation, then, as
x = 1 + t,
y = 2,
z = 2 − 2t.
3. Suppose you need to know an equation of the tangent plane to a surface S at the
point P (2, 1, 3). You don’t have an equation for S but you know that the curves
r1 (t) = h2 + 3t, 1 − t2 , 3 − 4t + t2 i
r2 (u) = h1 + u2 , 2u3 − 1, 2u + 1i
both lie on S. Find an equation of the tangent plane at P .
1
Solution. The key is that the tangent vectors of the two curves at the point P (2, 1, 3)
lie in the tangent plane. Note that r1 (0) = h2, 1, 3i and r2 (1) = h2, 1, 3i. Also,
r01 (t) = h3, −2t, −4 + 2ti,
r02 (u) = h2u, 6u, 2i,
so
r01 (0) = h3, 0, −4i,
r02 (1) = h2, 6, 2i,
and we get our normal vector to the tangent plane by taking a cross product. We get
n = r01 (0) × r02 (1) = h24, −14, 18i.
So the equation of the tangent plane is
24(x − 2) − 14(y − 1) + 18(z − 3) = 0,
or
24x − 14y + 18z − 88 = 0.
4. Are there any points on the hyperboloid x2 − y 2 − z 2 = 1 where the tangent plane is
parallel to the plane z = x + y?
Solution. For two planes to be parallel, their normal vectors must be parallel. The
normal vector to the tangent plane of the hyperboloid at the point (x, y, z) is
n1 = ∇F (x, y, z) = h2x, −2y, −2zi.
The normal vector to the given plane is
n2 = h1, 1, −1i,
which we obtain be rewriting the equation of the plane as x + y − z = 0. So, if the
planes are parallel, there’s some constant k so that
h2x, −2y, −2zi = kh1, 1, −1i = hk, k, −ki.
But for this to be true, we must have z = k/2 and, x = k/2, and y = −k/2, so our
point (x, y, z) = (k/2, −k/2, k/2). But this is impossible, since (x, y, z) must actually
lie on the hyperboloid, and substituting (x, y, z) into the left-hand side of the defining
equation of the hyperboloid gives
2 2 2
k
k
k
k2
− −
−
=− ,
2
2
2
4
and for this to be equal to 1, we’d need k 2 = −4, which has no real solutions. So there
are no points (x, y, z) at which the tangent plane of the hyperboloid is parallel to the
plane z = x + y.
2
5. One side of a triangle is increasing at a rate of 3 cm/s and a second side is decreasing
at a rate of 2 cm/s. If the area of the triangle remains constant, at what rate does the
angle between sides change when the first side is 20 cm long, the second side is 30 cm,
and the angle is π/6?
Solution. Call the sides x and y and the angle θ. Then
1
A = xy sin θ,
2
where A is the area. Taking derivatives, and observing that A does not change with
time, we have
dA
∂A dx ∂A dy ∂A dθ
0=
=
+
+
.
dt
∂x dt
∂y dt
∂θ dt
Solving for
dθ
,
dt
we get
∂A dx
dt
dθ
∂x
=−
dt
+
∂A dy
∂y dt
.
∂A
∂θ
Now we calculate the partials. (We could have done this before solving – it doesn’t
really matter.)
∂A
1
= y sin θ,
∂x
2
∂A
1
= x sin θ,
∂y
2
∂A
1
= xy cos θ.
∂θ
2
So, substituting, we have
y sin θ dx
+ x sin θ dy
dθ
dt
dt
=−
,
dt
xy cos θ
and so at the indicated point in time,
√
(30)( 21 )(3) + (20)( 12 )(−2)
dθ
3
√
=−
=−
≈ −0.05.
3
dt
36
(30)(20)
2
6. Suppose that the directional derivatives of f (x, y) are known at a given point in two
nonparallel directions given by unit vectors u and v. Is it possible to find ∇f at the
point? If so, how would you do it?
Solution. Write u = hu1 , u2 i and v = hv1 , v2 i. Then
Du f (x, y) = ∇f (x, y) · u = fx (x, y)u1 + fy (x, y)u2 ,
and
Dv f (x, y) = ∇f (x, y) · v = fx (x, y)v1 + fy (x, y)v2 ,
and these form a system of linear equations. You know u1 , u2 , v1 , v2 and the
directional derivatives, and you can solve for fx (x, y) and fy (x, y). Note that (x, y) is a
fixed point throughout the problem. Using standard methods, you find
fy (x, y) =
v1 Du f (x, y) − u1 Dv f (x, y)
,
u2 v1 − v2 u1
3
and a similar expression exists for fx (x, y). Note that u2 v1 − v2 u1 6= 0 precisely
because the vectors are nonparallel.
√
7. Show that the function f (x, y) = 3 xy is continuous and the partial derivatives fx and
fy exist at the origin, but the directional derivatives in all other directions do not exist.
√
Solution. Continuity should be clear; indeed, lim(x,y)→(0,0) 3 xy = 0 = f (0, 0). Note
that, like Problem 1, using the product and chain rules will not be useful here. We
must perform a direct calculation.
p
3
(h)(0) − 0
f (0 + h, 0) − f (0, 0)
fx (0, 0) = lim
= lim
= 0,
h→0
h→0
h
h
and
p
3
(0)(h) − 0
f (0, 0 + h) − f (0, 0)
= lim
= 0,
fy (0, 0) = lim
h→0
h→0
h
h
so fx (0, 0) and fy (0, 0) definitely exist. Now let u = hu1 , u2 i be any unit vector with
u1 , u2 6= 0. Then
p
√
3
3 u u
(hu1 )(hu2 )
f (0 + hu1 , 0 + hu2 ) − f (0, 0)
1 2
Du f (0, 0) = lim
= lim
= lim √
,
3
h→0
h to0
h→0
h
h
h
and this limit does not exist.
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