Lab Quiz 10 Solutions
Problem 1
The plane curve 2x3 − x2 y 2 = 3x − y − 7 can be written as the level curve f (x, y) = −7 for function
f (x, y) = 2x3 − x2 y 2 − 3x + y.
Then the gradient vector of f at P (1, −2) is perpendicular to the curve at this point since f (1, −2) = −7.
So
∇f (x, y) = (6x2 − 2xy 2 − 3)i + (−2yx2 + 1)j
and
∇f (1, −2)
=i−j
−5
is a normal vector to the curve at P (1, −2). We don’t have to divide by −5 because ∇f (1, −2) is a
∇f (1, −2) = −5i + 5j =⇒ N =
normal vector. Since we only need the vector to be perpendicular to the curve, however, we can take any
vector parallel to the gradient vector.
Now a tangent vector to the curve at P (1, −2) is found by using the formula in the textbook,
t = 5i + 5j
or taking any parallel vector to t,
T = i + j.
As before, knowing a point on a line and a direction vector of a line gives us the equation of a line. So
for the tangent line written in symmetric form,
x−1
y+2
=
=⇒ x − y − 3 = 0
1
1
and the normal line written in symmetric form,
x−1
y+2
=
=⇒ x + y + 1 = 0.
1
−1
Problem 2
The surface z = x2 + xy + y 2 − 6x + 2 can be written as the level surface f (x, y, z) = −2 for function
f (x, y, z) = x2 + xy + y 2 − 6x − z.
Then the gradient vector of f at P (4, −2, −10) is perpendicular to the surface at this point since
f (4, −2, −10) = −2.
1
So
∇f (x, y, z) = (2x + y − 6)i + (x + 2y)j − k
and
∇f (4, −2, −10) = −k
is a normal vector to the surface at P (4, −2, −10).
Since this normal vector is a direction vector of the normal line to the surface at P (4, −2, −10), then
scalar parametric equations for this line are
x(t) = 4,
y(t) = −2,
z(t) = −10 − t.
Also, knowing a point on a plane and a normal vector of a plane gives us the equation of a plane.
So the equation for the tangent plane to the surface at P (4, −2, −10) is
−1(z + 10) = 0 =⇒ z = −10.
The tangent plane is horizontal.
For a surface given by z = g(x, y) like we have here, the tangent plane and the scalar parametric
equations for the normal line to the surface can also be found directly using the formulas on page 900.
Problem 3
For f (x, y) = x2 + xy + y 2 − 6x + 2,
∇f (x, y) =
∂f
∂f
i+
j = (2x + y − 6)i + (x + 2y)j.
∂x
∂y
Since the stationary points are points (x, y) such that ∇f (x, y) = 0, then
∇f (x, y) = (2x + y − 6)i + (x + 2y)j = 0 =⇒ 2x + y − 6 = 0,
x + 2y = 0.
So,
2x + y = 6,
−2(x + 2y = 0) =⇒ −3y = 6 =⇒ y = −2 =⇒ x = 4
and so (4, −2) is the only stationary point for f (x, y).
Now
A=
∂ 2f
= 2,
∂x2
B=
∂ 2f
= 1,
∂x∂y
C=
∂ 2f
=2
∂y 2
so
D = AC − B 2 = 4 − 1 = 3.
Since D = 3 > 0 and A = 2 > 0 for any point (x, y), then by the second-partials test, f (x, y) has a
local minimum at point (4, −2) with a local extreme value of f (4, −2) = −10.
Problem 4
2
For f (x, y) = x3 + y 2 − 6xy + 6x + 3y − 2,
∇f (x, y) =
∂f
∂f
i+
j = (3x2 − 6y + 6)i + (2y − 6x + 3)j.
∂x
∂y
So
∇f (x, y) = (3x2 − 6y + 6)i + (2y − 6x + 3)j = 0 =⇒ 3x2 − 6y + 6 = 0,
2y − 6x + 3 = 0.
Solving,
3x2 − 6y + 6 = 0,
2y = 6x − 3 =⇒ x2 = 2y − 2 = 6x − 3 − 2 = 6x − 5
so
x2 = 6x − 5 =⇒ x2 − 6x + 5 = 0 =⇒ (x − 1)(x − 5) = 0 =⇒ x = 1, 5.
Then
2y = 6x − 3 =⇒ y =
6x − 3
2
so
3
27
x = 1 =⇒ y = , x = 5 =⇒ y = .
2
2
3
27
The stationary points for f (x, y) are then 1,
and 5,
.
2
2
Now
A=
∂ 2f
= 6x,
∂x2
B=
∂ 2f
= −6,
∂x∂y
C=
∂ 2f
=2
∂y 2
so
D = AC − B 2 = 12x − 36 = 12(x − 3).
3
For stationary point 1,
,
2
D = −24 < 0,
3
1,
.
2
so by the second-partials test, f (x, y) has a saddle point at
27
For stationary point 5,
,
2
D = 24 > 0, A = 30 > 0,
27
so by the second-partials test, f (x, y) has a local minimum at point 5,
with a local extreme value
2
27
117
=−
.
of f 5,
2
4
Problem 5
For f (x, y) = (x + y)(xy + 1),
∇f (x, y) =
∂f
∂f
i+
j = [(x + y)(y) + (xy + 1)]i + [(x + y)(x) + (xy + 1)]j.
∂x
∂y
= (2xy + y 2 + 1)i + (2xy + x2 + 1)j.
So
∇f (x, y) = (2xy + y 2 + 1)i + (2xy + x2 + 1)j = 0 =⇒ 2xy + y 2 + 1 = 0,
3
2xy + x2 + 1 = 0.
Solving,
2xy + y 2 + 1 = 0,
−(2xy + x2 + 1 = 0) =⇒ y 2 = x2 .
Assume x = y. Then
2xy + y 2 + 1 = 0 =⇒ 2y 2 + y 2 + 1 = 3y 2 + 1 6= 0 =⇒ x 6= y =⇒ x = −y.
So
2xy + y 2 + 1 = 0,
x = −y =⇒ −2y 2 + y 2 + 1 = 0 =⇒ y 2 = 1 =⇒ y = ±1.
The stationary points for f (x, y) are then (1, −1) and (−1, 1).
Now
A=
∂ 2f
= 2y,
∂x2
B=
∂ 2f
= 2x + 2y,
∂x∂y
C=
∂ 2f
= 2x
∂y 2
so
D = AC − B 2 = 4xy − (2x + 2y)2 .
Since D = −4 < 0 for both stationary points, then by the second-partials test, f (x, y) has saddle points
at (1, −1) and (−1, 1).
4