# MA2712 – Problems

## Transcription

MA2712 – Problems
```MA2712 – Problems
2016 | October | 25
Martins Bruveris
6
Gradients and Level Surfaces
55. True or false: ∇ f (a, b) is perpendicular to the graph of z = f (x, y) at the point (a, b)?
Solution. False.
The vector ∇ f (a, b) is perpendicular to the level curves of f (x, y), not to its graph.
56. Find ∇ f at the given point and plot it on the level surface f (x, y, z) = c passing through that point
(a) f (x, y, z) = x 2 + y 2 + z 2 at (0, 0, 1/2)
(b) f (x, y, z) = z − x 2 − y 2 at (0, 0, 1/2)
(c) f (x, y, z) = z − x + y at (−1/3, 1/3, 1/3)
(d) f (x, y, z) = x + y − z 2 at (0, 0, 0)
Solution. See the figures for plots of the level surfaces.
(a)
(b)
(c)
The vectors plotted in the figures are the normalized gradients
(a) The level surface
x 2 + y 2 + z2 =
p


2
1
and ∇ f 0, 0,
= k.
is a sphere of radius
2
2
(b) The level surface
z=
is a paraboloid, shifted by
(d)
1
∇ f at the given point.
k∇ f k
1
2
1
+ x2 + y2
2
1
in the z-direction and the gradient is
2


1
∇ f 0, 0,
= k.
2
(c) The level surface −x + y + z = 1 is a plane. To draw it one finds the points, where it crosses the
coordinate axes: the x-axis at −1, the y- and z-axes at 1. The gradient is


1 1 1
∇f
,− ,
= −i + j + k .
3 3 3
MA2712
2016 | October | 25
Problems
(d) The level surface can be written as the graph of the function x = z 2 − y. The gradient is
∇ f (0, 0, 0) = i + j.
57. Find a unit normal to the given surface at the given point.
(a) x yz = 8 at (1, 1, 8)
(b) x 2 y 2 + y − z + 1 = 0 at (0, 0, 1)
(c) cos(x y) = ez − 2 at (1, π, 0)
(d) e x yz = e at (1, 1, 1)
Solution. A normal vector to the surface f (x, y, z) = c at (x 0 , y0 , z0 ) is given by ∇ f (x 0 , y0 , z0 ). We
obtain a unit normal vector by rescaling the gradient to unit length. Note that there are two choices
for the unit normal vector, since the vectors n and −n both have the same length.
(a) With f (x, y, z) = x yz we have
∇ f (x, y, z) = yzi + xzj + x yk
∇ f (1, 1, 8) = 8i + 8j + k .
1
A unit normal vector is n = p
(8i + 8j + k).
129
(b) With f (x, y, z) = x 2 y 2 + y − z we have
∇ f (x, y, z) = 2x y 2 i + (2x 2 y + 1)j − k
∇ f (0, 0, 1) = j − k .
1
A unit normal vector is n = p (j − k).
2
(c) With f (x, y, z) = cos(x y) − ez we have
∇ f (x, y, z) = − sin(x y) ( yi + xj) − ez k
∇ f (1, π, 0) = −k .
A unit normal vector is n = k.
(d) With f (x, y, z) = e x yz we have
∇ f (x, y, z) = e x yz ( yzi + xzj + x yk)
∇ f (1, 1, 1) = i + j + k .
1
A unit normal vector is n = p (i + j + k).
3
58. Find the equation of the tangent plane to each surface at the indicated point.
p
(a) x 2 + y 2 + 3z 2 = 10; 1, 3, 1
(b) x yz 2 = 1; (1, 1, 1)


1
(c) x + 2 y + 3xz = 10; 1, 2,
3
(d) y 2 − x 2 = 3; (1, 2, 8)
2
2
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Problems
Solution. The tangent plane to the surface f (x) = c at the point x0 = (x 0 , y0 , z0 ) is the plane, which
passes through the point x0 and is orthogonal to ∇ f (x0 ). The equation of such a plane is
∇ f (x0 ) · (x − x0 ) = 0
⇔
f x (x0 ) · (x − x 0 ) + f y (x0 ) · ( y − y0 ) + fz (x0 ) · (z − z0 ) = 0 .
(a) With f (x, y, z) = x 2 + y 2 + 3z 2 we have
∇ f (x, y, z) = 2xi + 2 yj + 6zk
p
p
∇ f 1, 3, 1 = 2i + 2 3j + 6k ,
and so the tangent plane has the equation
p
p
2(x − 1) + 2 3( y − 3) + 6(z − 1) = 0
p
x + 3 y + 3z = 7 .
(b) With f (x, y, z) = x yz 2 we have
∇ f (x, y, z) = yz 2 i + xz 2 j + 2x yzk
∇ f (1, 1, 1) = i + j + 2k ,
and so the tangent plane has the equation
(x − 1) + ( y − 1) + 2(z − 1) = 0
x + y + 2z = 4 .
(c) With f (x, y, z) = x 2 + 2 y 2 + 3xz we have
∇ f (x, y, z) = (2x + 3z)i + 4 yj + 3xk


1
∇ f 1, 2,
= 3i + 8j + 3k ,
3
and so the tangent plane has the equation


1
3(x − 1) + 8( y − 2) + 3 z −
=0
3
3x + 8 y + 3z = 20 .
(d) With f (x, y, z) = y 2 − x 2 we have
∇ f (x, y, z) = −2xi + 2 yj
∇ f (1, 2, 8) = −2i + 4j ,
and so the tangent plane has the equation
−2(x − 1) + 4( y − 2) + 0 · (z − 8) = 0
−x + 2 y = 3 .
p 59. Suppose that a particle is ejected from the surface x 2 + y 2 − z 2 = −1 at the point 1, 1, 3 in a
direction normal to the surface at time t = 0 with a speed of 10 units per second. When and where
does it cross the x y-plane?
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MA2712
2016 | October | 25
Problems
p Solution. A normal vector to the surface x 2 + y 2 − z 2 = −1 at the point 1, 1, 3 is given by
p ∇ f 1, 1, 3 , where f is the function f (x, y, z) = x 2 + y 2 − z 2 . The gradient is
p p
∇ f (x, y, z) = 2xi + 2 yj − 2zk ,
∇ f 1, 1, 3 = 2i + 2j − 2 3k .
A unit normal vector is given by
p 1
n= p
i + j − 3k .
11
p Denote by r0 = 1, 1, 3 the starting point. Then the path of the particle is described by the curve
σ(t) = r0 + 10tn ,
and we want to know, when this curve crosses the x y-plane. It crosses the x y-plane, when the
z-coordinate equals 0, leading to the equation
p
p
10 3
3− p
t = 0,
11
p
11
. To find out, where the particle crosses the x y-plane, we evaluate
which has the solution t =
10
p 11
= (2, 2, 0) .
σ
10
p
11
Thus the particle crosses the x y-plane at time
at the point (2, 2, 0).
10
60. If f (x, y) = x y, find the gradient vector ∇ f (3, 2) and use it to find the tangent line to the level curve
f (x, y) = 6 at the point (3, 2). Sketch the level curve, the tangent line, and the gradient vector.
Solution. Even though this function depends only on two variables, instead of three, we can use the
same methods to compute the gradient and tangent line. The
y
∇ f (x, y) = yi + xj
xy =6
∇ f (3, 2) = 2i + 3j .
∇ f (3, 2)
The tangent line is the line through (3, 2) that is orthogonal to
∇ f (3, 2). It has the equation
f x (3, 2)(x − 3) + f y (3, 2)( y − 2) = 0
2
2x + 3 y = 12 .
2x + 3 y = 12
3
x
61. At what point on the ellipsoid x 2 + y 2 +2z 2 = 1 is the tangent plane parallel to the plane x +2 y +z = 1?
Solution. First we compute the tangent plane to the ellipsoid at a general point (x 0 , y0 , z0 ). The
ellipsoid is the level set of f (x, y, z) = x 2 + y 2 + 2z 2 for the value 1. Thus
∇ f (x, y, z) = 2xi + 2 yj + 4zk ,
and the tangent plane is given by
2x 0 x + 2 y0 y + 4z0 z = 2x 02 + 2 y02 + 4z02
2x 0 x + 2 y0 y + 4z0 z = 2(x 02 + y02 + 2z02 )
x 0 x + y0 y + 2z0 z = 1 ,
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Problems
because the point (x 0 , y0 , z0 ) is assumed to lie on the ellipsoid and thus satisfies x 02 + y02 + 2z02 = 1.
The question become, for which (x 0 , y0 , z0 ) are the planes x 0 x + y0 y +2z0 z = 1 and x +2 y +z = 1
parallel. Two planes are parallel, if the normal vectors are parallel. So there must exist λ 6= 0, such
that
x0 = λ
y0 = 2λ
2z0 = λ ,
and because x 02 + y02 + 2z02 = 1, also
p
1 2
11 2
22
λ + 4λ + λ = 1 ⇔
λ =1 ⇔ λ=±
.
2
2
11
p
p
p
p p
p 22 2 22 22
22 2 22
22
,
,
and −
,−
,−
.
Thus we find the two points
11
11
22
11
11
22
2
2
62. Show that every plane that is tangent to the cone x 2 + y 2 = z 2 passes through the origin.
Solution. The cone is the zero level set of the function f (x, y, z) = x 2 + y 2 − z 2 , whose gradient is
∇ f (x, y, z) = 2xi + 2 yj − 2zk .
The tangent plane at a point (x 0 , y0 , z0 ) is
2x 0 x + 2 y0 y − 2z0 z = 2x 02 + 2 y02 − 2z02
x 0 x + y0 y + z0 z = 0 ,
because the point (x 0 , y0 , z0 ) is assumed to lie on the cone. The right hand side of the equation
x 0 x + y0 y + z0 z = 0 vanishes and therefore the tangent plane through (x 0 , y0 , z0 ) passes through the
origin. Since (x 0 , y0 , z0 ) was an arbitrary point, every tangent plane to the cone passes through the
origin.
63. Where does the normal line to the paraboloid z = x 2 + y 2 at the point (1, 1, 2) intersect the paraboloid
a second time?
Solution. The paraboloid is the zero level set of the function f (x, y, z) = x 2 + y 2 − z. Its gradient is
∇ f (x, y, z) = 2xi + 2 yj − k
The normal line can be parametrized by
 
 
1
2
 
 
σ(t) = 1 + t  2 
2
−1
∇ f (1, 1, 2) = 2i + 2j − k .
⇔
σ(t) = (1 + 2t, 1 + 2t, 2 − t) .
It intersect the paraboloid, when f (σ(t)) = 0, meaning that t satisfies
(1 + 2t)2 + (1 + 2t 2 ) − (2 − t) = 0
8t 2 + 9t = 0
⇔
When t = 0 we recover the point (1, 1, 2) and when t = −
Martins Bruveris
9
t = 0 or t = − .
8


9
5 5 25
we find the second point − , − ,
.
8
4 4 8
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Problems
64. Show that the sum of the x-, y- and z-intercepts of any tangent plane to the surface
p
x+
p
p
p
y+ z = c
is a constant.
Hint: The x-intercept is where the tangent plane meets the x-axis.
p
p
p
Solution. The gradient of the function f (x, y, z) = x + y + z is
∇ f (x, y, z =
1 1
1 1
1 1
p i + p j + p k,
2 x
2 y
2 z
and the tangent plane at the point (x 0 , y0 , z0 ) is
1 x
1 y
1 z
1 x0
1 y0
1 z0
p + p + p = p + p + p
2 x 0 2 y0 2 z0
2 x 0 2 y0 2 z0
⇔
p
y
x
z
p +p +p = c
x0
y0
z0
What are the intercepts? To find the x-intercept, we set y = z = 0 and obtain x =
p p
p p
the y-intercept is y = c y0 and the z-intercept is z = c z0 . Their sum is
p p
c x 0 . Similarly,
p p p
p
p p
p p
p p
p p
c x 0 + c y0 + c z0 = c
x 0 + y0 + z0 = c · c = c ,
because the point (x 0 , y0 , z0 ) satisfies f (x 0 , y0 , z0 ) =
p
c. Thus the sum of the intercepts of any
tangent plane does not depend on the point, where the tangent plane is computed.
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