Math 101C Ch. 12 Test
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1. Find the equation of the plane perpendicular to the line x = −2t, y = 2 + 3t, z = 4 − t and through
the point (3, 5, 0).
Solution: (See online HW 12.1.3)
−2( x − 3) + 3(y − 5) − (z − 0) = 0 or − 2x + 3y − z = 9
2. Find the equation of the plane through the points P = (3, 2, 0), Q = (1, 0, 2) and R = (3, 4, 1).
Solution: (See online HW 12.1.7)
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# » #»
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PQ = h−2, −2, 2i PR = h0, 2, 1i, N = PQ × PR = h−6, 2, −4i
This gives the plane − 2( x − 3) + 2(y − 2) − 4z = 0or 3x − y + 2z = 7
3. Sketch the portion of the plane 2x + 3y + 4z = 24 that lies in the first octant.
Solution: (See online HW 12.1.28.a) The plane should pass through the points (12, 0, 0), (0, 8, 0), and
(0, 0, 6).
4. Find the intersection of the planes 3x + 3y + 4z = 24 and x − 3y − 2z = 0.
Solution: (see online HW 12.1.11) Adding the equations gives 4x + 2z = 24 so z = 12 − 2x. Substituting back into the equation of the plane give the equation of the line. Four different versions are given
below.






 x = −1t + 6
 x = 3 t + 24


x
=
t




5
5
2



 x = 6t + 6
5
y =−t+8
y=t
y = 10t + 2
or
or
or
y = − t+2








6


6
12
z = − 2t + 12
z = −2t + 12

z = − t+


z
=
t
5
5
Math 101C
Ch. 12 Test
Page 2 of 4
5. Sketch the graph of the following. All should be drawn as three dimensional graphs.
(a) z =
p
(d) x2 + z2 = 1, Cylinder around the y axis.
x 2 + y2
Solution:
Solution:
(b) x2 + y2 = z2 + 1
(e) z + y2 + x2 = 0
Solution:
Solution:
(c) x2 + y2 = z2 − 1
(f) z = sin x
Solution:
Solution:
6. The portion of the line y = 6 − 2x that lies in the first quadrant of the xy plane is rotated around the y
axis. Find the equation of the resulting surface.
Solution: (See online HW 12.1.27) Since the circles of rotation are around the y axis, the equation must
6−y
be in the form x2 + z2 = R2 . The radius is given by x =
. Therefore the surface is
2
6−y 2
x 2 + z2 =
2
7. Sketch the level curves of z = 2− x
with which z value.
2 − y2
when z = 1 and z =
1
. Be sure to label which level curve goes
4
Math 101C
Ch. 12 Test
Page 3 of 4
Solution: (See online HW 12.4.17.d) z = 1 has a level curve that is a point at the origin. z =
circle centered at the origin with radius 2.
8. Determine the domain of the function z = p
1
4 − x 2 − y2
1
is a
4
.
Solution: The domain is the region inside the circle x2 + y2 = 4 but not including the circle itself.
9. Determine the equation of the tangent line to the level curve of z = 24 − x2 − 2y2 at the point (2, 3, 2).
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Solution: (from online HW 12.6.14) The normal to the level curve is given by the gradient ∇z =
h−4, −12i so the tangent vector to the curve is given by h12, −4i. This gives the tangent line as
x = 2 + 12t, y = 3 − 4t.
10. Determine the maximum value of the directional derivative of f ( x, y) = 24 − x2 − 2y2 at the point
(2, 3, 2).
Solution: (See 12.6.12.d) The maximum value of the directional derivative is
q
√
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||∇z|| = (−4)2 + (−12)2 = 4 10
11. Find the equation of the line normal to the level surface of w = x2 + y2 − z when w = 0 and you are at
the point ( x, y, z) = (3, 4, 25).
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Solution: The gradient ∇w is perpendicular to the surface so the normal vector is ∇w(3, 4, 25) =
h6, 8, −1i. This give the equation of the line as x = 3 + 6t, y = 4 + 8t, z = 25 − t.
12. For xyz2 − 3xz − xy = 7, find
∂z
.
∂x
Solution: (See online HW 12.5.12)
∂z
Fx
−yz2 + 3z + y
=− =
∂x
Fz
2xy − 3x
13. Consider the surface z = 24 − x2 − 2y2 and the path C in the xy plane given by x = cos t, y = sin t on
the interval 0 ≤ t ≤ 2π. Imagine that you are walking on the surface directly above the curve C in the
direction of increasing values of t. Find the values of t for which you are walking uphill.
Solution: (See online HW 12.5.13) Substituting the values of x and y into the surface gives
z = 24 − cos2 t − 2 sin2 t = 23 − sin2 t
Using any of several methods (such as the derivative being positive), this is shown to increase on the
intervals
π
3π
< t < π and
< t < 2π
2
2
Math 101C
Ch. 12 Test
Page 4 of 4
14. Consider the paraboloid z = x2 + y2 and the plane z = 2x + y + 15 which intersects the paraboloid in
a curve C. Find the equation of the tangent line to C at (3, 4, 25).
Solution: (From online HW 12.7.13) The tangent vector to the intersection is found by taking the
cross product of the normal to the paraboloid (n#»1 ) and the normal to the curve of intersection (n#»2 ).
The normal to the paraboloid is given by the gradient of F ( x, y, z) = x2 + y2 − z
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n1 = ∇ F = h6, 8, −1i
The curve C is found by equating the two surfaces. Therefore C is given by
x2 + y2 = 2x + y + 15
or
f ( x, y) = x2 − 2x + y2 − y − 15 = 0
Therefore, the normal to the curve is given by the gradient of f ( x, y)
n2 = h2x − 2, 2y − 1, 0i|(3,4) = h4, 7, 0i
Thus, the tangent vector to the curve of intersection is n#»1 × n#»2 = h7, −4, 10i. Finally, using the point
(3, 4, 25), this gives the equation of the tangent line as
x = 7t + 3, y = −4t + 4, z = 10t + 25