C Roettger, Spring 13
Math 265 Exam 1A – Solutions
Problem 1 Look at the surface above.
a) Which equation belongs to this surface? Choose one and circle it.
x = z2 + y2
x2 + 2y 2 + 4z 2 = 10
x2 + y 2 + 1 = z 2
x2 + y 2 = z 2
b) A spaceship travels on the surface from part a). At time t, the spaceship’s
position is [x, y, z] with x = t, y = 1 − t and z > 0. Find the z-coordinate of
the spaceship from the equation for the surface.
c) Find the velocity vector of the spaceship at time t = 2.
2
2
2
Solution. a) This is an ellipsoid, the correct
p equation is x +2y +4z = 10.
b) From the equation for the surface, z = 10 − x2 − 2y 2 /2. We write this
as a function of t,
z(t) =
Then for c),
1p
1√
10 − t2 − 2(1 − t)2 =
8 − 3t2 + 4t
2
2
√
2
v = 1, −1, (2 − 3t)/(2 8 − 3t + 4t)
At t = 2, this equals
v(2) = (1, −1, −1)
(so this would be the spaceship’s direction if the spaceship was kept on the
surface by some kind of force, and suddenly, at t = 2, the force stops/eg the
engine is switched off).
Problem 2 Consider the plane given by 3x + 4y − 2z = 0.
a) Find the distance of this plane from the origin.
b) Find a point-vector equation of a line perpendicular to this plane, passing
through the point (2, 3, 5).
Solution. a) This can be done with a projection, or a formula from the
book. The easiest way is however to observe that the origin is in this plane,
so the distance is zero.
If you did use the book formula: take the normal vector n = (3, 4, −2) to
the given plane. Then take any point P on the plane, eg P = (0, 1, 2). The
distance of the plane from the origin is
d=
0
|OP .n|
= √ = 0.
||n||
29
b) The direction vector of this line can be chosen as n = (3, 4, −2). The
easiest way to chose a point on the line is P = (2, 3, 5). So we get the vector
equation
r(t) = (2, 3, 5) + t (3, 4, −2) .
Problem 3 A point travels on the helix
r(t) = 3 cos(t)i + 3 sin(t)j + 5tk.
a) Find the velocity and acceleration vectors v, a at time t.
b) At what time(s) is the velocity vector perpendicular to the plane 3x−3z =
0?
Solution. a)
v = −3 sin ti + 3 cos tj + 5k
a = −3 cos ti − 3 sin tj
b) It is perpendicular to the given plane exactly if it is parallel to the normal
vector n = 3i − 3k. This can never happen, because then we would need
cos t = 0, which implies sin t = ±1, and neither of the two resulting vectors
−3i + 5k, 3i + 5k is a multiple of n.
You could also check that the cross product of v and n is never zero to show
this.
Problem 4 Find the area of the triangle given by A = (2, 3, 5), B =
(4, 4, 5), C = (0, 2, 4).
Solution. First, determine two vectors describing two sides of the triangle.
u = AB = (2, 1, 0)
v = AC = (−2, −1, −1)
The area can then be found with the cross product
√
||u × v||
||(−1, 2, 0)||
5
A=
=
=
2
2
2
Problem 5 Consider the curve in space, given for t in [−5, 5] by
r(t) = t4 − 32t, t2 − 4t, t3 − 12t
a) Find the velocity vector v and the acceleration vector a as functions of t.
b) Find all values of t where this curve is not smooth, or explain why there
are none.
c) Set up, but do NOT evaluate, the integral for the arc length of this curve
between t = 1 and t = 5.
Solution. a) v(t) = (4t3 − 32, 2t − 4, 3t2 − 12) and a = (12t2 , 2, 6t).
b) The second component of v is zero only for t = 2. The other components
of v are zero there, too. So the curve is smooth for all values of t except
t = 2. Can you spot the point on the curve where t = 2?
c) The arc length is
Z 5
Z 5p
(4t3 − 32)2 + (2t − 4)2 + (3t2 − 12)2 dt
A=
||v|| dt =
1
which is best left alone.
1