Challenge Problems 6
Math126 E. 07/21/2011
Problem 1. [Midterm 2, Pevtsova, Winter 2007, 2] Let r(t) =< t, t2 , t3 >. Find the equation of the
normal plane:
(a) at t = 1;
(b) at r(t) =< −1, 1, −1 >;
(c) find the parametric equations of the line of intersection of these two normal planes.
Problem 2. [Midterm 2, Pevtsova, Winter 2007, 4] (a) Find the velocity and position vectors v(t),
r(t) of a particle if a(t) =< 2, cos t, sin t >, and at the moment t = 0 we have v(0) =< 0, 0, −1 >,
r(0) =< 1, 1, 0 >.
(b) Find the curvature of r(t) at t = 1.
Problem 3. [Final Exam, Spring 2010, 2a] The location of a particle is given by the vector function
r(t) =< 3t − 6, 2t3 − 5t, −t2 + 11 > .
Find the speed of the particle at the instant when it passes through the yz-plane.
Problem 4. [Midterm 2, Milakis, Spring 2007, 5] Let r(t) =< t2 , 2t, t >.
(a) Find the unit tangent vector T, the unit normal vector N and the binormal vector B at the point
where T is parallel to the plane x + y + z = 0.
(b) Find the curvature of r(t) at the point identified in part (a).
Problem 5. [Final Exam, Spring 2007, 6a] An object is moving so that its position at time t (where
t > 0) is given by the vector function r(t) =< t, 1/t, t2 >. Find all values of t at which its acceleration
vector is orthogonal to its velocity vector.
Problem 6. [Midterm 2, Bekyel, Spring 2008, 2] For the curve given by
r(t) =< 4 sin t, 3t, 4 cos t >
(a) Find the unit tangent vector T(t).
(b) Find the unit normal vector N(t).
√
(c) Find parametric equations for the tangent line to the curve at the point (2,
π/2,
2
3).
√
(d) Find the equation of the normal plane to the curve at the point (2, π/2, 2 3).
1
Solutions
Problem 1. A normal plane at any point of a curve is the plane which passes through this point
and has a tangent vector to this curve as its normal vector. But r0 (t) =< 1, 2t, 3t2 >. Hence:
(a) at the point where t = 1 (this point is (1, 1, 1)) we have r0 (1) =< 1, 2, 3 >, and the equation of
the plane is
1 · (x − 1) + 2 · (y − 1) + 3 · (z − 1) = 0, x + 2y + 3z = 6.
(b) at the point (−1, 1, −1) (where t = −1) we have r0 (1) =< 1, −2, 3 >, and the equation of the
plane is
1 · (x + 1) + (−2) · (y − 1) + 3 · (z + 1) = 0, x − 2y + 3z + 6 = 0.
(c) The normal vectors to these planes are n1 =< 1, 2, 3 > and n2 =< 1, −2, 3 >, so the directional
vector of the intersection line is given by n1 × n2 =< 12, 0, −4 >.
Indeed, this line lies on the first plane; therefore, it is orthogonal to the first normal vector n1 .
Similarly, this line lies on the second plane; therefore, it is orthogonal to n2 . Recall that the cross
product n1 × n2 is also orthogonal to n1 and n2 ; and so we can take this cross product as a directional
vector of this line.
But < 12, 0, −4 >= 4 < 3, 0, −1 >; so we can also take < 3, 0, −1 > as a directional vector. Let us
find any point which lies on this line, or, in other words, satisfies the equations
x + 2y + 3z = 6, x − 2y + 3z + 6 = 0.
Set up z = 0. Then we have:
x + 2y = 6, x − 2y = −6.
Sum these equations: 2x = 0, x = 0, 2y = 6 − x = 0, y = 3. So this point is (0, 3, 0). And the
parametric equation of the line of intersection are
x = 3t, y = 3, z = −t.
Problem 2. (a) Use the fundamental theorem of calculus for the vector function r0 (t):
r0 (t) =
Zt
r00 (s)ds + r0 (0) =
0
Zt
Zt
(2i + cos sj + sin sk)ds − k = i
0
Zt
2ds + j
0
Zt
sin sds − k =
cos sds + k
0
0
= 2ti + sin tj + (1 − cos t)k − k = 2ti + sin tj − cos tk.
Similarly,
Zt
r(t) =
r0 (s)ds+r(0) =
0
Zt
Zt
(2si+sin sj−cos sk)ds+i+j = i
0
Zt
2sds+j
0
Zt
sin sds−k
0
cos sds+i+j =
0
s=t
s=t
2
= i s2 s=0 + j (− cos s)|s=0
− k sin s|s=t
s=0 + i + j = (t + 1)i + (2 − cos t)j − sin tk.
(b) The curvature:
r0 (1) = 2i + sin 1j − cos 1k =< 2, sin 1, − cos 1 >, r00 (1) =< 2, cos 1, sin 1 >,
r0 (1) × r00 (1) = i − 2(cos 1 + sin 1)j + 2(cos 1 − sin 1)k,
p
|r0 (1) × r00 (1)| = 12 + 4(cos 1 + sin 1)2 + 4(cos 1 − sin 1)2 =
q
= 1 + 4(cos2 1 + 2 sin 1 cos 1 + sin2 1) + 4(cos2 1 − 2 sin 1 cos 1 + sin2 1) =
2
=
√
1 + 4 + 8 sin 1 cos 1 + 4 − 8 sin 1 cos 1 = 9 = 3,
p
√
√
|r0 (1)| = 22 + sin2 1 + cos2 1 = 4 + 1 = 5,
√
and the curvature at the point t = 1 is equal to
√
|r0 (1) × r00 (1)|
3 5
3
.
=√ 3 =
|r0 (1)|3
25
5
Problem 3. The particle passes through yz-plane when x = 0, i.e. 3t − 6 = 0, t = 2. The velocity
is
r0 (t) =< 3, 6t2 − 5, −2t > .
So r0 (2) =< 3, 24 −p
5, −4 >=< 3, 19, −4 √
>, and the speed (=the absolute value of velocity) at this
2
2
2
moment is equal to 3 + 19 + (−4) = 386.
Problem 4. (a) We shall
use formulas from
the end of Section 13.3, Stewart. First of all, r0 (t) =<
p
√
2t, 2, 1 >. Hence |r0 (t)| = (2t)2 + 22 + 12 = 4t2 + 5,
T(t) =
r(t)
2t
2
1
√
√
√
=<
,
,
>
|r0 (t)|
4t2 + 5 4t2 + 5 4t2 + 5
This vector is parallel to r. Hence it is parallel to the plane x + y + z = 0 if and only if the vector r is
parallel to this plane. And this condition is equivalent to the following: r is orthogonal to the vector
< 1, 1, 1 >, which is normal to this plane. These vectors are orthogonal if and only if their dot product
equals 0, i.e.
2t · 1 + 2 · 1 + 1 · 1 = 0, 2t + 3 = 0, t = −3/2.
Let us compute the normal vector:
√
√
√
2 4t2 + 5 − 2t 2√4t8t2 +5
d
(2t)0 4t2 + 5 − 2t( 4t2 + 5)0
2t
√
=
=
=
√
2
dt
4t2 + 5
4t2 + 5
4t2 + 5
10
2(4t2 + 5) − 8t2
=
,
2
3/2
2
(4t + 5)
(4t + 5)3/2
d
1
1
(4t2 + 5)0
4t
√
= −
·
=− 2
.
2
3/2
2
dt
2
(4t + 5)
(4t + 5)3/2
4t + 5
Hence, differentiating each component, we obtain:
=
T0 (t) =<
10
8t
4t
,
−
,
−
>.
(4t2 + 5)3/2 (4t2 + 5)3/2 (4t2 + 5)3/2
At the point t = −3/2, we have 4t2 + 5 = 4(−3/2)2 + 5 = 14, and
−3
2
1
1
T(−3/2) =< √ , √ , √ >= √ < −3, 2, 1 >,
14 14 14
14
T0 (−3/2) =<
10
12
6
, 3/2 , 3/2 >= 14−3/2 < 10, 12, 6 > .
3/2
14
14
14
Hence
N(−3/2) =
And
T0 (−3/2)
< 10, 12, 6 >
< 10, 12, 6 >
1
1
=
=√
=√
< 10, 12, 6 >= √ < 5, 6, 3 > .
0
|T (−3/2)|
| < 10, 12, 6 > |
102 + 122 + 62
280
70
1
B(−3/2) = T(−3/2) × N(−3/2) = √ √ < −3, 2, 1 > × < 5, 6, 3 >=
14 70
3
1
1
√ < 0, 14, −28 >= √ < 0, 1, −2 > .
14 5
5
=
(b) The curvature is equal to
r
√
√
√
|T0 (−3/2)|
14−3/2 102 + 122 + 62
14−3/2 280
10
5
1
p
√
=
=
=
.
=
|r0 (−3/2)|
143/2
14 14
14
4(−3/2)2 + 5
Problem 5. We have:
1
2
r0 (t) =< 1, − 2 , 2t >, r00 (t) =< 0, 3 , 2 > .
t
t
So these vectors are orthogonal if and only if
r0 (t) · r00 (t) = 0, −
1
2
1
6
√
,
t
=
±
+
4t
=
0,
t
=
.
6
t5
2
2
√
But t > 0, so t = 1/ 6 2.
√
√
Problem 6. (a) r0 (t) =< 4 cos t, 3, −4 sin t >, |r0 (t)| = 16 cos2 t + 9 + 16 sin2 t = 16 + 9 = 5.
Therefore,
r0 (t)
4
3 4
T(t) = 0
=< cos t, , − sin t > .
|r (t)|
5
5 5
(b)
s 2
2
4
4
4
4
4
0
0
2
sin t +
cos2 t = .
T (t) =< − sin t, 0, − cos t >, |T (t)| =
5
5
5
5
5
Therefore,
N(t) =
T0 (t)
=< − sin t, 0, − cos t > .
|T0 (t)|
(c) This point corresponds to the value t = π/6√(compare the second components: 3t = π/2, t = π/6).
The tangent vector at this point is r0 (π/6) =<√
2 3, 3, −2 >. This is a directional vector of this tangent
line, which passes through the point (2, π/2, 2 3). The parametric equation of this tangent line is:
√
√
x = 2 + 2 3t, y = π/2 + 3t, z = 2 3 − 2t.
√
(d) This plane passes through the point (2, π/2, 2 3) and has r0 (π/6) as a normal vector. So the
equation of this plane is
√
√
2 3(x − 2) + 3(y − π/2) − 2(z − 2 3) = 0.
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