MATH 203
Lab 4 Solutions
Spring 2005
(1) Find the point of intersection of the tangent lines to the curve
r(t) =< sin πt, 2 sin πt, cos πt >
at the points t = 0 and t = 1/2. Illustrate by graphic the curve and both tangent lines.
r0 (t) =< π cos πt, 2π cos πt, −π sin πt >.
At t = 0, we have r(0) =< 0, 0, 1 > and r0 (0) =< π, 2π, 0 >. The equation of tangent line is hence
a(u) =< 0, 0, 1 > +u < π, 2π, 0 > .
At t = 1/2, we have r(1/2) =< 1, 2, 0 > and r0 (1/2) =< 0, 0, −π/2 >. The equation of tangent line is
b(v) =< 1, 2, 0 > +v < 0, 0, −π/2 > .
If the lines intersect, then there is a common solution to u, v for the lines a, b. That is, for the x, y
and z coordinates,
0 + πu
0 + 2πu
1 + 0u
= 1 + 0v
= 2 + 0v
= 0 + −π/2v
Hence u = 1/π and u = −2/π. Substituting, the common point is (1, 2, 1).
(2) Consider the curve
r(t) = cos(4t)i + sin(4t)j + 4tk
(a) Find the length of the curve for 0 ≤ t ≤ π/2.
r0 (t) = −4 sin(4t)i + 4 cos(4t)j + 4k. Hence
Z
π/2
p
√
(−4 sin(4t))2 + (4 cos(4t))2 + 42 dt = 2 2π
0
(b) Reparametrize the curve with respect to arc length measured from the point t = 0 in the direction
of increasing t.
The length function is
Z t
√
s(t) =
kr0 (t)kt du = 4 2t.
0
Hence reparametrize as
√
√
√
r(s) = cos(s/ 2)i + sin(s/ 2)j + s/ 2k.
(c) Find the curvature of the curve. Use the curvature to describe the behavior of the curve as t
increases.
We have
r0 (t)
1
1
1
= − √ sin(4t)i + √ cos(4t)j + √ k.
T(t) = 0
kr (t)k
2
2
2
Hence
√
√
T0 (t) = −2 2 cos(4t)i − 2 2 sin(4t)j.
MATH 203
Lab 4 Solutions
and
Spring 2005
√
kT0 (t)k = 2 2
Hence
κ(t) =
kT0 (t)k
1
= .
kr0 (t)k
2
The curvature is constant as t increases.
(3) A circle of radius r rolls along the x−axis in the positive direction, rotating at the rate of 1 radian per
unit of time. Let P be the point on the circle that is at the origin at time 0. The spacecurve p(t) that
describes the movement of P can be derived from the following figure:
y
p(t)
a
O
C
t
b
q
x
rt
(a) Verify that
p(t) = a + b =< r(t − sin t), r(1 − cos t) >
Note that a =< rt, r > and b =< r cos q, r sin q >. Since q + r = 3π/2. Hence
b =< r cos(3π/2 − r), r sin(3π/2 − r) >=< −r sin t, −r cos t > .
Hence result.
(b) Graph several arches of the cycloid using Maple.
(c) If p represents the position vector, find the velocity and acceleration vectors at t = 0, 1, 2, 3, then
graph them with Maple on the cycloid graph.
v(t) = p0 (t) =< r(1 − cos t), r sin t >
a(t) = p00 (t) =< r sin t, r cos t >
(d) Find the distance travelled by P along one arch of the cycloid. What does the length tell you
with respect to r?
Z
L=
2π
kp0 (t)k = 8r
0
(Read page 470, example 4, substitution is the half angle formula 1 − cos t = 2 sin2 (t/2).)
MATH 203
Lab 4 Solutions
Spring 2005
(e) Find the curvature of the path with respect to t. Is the curvature dependent on r?
Note that
i
j
k p0 (t) × p00 (t) = r(1 − cos t) r sin t 0 = r2 (cos t − 1)k
r sin t
r cos t 0 kp0 (t) × p00 (t)k = r2 (1 − cos t).
and
kp0 (t)k3 = (2r2 (1 − cos t))3/2 .
Hence
1
κ(t) = √ √
.
2 2r 1 − cos t
Hence, the curvature decreases as r increases.
(4) The velocity of a particle moving in space is given by
v(t) = cos ti − sin tj + k
(a) Find the position function of the particle if r(0) = 2i + k.
Integrate on both sides
Z
r(t) = cos ti − sin tj + k dt = sin ti + cos tj + tk + C
Since r(0) = 0i + j + 0k + c = 2i + k, we have C = 2i − j + k. Therefore
r(t) = (2 + sin t)i + (−1 + cos t)j + (1 + t)k.
(b) What is the position of the aprticle when t = π.
r(π) = 2i − 2j + (π + 1)k.
(c) Determine the curve which is the projection of the path onto the xy−plane, and then give a sketch
of the path of the article.
Set z coordinate to be 0, we get the path
s(t) = (2 + sin t)i + (−1 + cos t)j
Let x = 2 + sin t and y = −1 + cos t. We have
(x − 2)2 + (y + 1)2 = 1.
Hence the path is a circle with center (2, −1) and radius 1 on the xy− plane.
(d) Find the distance travelled by the particle on the interval [0, π]
The distance travelled is the arc length of r(t) for t from 0 to π, which is
Z π
√
L=
kr0 (t)k dt = 2π.
0
Note that this is different from the displacement, which is
Z π
v(t) dt = −2j + πk.
0
(What does each of the two quantities mean?)
MATH 203
Lab 4 Solutions
Spring 2005
(e) Find the curvature of the path.
T(t) =
Hence
r0 (t)
1
= √ (cos ti − sin tj + k)
kr0 (t)k
2
1
T0 (t) = √ (− sin ti − cos tj),
2
Therefore,
kT0 (t)k = 1.
κ(t) =
kT0 (t)k
1
=√ .
0
kr (t)k
2
y 2
z 2
(5) A particle moves around the ellipse
+
=1
3
2
in the yz−plane in such a way that its position at time t is given by r(t) = 3 cos tj + 2 sin tk. Find the
maximum and minimum speed of the particle.
r0 (t) = −3 sin tj + 2 cos tk
Let v(t) be the speed function of the particle at time t, then
p
p
v(t) = kr0 (t)k = 9 sin2 t + 4 cos2 t = 4 + 5 sin2 t
To find the maximum and minimum, we go back to single variable calculus,
5 sin(2t)
v 0 (t) = p
2 4 + 5 sin2 t
Then v 0 (t) = 0 when t = kπ/2, where k is any integer.
Further analysis shows that v(t) obtains a minimum value of 2 at t = 0, π, 2π, . . . and a maximum
value of 3 at t = π/2, 3π/2, 5π/2, . . .