2E1 (Timoney) Tutorial sheet 1
[Tutorials October 11 – 12, 2006]
Name: Solutions
Consider the vector function (with values in R3 )
x(t) = (cos t2 , sin t2 , t2 )
1. Show that the image curve lies in the set
{(x, y, z) ∈ R3 : x2 + y 2 = 1}
and sketch the set in R3 .
Solution: Taking x = cos t2 = cos(t2 ), y = sin t2 and z = t2 with any value of t we find
that x2 + y 2 = cos2 t2 + sin2 t2 = 1 always. (Reason: cos2 θ + sin2 θ = 1 for all θ and here
we are taking θ = t2 ).
For the sketch we should get a cylinder of radius 1 centered along the z-axis. A sketch
(also showing the curve wrapping around it) is included at the end.
2. Find the derivative dx/dt and the unit tangent vector T.
Solution:
dx
=
dt
d
d
d
cos t2 , sin t2 , t2
dt
dt
dt
= (−2t sin t2 , 2t cos t2 , 2t)
The maginitude of this vector is
p
q
√
dx = 4t2 sin2 t2 + 4t2 cos2 t2 + 4t2 = 4t2 (sin2 t2 + cos2 t2 ) + 4t2 = 8t2 .
dt √
√
√
This simplifies to 2 2t —- well actually
it
simplifies
to
2
2|t|
and
2
2t is only correct
√
1
for t ≥ 0, but we will stick to the 2 2t to make life easier. So the unit tangent vector is
dx
1
1
1
1
2
2
2
2
dt
√
√
√
√
T= dx = 2t 2 (−2t sin t , 2t cos t , 2t) = − 2 sin t , 2 cos t , 2
dt
3. Find the vector dT/dt and show that it is perpendicular to T.
Solution:
d
1
d 1
2
2 d 1
− √ sin t , √ cos t , √
dt
dt 2
dt 2
2
√
√
2
2
= (− 2t cos t , − 2t sin t , 0)
dT
=
dt
1
For t < 0 the answer we get for T is minus the right answer. For t = 0, there is really no answer for T as we
cannot find a unit vector in the direction of the zero vector.
To show the vectors are perpendicular, show that their dot product is 0. To take the dot
product we use (x1 , x2 , x3 ) · (y1 , y2 , y3 ) = x1 y1 + x2 y2 + x3 y3 and we find
T·
Richard M. Timoney
dT
= +t sin t2 cos t2 − t cos t2 sin t2 + 0 = 0.
dt