Page 374 - Advanced engineering mathematics

354 CHAPTER 11 Vector Differential Calculus

Since t > 0, a N = t. The acceleration may be written as
√
a = 5T + tN.
If we know a N and v, it is easy to compute the curvature, since
2
2
a N = t = κv = 5t κ,
implying that κ = 1/5t, as we found in Example 11.3.

SECTION 11.2 PROBLEMS

2
2
2
In each of Problems 1 through 10, a position vector is 9. F = αt i + βt j + γ t k
given. Determine the velocity, speed, acceleration, cur-
vature, and the tangential and normal components of the 10. F = 3t cos(t)j − 3t sin(t)k
acceleration.
11. Show that any straight line has curvature zero. Con-
2
1. F = 3ti − 2j + t k versely, if a smooth curve has curvature zero, then it
must be a straight line. Hint: For the first part, recall
2. F = t sin(t)i + t cos(t)j + k
that any straight line has a position vector F(t) =
3. F = 2ti − 2tj + tk (a + bt)i + (d + ct)j + (h + kt)k. For the converse,

t
t
4. F = e sin(t)i − j + e cos(t)k if κ = 0, then T (t) = O.
−t
5. F = 3e (i + j − 2k) 12. Show that the curvature of a circle is constant. Hint:If
6. F = α cos(t)i + βtj + α sin(t)k the radius is r, show that the curvature is 1/r.

||F (t) × F (t)||
7. F = 2sinh(t)j − 2cosh(t)k 13. Show that κ(t) = .

||F (t)|| 3
8. F = ln(t)(i − j + 2k)
11.3 Vector Fields and Streamlines
Vector functions F(x, y) in two variables, and G(x, y, z) in three variables, are called vec-
tor fields. At each point where the vector field is defined, we can draw an arrow representing
the vector at that point. This suggests fields of arrows "growing" out of points in regions of
the plane or 3-space.

Take partial derivatives of vector fields by differentiating each component. For example, if
2
z
F(x, y, z) = cos(x + y + 2z)i − xyzj + xye k,
then
∂F
z
2
=−sin(x + y + 2z)i − yzj + ye k,
∂x
∂F
2
z
=−2y sin(x + y + 2z)i − xzj + xe k,
∂y
and
∂F
z
2
=−2sin(x + y + 2z)i − xyj + xye k.
∂z
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October 15, 2010 16:11 THM/NEIL Page-354 27410_11_ch11_p343-366