find the length of the indicated portion of the curve r(t) =(e^t cost) i+(e^t sint) j+ e^t k, -ln 4<= t <= 0
Answers
Answer:
Suppose that C is a curve in xy-plane given by the equations x = x(t) and y = y(t) on the
interval a ≤ t ≤ b. Recall that the length element ds is given by ds =
q
(x
0
(t))2 + (y
0
(t))2 dt.
Let z = f(x, y) be a surface. The line inte-
gral of C with respect to arc length is
Z
C
f(x, y)ds =
Z b
a
f(x(t), y(t)) q
(x
0
(t))2 + (y
0
(t))2 dt
This integral represents the area of the surface
between the curve C in the xy-plane and the pro-
jection of the curve C on the surface z = f(x, y).
This area is represented by the “curtain” in or-
ange in the figure on the right.
Three dimensional curves. Suppose that C is a curve in space given by the equations x = x(t),
y = y(t), and z = z(t) on the interval a ≤ t ≤ b. Recall that the length element ds is given by
ds = |~r(t)| =
q
(x
0
(t))2 + (y
0
(t))2 + (z
0
(t))2 dt
Let f(x, y, z) be a function. The line integral of C with respect to arc length is
Z
C
f(x, y, z) ds =
Z b
a
f(x(t), y(t), z(t)) q
(x
0
(t))2 + (y
0
(t))2 + (z
0
(t))2 dt
Note that the quotient ds
dt =
q
(x
0
(t))2 + (y
0
(t))2 + (z
0
(t))2
is always positive because the right side of
the equation is always positive. Thus, if the lower t-value is used as the lower bound of the integral
and the larger t-value as the upper, that ensures that dt is positive and, hence, ds is positive also.
Three applications. (1) The total length L of a curve C parametrized by ~r(t) = hx(t), y(t), z(t)i
can be found by integrating ds from the beginning to the end of C.
L =
Z
C
ds =
Z b
a
q
(x
0
(t))2 + (y
0
(t))2 + (z
0
(t))2 dt
(2) If C is the trajectory of an object and it is parametrized by ~r(t) = hx(t), y(t), z(t)i, the
quotient
ds
dt =
q
(x
0
(t))2 + (y
0
(t))2 + (z
0
(t))2
computes the speed of the object at time t. Thus, the integral R b
a
ds computes the total distance
traveled from the time t = a to the time t = b