A cable hanging freely between two vertical support beams forms a curve called a catenary. The shape of a catenary resembles a parabola but mathematically the two functions are quite different. a . On a graphing utility, graph a catenary defined by y = 1 2 e x + e − x and graph the parabola defined by y = x 2 + 1 . b. A catenary and a parabola are so similar in shape that we can often use a parabolic curve to approximate the shape of a catenary. For example, a bridge has cables suspended from a larger approximately parabolic cable. Take the origin at a point on the road directly below the vertex and write an equation of the parabolic cable. c. Determine the focal length of the parabolic cable. d. Determine the length of the vertical support cable 100 ft from the vertex. Round to the nearest tenth of a foot.
A cable hanging freely between two vertical support beams forms a curve called a catenary. The shape of a catenary resembles a parabola but mathematically the two functions are quite different. a . On a graphing utility, graph a catenary defined by y = 1 2 e x + e − x and graph the parabola defined by y = x 2 + 1 . b. A catenary and a parabola are so similar in shape that we can often use a parabolic curve to approximate the shape of a catenary. For example, a bridge has cables suspended from a larger approximately parabolic cable. Take the origin at a point on the road directly below the vertex and write an equation of the parabolic cable. c. Determine the focal length of the parabolic cable. d. Determine the length of the vertical support cable 100 ft from the vertex. Round to the nearest tenth of a foot.
Solution Summary: The author explains how to graph a catenary and parabola using the Ti-83 graphing calculator.
A cable hanging freely between two vertical support beams forms a curve called a catenary. The shape of a catenary resembles a parabola but mathematically the two functions are quite different.
a. On a graphing utility, graph a catenary defined by
y
=
1
2
e
x
+
e
−
x
and graph the parabola defined by
y
=
x
2
+
1
.
b. A catenary and a parabola are so similar in shape that we can often use a parabolic curve to approximate the shape of a catenary. For example, a bridge has cables suspended from a larger approximately parabolic cable. Take the origin at a point on the road directly below the vertex and write an equation of the parabolic cable.
c. Determine the focal length of the parabolic cable.
d. Determine the length of the vertical support cable
100
ft
from the vertex. Round to the nearest tenth of a foot.
Write the given third order linear equation as an equivalent system of first order equations with initial values.
Use
Y1 = Y, Y2 = y', and y3 = y".
-
-
√ (3t¹ + 3 − t³)y" — y" + (3t² + 3)y' + (3t — 3t¹) y = 1 − 3t²
\y(3) = 1, y′(3) = −2, y″(3) = −3
(8) - (888) -
with initial values
Y
=
If you don't get this in 3 tries, you can get a hint.
Question 2
1 pts
Let A be the value of the triple integral
SSS.
(x³ y² z) dV where D is the region
D
bounded by the planes 3z + 5y = 15, 4z — 5y = 20, x = 0, x = 1, and z = 0.
Then the value of sin(3A) is
-0.003
0.496
-0.408
-0.420
0.384
-0.162
0.367
0.364
Question 1
Let A be the value of the triple integral SSS₂ (x + 22)
=
1 pts
dV where D is the
region in
0, y = 2, y = 2x, z = 0, and
the first octant bounded by the planes x
z = 1 + 2x + y. Then the value of cos(A/4) is
-0.411
0.709
0.067
-0.841
0.578
-0.913
-0.908
-0.120
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