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.
The bracket BCD is hinged at C and attached to a control cable at B. Let F₁ = 275 N and F2 = 275 N.
F1
B
a=0.18 m
C
A
0.4 m
-0.4 m-
0.24 m
Determine the reaction at C.
The reaction at C
N Z
F2
D
The correct answer is C,i know that we need to use stokes theorem and parametrize the equations then write the equation F with respect to the curve but i cant seem to find a way to do it, the integral should be from 0 to 2pi but i might be wrongcould you show me the steps to get to 18pi
A 10-ft boom is acted upon by the 810-lb force as shown in the figure.
D
6 ft
6 ft
E
B
7 ft
C
6 ft
4 ft
W
Determine the tension in each cable and the reaction at the ball-and-socket joint at A.
The tension in cable BD is
lb.
The tension in cable BE is
lb.
The reaction at A is (
lb) i +
Ib) j. (Include a minus sign if necessary.)
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