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The gateway arch was designed by eero saarinen and was constructed using the equation

y=211. 49-20. 6cosh(0. 03291765x)

for the central curve of the arch, where x and y are in meters and the absolute value of x is less than or equal to 91. 20


If the height is 60 m at the points (-81, 60) and (81, 60), what is the slope of the arch at these points?



Answer :

The slope of the arch at these points is for the equation  y=211. 49-20. 6cosh(0. 03291765x) is  4.852 and -4.852

The height function is  given by  y=211.49- 20. 6 cos h(0. 03291765x), the points  given to us are   (-81, 60) and    (81, 60), now differentiating the height  function  with respect to x, we get:

 y'(x)  =0 +20.6(0. 03291765)sinh(0. 03291765x)

         =   0.67810359 sinh(0. 03291765x)

the slope will be, m =-0.67810359 sinh(0. 03291765x),now  putting  the smallest value of x =-81 ,

the slope will be y'(-81)= -0.67810359 sinh(0. 03291765*-81)

                =0.67810359 sinh(-2.666)

                                    = - 0.67810359*  -7.1563 = 4.852

the slope for   y'(81)= 0.67810359 sinh(2.666)

                                   =- 0.67810359*7.15639 =-4.852

To know more about slope refer to the  link  https://brainly.com/question/3605446?referrer=searchResults.

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