Answered: The Gateway Arch in St. Louis was…

Algebra: Structure And Method, Book 1

(REV)00th Edition

ISBN:9780395977224

Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Chapter12: Quadratic Functions

Section12.8: Joint And Combined Variation

Problem 2P

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A graphing calculator is recommended.

The Gateway Arch in St. Louis was designed by Eero Saarinen and was constructed using the equation

y = 211.49 − 20.96 cosh(0.03291765x)

for the central curve of the arch, where x and y are in meters and

|x| ≤ 91.20.

a-What is the height (in m) of the arch at its center? (Round your answer to two decimal places.)

b-At what points is the height 50 m? (Round your answers to two decimal places.)

smaller x-value (x, y)=

larger x-value (x, y)=

c-What is the slope of the arch at the points in part b? (Round your answers to one decimal place.)

at the point with smaller x-value
at the point with larger x-value

Expert Solution

Step 1

Step:-1

Given equation is $y = 211.49 - 20.96 \cos h (0.03291765 x)$ and x, y are in meters and $|x| \leq 91.20$

Sketching the graph using given equation, we get

This is parabola opens downward.

Part (a):-

The height of arch at its center is

$y = 211.49 - 20.96 \cos h (0.03291765 x) put x=0 for height y = 211.49 - 20.96 \cos h (0.03291765 \times 0) y = 211.49 - 20.96 \cos h (0) \cos h (0) = 1 then y = 211.49 - 20.96 \times 1 = 190.53 y = 190.53$

So, height is 190.63 meter.

Using graph, we can see the height of arch is 190.53 meter.

Answer:-

height = 190.63 meter.

Step:-2

Part (b):-

Given that height is 50 m.

Now, we have to find the value of x for which arch attains height 50 m.

Given equation is

$y = 211.49 - 20.96 \cos h (0.03291765 x) put y=50 for values of x, \Rightarrow 50 = 211.49 - 20.96 \cos h (0.03291765 x) \Rightarrow 50 - 211.49 = - 20.96 \cos h (0.03291765 x) \Rightarrow - 161.49 = - 20.96 \cos h (0.03291765 x) \Rightarrow 161.49 = 20.96 \cos h (0.03291765 x) - - - - - - (1) Note: - \cos h (u) = \frac{e^{u} + e^{- u}}{2}, then \cos h (0.03291765 x) = \frac{e^{(0.03291765 x)} + e^{- (0.03291765 x)}}{2} Take v = (0.03291765 x) t h e n \cos h (v) = \frac{e^{v} + e^{- v}}{2} = \frac{e^{v} + \frac{1}{e^{v}}}{2} Again take e^{v} = t then \cos h (v) = \frac{e^{v} + e^{- v}}{2} = \frac{e^{v} + \frac{1}{e^{v}}}{2} = \frac{t + \frac{1}{t}}{2} = \frac{t^{2} + 1}{2 t}$

Step:-3

by (1), we get

$161.49 = 20.96 \cos h (0.03291765 x) \Rightarrow 161.49 = 20.96 (\frac{t^{2} + 1}{2 t}) \Rightarrow 161.49 = 10.48 (\frac{t^{2} + 1}{t}) \Rightarrow 15.4094 = \frac{t^{2} + 1}{t} \Rightarrow 15.4094 t = t^{2} + 1 \Rightarrow t^{2} - 15.4094 t + 1 = 0$

Step:-4

Applying quadratic formula, we get

$t = \frac{15.4094 \pm \sqrt{{(15.4094)}^{2} - 4 \times 1 \times 1}}{2 \times 1} t = \frac{15.4094 \pm \sqrt{233.4481}}{2} t = \frac{15.4094 \pm 15.2790}{2} = 0.0652, 15.3442 t = 0.0652, 15.3442 but we assume e^{v} = t \Rightarrow v = \ln (t) \Rightarrow 0.03291765 x = \ln (t) \Rightarrow x = \frac{\ln (t)}{0.03291765} when t = 0.0652 then \Rightarrow x = - 82.9432 \approx - 82.94 \Rightarrow x = - 82.94 when t = 15.3442 then \Rightarrow x = 82.9566 \approx 82.96 \Rightarrow x = 82.96$

Answer:-

Smaller x-value $(x, y) = (- 82.94, 50)$

Larger x-value $(x, y) = (82.96, 50)$

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