1. ## hyperbolic functions #1

I need help with these two problems in a hurry. Thanks very much!!
1. The Gateway Arch in St. Louis was designed by Eero Saarinen and was constructed using the equation: y=211.49 -20.96 cosh .03291765x for the central curve of the arch, where x and y are measured in meters and lxl < or equal to 91.20.
so the question is how do I graph the central curve.
2. Verify that the function y=f(x)= T/pg cosh (pgx/T) is the solution of the differential equation d^2y/dx^2 = pg/T squareroot 1 + (dy/dx)^2

2. Originally Posted by jsu03
2. Verify that the function y=f(x)= T/pg cosh (pgx/T) is the solution of the differential equation d^2y/dx^2 = pg/T squareroot 1 + (dy/dx)^2
$\displaystyle y = \frac{T}{pg} \cosh \frac{pgx}{T}$

$\displaystyle y' = \frac{T}{pg} \cdot \frac{pg}{T} \sinh \frac{pgx}{T} = \sinh \frac{pgx}{T}$

$\displaystyle y'' = \frac{pg}{T}\cosh \frac{pgx}{T}$

Now, $\displaystyle \frac{pg}{T}\cosh \frac{pgx}{T} = \frac{pg}{T} \sqrt{1+\sinh^2 \frac{pgx}{T} }$ it follows that $\displaystyle y'' = \frac{pg}{T}\sqrt{1+(y')^2}$.