Answer to Question #206415 in Algebra for Ola

Question

Marble Arch in London is a monument with three arches – one large and two smaller.

The smaller arches are each 3.66m high at their apex and 1.83 wide at the base.

By modelling these arches as parabolas, taking the bottom left point of the arch to be (0, 0), find an equation describing the arch.

When on parade, the highest point of the standard carried by a man on horseback is 2.8m above the ground. If the standard would pass 30cm horizontally from the left- most point of the arch, would it fit underneath?

 

Answer 1

1. Assume that the equation of the parabola is

$$y=-ax^2+bx+c$$

$y(0)=0:$

$$0=c=>y=-ax^2+bx$$

$$\dfrac{b}{2a}=\dfrac{1.83}{2}=>b=1.83a$$

$y(0.915)=3.66:$

$$3.66=-a(0.915)^2+1.83a(0.915)$$

$$a=\dfrac{4}{0.915}$$

$$b=8$$

The equation describing the arch is

$$y=-\dfrac{4}{0.915}x^2+8x$$

2.

$$y(0.3)=-\dfrac{4}{0.915}(0.3)^2+8(0.3)\approx2.007<2.8$$

No. The standard would not fit underneath.