Question

An architect has designed two tunnels… Question and rubric listed below. Thanks for your help!

Answer 1

Given:

An architect has designed two tunnels.

Tunnel A is modeled by:

`x^2+y^2+30x+56=0`

Tunnel B is modeled by:

`x^2-30x+16y-95=0`

Part A: write the equation of tunnel A in standard from

So, we will complete the square for the two variables x, and y as follows:

`\begin{gathered} (x^2+30x)+y^2=-56 \\ (x^2+30x+225)+y^2=-56+225 \\ (x+15)^2+y^2=169 \\ (x+15)^2+y^2=13^2 \end{gathered}`

So, the answer to part A:

The equation will be: (x+15)² + y² = 13²

The conic section is: Circle with canter (-15, 0) and radius = 13 feet

Part B: Write the equation of tunnel B in standard from

So, we will complete the square for the variable x as follows:

`\begin{gathered} (x^2-30x)=-16y+95 \\ (x^2-30x+225)=-16y+95+225 \\ (x-15)^2=-16y+320 \\ (x-15)^2=-16(y-20) \\ (x-15)^2=-4(4)(y-20) \end{gathered}`

So, the answer to part B:

The equation will be: (x-15)² = -4(4)(y-20)

The conic section is: a Parabole with the vertex at (15, 20)

Part C: Determine the maximum height of each tunnel.

Tunnel A:

`(x+15)^2+y^2=13^2`

So, the radius = 13, so, the diameter = 2 * 13 = 26

So, the maximum height = 13 feet

The truck has a height = of 13.5 feet

so, it will not be able to pass through the tunnel A

Tunnel B:

`(x-15)^2=-16(y-20)`

when y = 0, solve for x:

`\begin{gathered} (x-15)^2=-16*-20 \\ (x-15)^2=320 \\ x-15=\pm√(320) \\ x=15\pm√(320)=32.88,-2.88 \end{gathered}`

When x = 15, we will find the maximum height

`\begin{gathered} 0=-16(y-20) \\ y-20=0 \\ y=20 \end{gathered}`

So, the maximum height = 20 feet

the truck will be able to pass through the tunnel