MHB Ale's question at Yahoo Answers regarding a tunnel with parabolic cross-section

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A parabolic tunnel is proposed to accommodate two lanes of truck traffic, with trucks measuring 5m wide and 7m high. To ensure safe passage, the design includes an additional 1m of clearance on either side and above the trucks. The parabolic equation derived for the tunnel's cross-section is P(x) = -1/13x^2 + 140/13, which passes through the points (±6, 8) and (±7, 7). This equation ensures that the tunnel's shape provides adequate space for vehicles without risk of collision. Further algebraic problems are welcomed for discussion in the forum.
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Here is the question:

This is another question about parabola?

This is another question about parabola.

A tunnel is to built to allow 2 lanes of traffic to pass from one side of a mountain to the other side. The largest vehicles are trucks, which can be considered as rectangles 5m wide and 7m high.
Investigate the cross-section of a parabolic tunnel and find a possible equation to represent it. Allow some space so trucks do not bump into each other or the sides of the tunnel.

Here is a link to the question:

This is another question about parabola? - Yahoo! Answers

I have posted a link there so the OP can find my response.
 
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Re: ale's question at Yahoo! Ansers regarding a tunnel with parabolilc cross-section

Hello ale,

Let's orient the origin of our coordinate system such that it is at the mid-point of the base of the cross-section, with the vertex of the parabola directly above it.

Let's allow 1 m on either side of the trucks and above them as in the diagram:

https://www.physicsforums.com/attachments/832._xfImport

We see we will require the parabola to pass through the points:

$$(\pm6,8),\,(\pm7,7)$$

Now, letting the parabola be:

$$P(x)=ax^2+bx+c$$

we obtain the linear system:

(1) $$P(6)=36a+6b+c=8$$

(2) $$P(-6)=36a-6b+c=8$$

(3) $$P(7)=49a+7b+c=7$$

(4) $$P(-7)=49a-7b+c=7$$

Adding (1) and (2) and (3) and (4) to eliminate $b$, we get:

(5) $$36a+c=8$$

(6) $$49a+c=7$$

Subtracting the (5) from (6) to eliminate $c$, we find:

$$13a=-1\,\therefore\,a=-\frac{1}{13}$$

and so substituting into (5), we have:

$$36\left(-\frac{1}{13} \right)+c=8\,\therefore\,c=\frac{140}{13}$$

and then substituting into (1) to find $b$, we get:

$$36\left(-\frac{1}{13} \right)+6b+8+36\left(\frac{1}{13} \right)=8\,\therefore\,b=0$$

And so we have found:

$$P(x)=-\frac{1}{13}x^2+\frac{140}{13}=\frac{140-x^2}{13}$$

To ale and any other guests viewing this topic, I invite and encourage you to post other algebra problems here in our http://www.mathhelpboards.com/f2/ forum.

Best Regards,

Mark.
 

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Re: ale's question at Yahoo! Ansers regarding a tunnel with parabolilc cross-section

MarkFL said:
Hello ale,

Let's orient the origin of our coordinate system such that it is at the mid-point of the base of the cross-section, with the vertex of the parabola directly above it.

Let's allow 1 m on either side of the trucks and above them as in the diagram:

View attachment 832

We see we will require the parabola to pass through the points:

$$(\pm6,8),\,(\pm7,7)$$

Now, letting the parabola be:

$$P(x)=ax^2+bx+c$$

we obtain the linear system:

(1) $$P(6)=36a+6b+c=8$$

(2) $$P(-6)=36a-6b+c=8$$

(3) $$P(7)=49a+7b+c=7$$

(4) $$P(-7)=49a-7b+c=7$$

Adding (1) and (2) and (3) and (4) to eliminate $b$, we get:

(5) $$36a+c=8$$

(6) $$49a+c=7$$

Subtracting the (5) from (6) to eliminate $c$, we find:

$$13a=-1\,\therefore\,a=-\frac{1}{13}$$

and so substituting into (5), we have:

$$36\left(-\frac{1}{13} \right)+c=8\,\therefore\,c=\frac{140}{13}$$

and then substituting into (1) to find $b$, we get:

$$36\left(-\frac{1}{13} \right)+6b+8+36\left(\frac{1}{13} \right)=8\,\therefore\,b=0$$

And so we have found:

$$P(x)=-\frac{1}{13}x^2+\frac{140}{13}=\frac{140-x^2}{13}$$

To ale and any other guests viewing this topic, I invite and encourage you to post other algebra problems here in our http://www.mathhelpboards.com/f2/ forum.

Best Regards,

Mark.

thank u :D
 
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