# Integration modelling task.

• November 1st 2010, 11:03 PM
MickG86
Hello,
I got this problem here which I can't really solve (Worried)

A Contractor is given the task of calculating the cost of a tunnel through a hill. The shape of the tunnel's cross section is made up of two curves as shown in the diagram. The upper curve is parabolic while the lower curve is sinusoidal (half cycle of sine curve 2 m deep). The tunnel is 10 m wide 6 m high at the wiedest points as shown in the diagram. Attachment 19556

The Task: Given that the tunnel is 1.25 km long and cast $120/m^3 to excavate, estimate the cost of excavation to the nearest$ 1000.

Thanks in advance (Bow)
• November 1st 2010, 11:44 PM
Unknown008
Well, you can find the cross sectional area of the two curves.

The sine curve has equation:

$y = -\sin\left(\dfrac{\pi x}{10}\right)$

The parabola has equation

y = ax(x-12)

Putting (6, 4), we get a = -1/9

Hence, equation of parabola: $y = -\dfrac{x}{9} (x - 12) = -\dfrac{x^2}{9} + \dfrac{4x}{3}$

Find the Area using:

$\displaystyle \int^{12}_0 -(-\sin\left(\dfrac{\pi x}{10}\right)) + (-\dfrac{x^2}{9} + \dfrac{4x}{3})\ dx$

Then, find the volume using

$V = Ad$

A is the cross sectional area and d is the depth of the prism.

The price will become trivial then.
• November 2nd 2010, 01:10 AM
MickG86
Thanks for the help

i understand the fuction for the sine curve but the parabola equation makes no sense to me?! isn't a parabola function y=a(x-h)^2+k

regards
• November 2nd 2010, 01:25 AM
Unknown008
That too, but I prefered using the form

$y = a(x-b)(x-c)$

where b and c are the roots of the parabola.

Since the roots occur at x = 0 and x = 12, this becomes y = a(x -0)(x - 12) = ax(x-12)

You can go with y = a(x-h)^2 +k too.

From the graph, we know that h = 6 and k = 4 then we use the point (0, 0)

0 = a(0 - 6)^2 + 4

a = -4/36 = -1/9

Which gives the same answer (Smile)
• November 2nd 2010, 02:26 AM
MickG86
ah ok...makes more sense to me now thanks (Itwasntme)