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Ymorist [56]
3 years ago
12

The profile of the cables on a suspension bridge may be modeled by a parabola. The central span of the bridge is 1210 m long and

128 m high. The parabola y equals 0.00035 x squared gives a good fit to the shape of the​ cables, where StartAbsoluteValue x EndAbsoluteValue less than or equals 605​, and x and y are measured in meters. Approximate the length of the cables that stretch between the tops of the two towers.
Mathematics
1 answer:
brilliants [131]3 years ago
7 0

Answer:

The approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

Step-by-step explanation:

The equation of the parabola is:

y=0.00035x^{2}

Compute the first order derivative of <em>y</em> as follows:

 y=0.00035x^{2}

\frac{\text{d}y}{\text{dx}}=\frac{\text{d}}{\text{dx}}[0.00035x^{2}]

    =2\cdot 0.00035x\\\\=0.0007x

Now, it is provided that |<em>x </em>| ≤ 605.

⇒ -605 ≤ <em>x</em> ≤ 605

Compute the arc length as follows:

\text{Arc Length}=\int\limits^{x}_{-x} {1+(\frac{\text{dy}}{\text{dx}})^{2}} \, dx

                  =\int\limits^{605}_{-605} {\sqrt{1+(0.0007x)^{2}}} \, dx \\\\={\displaystyle\int\limits^{605}_{-605}}\sqrt{\dfrac{49x^2}{100000000}+1}\,\mathrm{d}x\\\\={\dfrac{1}{10000}}}{\displaystyle\int\limits^{605}_{-605}}\sqrt{49x^2+100000000}\,\mathrm{d}x\\\\

Now, let

x=\dfrac{10000\tan\left(u\right)}{7}\\\\\Rightarrow u=\arctan\left(\dfrac{7x}{10000}\right)\\\\\Rightarrow \mathrm{d}x=\dfrac{10000\sec^2\left(u\right)}{7}\,\mathrm{d}u

\int dx={\displaystyle\int\limits}\dfrac{10000\sec^2\left(u\right)\sqrt{100000000\tan^2\left(u\right)+100000000}}{7}\,\mathrm{d}u

                  ={\dfrac{100000000}{7}}}{\displaystyle\int}\sec^3\left(u\right)\,\mathrm{d}u\\\\=\dfrac{50000000\ln\left(\tan\left(u\right)+\sec\left(u\right)\right)}{7}+\dfrac{50000000\sec\left(u\right)\tan\left(u\right)}{7}\\\\=\dfrac{50000000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+5000x\sqrt{\dfrac{49x^2}{100000000}+1}

Plug in the solved integrals in Arc Length and solve as follows:

\text{Arc Length}=\dfrac{5000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+\dfrac{x\sqrt{\frac{49x^2}{100000000}+1}}{2}|_{limits^{605}_{-605}}\\\\

                  =1245.253707795227\\\\\approx 1245.25

Thus, the approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

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The sum of the ages of Berma, her mother Rinna and her father Erwin is 80. Two years from now, Rinna’s age will be 13 less than
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Answer: Berma is 5 years old

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Erwin is 37 years old

Step-by-step explanation:

Let x represent Berma's age

Let y represent Rinna's age

Let z represent Erwin's age

Since the sum of their ages is 80,

x + y + z = 80 - - - - - - -1

Two years from now, Rinna’s age will be 13 less than the sum of Erwin’s age and twice Berma’s age. This means that

y +2 = [ (z+2) + 2(x+2) ] - 13

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2x - y + z = 13 + 2 - 4 -2

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Three years ago, 15 times Berma’s age was 5 less than the age of Rinna. It means that

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15x - y = 37 - - - - - - - -3

From equation 3, y = 15x - 37

Substituting y = 15x - 37 into equation 1 and equation 2, it becomes

x + 15x - 37 + z = 80

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subtracting equation 5 from equation 4,

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x = 145/29 = 5

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y = 15×5 -37

y = 38

Substituting x= 5 and y = 38 into equation 1, it becomes

5 + 38 + z = 80

z = 80 - 43

z = 37

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