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2 years ago

8

Suppose an architect draws a segment on a scale drawing with the end points (0,0) and (3⁄4,9⁄10). The same segment on the actual

structure has the end points (0,0) and (30,36). What proportion could model this situation?

1 answer:

Kipish [7]2 years ago

6 0

Segment drawn on scale having end points O (0,0) and A $(\frac{3}{4},\frac{9}{10})$ is a line segment.

O A= $\sqrt{[\frac{3}{4} -0]^{2} +[\frac{9}{10} -0]^{2}\\\\$

O A = $= \sqrt{\frac{9}{16}+\frac{81}{100}$

= $\sqrt\frac{549}{400}$

= $\frac{\sqrt{549}}{20}$

Now , the same segment actual structure having end points O'(0,0) and B (30,36) is also a line segment.

O'B= $\sqrt{(30-0)^{2}+(36-0)^{2}$

= $\sqrt{900+1296}$

= $\sqrt {2196}$

= 2√549

$\frac{\text{Actual length}}{\text{Length on scale}}=\frac{OB'}{OA}=\frac{2\sqrt549}{\frac{\sqrt549}{20}}=40$ [Cancelling √549 from numerator and denominator]

So, Actual length = 40 × Length on scale

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An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t

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The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

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Split up the interval $[a,b]$ into $n$ equal subintervals,

$[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]$

where $a=x_0$ and $b=x_n$ . Each subinterval has measure (width) $\dfrac{a-b}n$ .

Now denote the left- and right-endpoint approximations by $L$ and $R$ , respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are $\{x_0,x_1,\cdots,x_{n-1}\}$ . Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, $\{x_1,x_2,\cdots,x_n\}$ .

So, you have

$L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)$
$R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)$

Now let $T$ denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

$T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)$

Factoring out $\dfrac12$ and regrouping the terms, you have

$T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)$

which is equivalent to

$T=\dfrac12\left(L+R)$

and is the average of $L$ and $R$ .

So the trapezoidal approximation for your problem should be $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$

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