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konstantin123 [22]
3 years ago
5

For the data points (1,1),(2,1/2),(3,1/3),(4,1/4), finde the natural cubic spline.

Engineering
1 answer:
kow [346]3 years ago
7 0

Answer:

y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12

Explanation:

A cubic has the form:

y = ax³ + bx² + cx + d

Given four points, we can write a system of equations:

1 = a + b + c + d

1/2 = 8a + 4b + 2c + d

1/3 = 27a + 9b + 3c + d

1/4 = 64a + 16b + 4c + d

Solving this algebraically would be time-consuming, but we can use matrices to make it easy.

\left[\begin{array}{cccc}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{array}\right]\left[\begin{array}{cccc}a\\b\\c\\d\end{array}\right]=\left[\begin{array}{cccc}1\\1/2\\1/3\\1/4\end{array}\right]

First, we find the inverse of the coefficient matrix.  This is messy to do by hand, so let's use a calculator:

\left[\begin{array}{cccc}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{array}\right] ^{-1} =-\frac{1}{12}\left[\begin{array}{cccc}2&-6&6&-2\\-18&48&-42&12\\52&-114&84&-22\\-48&72&-48&12\end{array}\right]

Now we multiply by the solution matrix (again using a calculator):

-\frac{1}{12} \left[\begin{array}{cccc}2&-6&6&-2\\-18&48&-42&12\\52&-114&84&-22\\-48&72&-48&12\end{array}\right]\left[\begin{array}{cccc}1\\1/2\\1/3\\1/4\end{array}\right] =\left[\begin{array}{cccc}-1/24\\5/12\\-35/24\\25/12\end{array}\right]

So the cubic is:

y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12

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baherus [9]

Answer:

88.18 W

Explanation:

The weight of the boy is given as 108 lb

Change to kg =108*0.453592= 48.988 kg = 49 kg

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F=49*9.81*sin 3.43° =28.17

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4 0
3 years ago
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BabaBlast [244]

Answer:

a)

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5 0
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If you are interested only in the temperature range of 20° to 40°C and the ADC has a 0 to 3V input range, design a signal condit
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Explanation:

Temperature range → 0 to 80'c

respective voltage output → 0.2v to 0.5v

required temperature range 20'c to 40'c

Where T = 20'c respective voltage

\begin{aligned}v_{20} &=0.2+\frac{0.5-0.8}{80} \times 20 \\&=0.2+\frac{0.3}{80} \times 20 \\V_{20} &=0.275 v\end{aligned}

\begin{aligned}\text { when } T=40^{\circ} C & \text { . } \\v_{40} &=0.2+\frac{0.5-0.2}{80} \times 40 \\&=0.35 V\end{aligned}

Therefore, Sensor output changes from 0.275v to 0.35volts for the ADC the required i/p should cover the dynamic range of ADC (ie - 0v to 3v)

so we have to design a circuit which transfers input voltage 0.275volts - 0.35v to 0 - 3v

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\begin{array}{l}v_{0}=\left(v_{i n}-0.275\right) G \\\sigma=\ldots \frac{3-0}{0.35-0.275}=3 / 0.075=40 \\v_{0}=\left(v_{i n}-0.275\right) 40\end{array}

The simplest circuit will be a op-amp

NOTE: Refer the figure attached

Vs is sensor output

Vr is the reference volt, Vr = 0.275v

\begin{aligned}v_{0}=& v_{s}-v_{v}\left(1+\frac{R_{2}}{R_{1}}\right) \\\Rightarrow & \frac{1+\frac{R_{2}}{R_{1}}}{2}=40 \\& \frac{R_{2}}{R_{1}}=39 \quad \Rightarrow\end{aligned}

choose R2, R1 such that it will maintain required  ratio

The output Vo can be connected to voltage buffer if you required better isolation.

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oksano4ka [1.4K]

Answer:

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