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mylen [45]
3 years ago
12

I have to find the critical points of this function of two variables

4%2By%5E4-16%28x%2By%29%5E2%5C%5C" id="TexFormula1" title="\\f(x,y)=x^4+y^4-16(x+y)^2\\" alt="\\f(x,y)=x^4+y^4-16(x+y)^2\\" align="absmiddle" class="latex-formula">, for this we use the gradient method which equals to ZERO the first partial derivatives. After clearing both equations we should find the following points: P (__, 0), Q (4, __), R (4, __) My problem is that I cannot clear the following equations:
\\1)  4x^{3} -32x -32y = 0\\2) 4y^{3} -32x-32y = 0
Is there any way to deal with these equations of degree n?
Engineering
1 answer:
liraira [26]3 years ago
4 0

Answer:

no i dont think there is

Explanation:

because theres not

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Cars going straight or turning right have the right-of-way before you when you're making this type of turn
Paladinen [302]

Answer:

Yes. YES yes yes. Unless you are in Australia or something.

4 0
3 years ago
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
mario62 [17]

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

Therefore, the formula for the circuit will be

\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.

3 0
3 years ago
British standered institution
Scorpion4ik [409]

Answer:

The British Standards Institution, is the national standards body of the United Kingdom. BSI produces technical standards on a wide range of products and services and also supplies certification and standards-related services to businesses

6 0
3 years ago
A 3-ft-diameter vertical cylindrical tank open to the atmosphere contains 1-ft-high water. The tank is now rotated about the cen
arlik [135]

Answer:

The angular velocity is 7.56 rad/s

the maximum water height is 2 ft

Explanation:

The z-position as a function of r is equal to

z_{s(r)} =h_{0} -\frac{w^{2}(R^{2}-2r^{2}   }{4g} (eq. 1)

where

h0 = initial height = 1 ft

w = angular velocity

R = radius of the cylinder = 1.5 ft

zs(r) = 0 when the free surface is lowest at the centre

Replacing and clearing w

w=\sqrt{\frac{4gh_{0} }{R^{2} } } =\sqrt{\frac{4*32.17*1}{1.5^{2} } } =7.56rad/s

if you consider the equation 1 for the free surface at the edge is equal to

z_{s(R)} =h_{0} +\frac{w^{2}R^{2}   }{4g} =1+\frac{(7.56^{2})*(1.5^{2} ) }{4*32.17} =1.99ft=2ft

7 0
3 years ago
What is the difference between a refrigeration cycle and a heat pump cycle?
sukhopar [10]

Answer:

In refrigeration cycle heat transfer from inside refrigeration

In heat pump cycle heat transfer from environment

Explanation:

heat cycle is mechanical process use for cool the temperature but

In refrigeration heat transfer from inside of refrigeration that decrease temperature of refrigerator and in heat pump it decrease temperature negligible as compare to refrigerator

5 0
3 years ago
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