Answer:
![D_uT(3,1)=-\frac{44}{9}*\frac{1}{\sqrt{2} } \approx-3.46](https://tex.z-dn.net/?f=D_uT%283%2C1%29%3D-%5Cfrac%7B44%7D%7B9%7D%2A%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%20%7D%20%5Capprox-3.46)
Step-by-step explanation:
To find the rate of change of temperature with respect to distance at the point (3, 1) in the x-direction and the y-direction we need to find the Directional Derivative of T(x,y). The definition of the directional derivative is given by:
![D_uT(x,y)=T_x(x,y)i+T_y(x,y)j](https://tex.z-dn.net/?f=D_uT%28x%2Cy%29%3DT_x%28x%2Cy%29i%2BT_y%28x%2Cy%29j)
Where i and j are the rectangular components of a unit vector. In this case, the problem don't give us additional information, so let's asume:
![i=\frac{1}{\sqrt{2} }](https://tex.z-dn.net/?f=i%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%20%7D)
![j=\frac{1}{\sqrt{2} }](https://tex.z-dn.net/?f=j%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%20%7D)
So, we need to find the partial derivative with respect to x and y:
In order to do the things easier let's make the next substitution:
![u=2+x^2+y^2](https://tex.z-dn.net/?f=u%3D2%2Bx%5E2%2By%5E2)
and express T(x,y) as:
![T(x,y)=88*u^{-1}](https://tex.z-dn.net/?f=T%28x%2Cy%29%3D88%2Au%5E%7B-1%7D)
The partial derivative with respect to x is:
Using the chain rule:
![\frac{\partial u}{\partial x}=2x](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D%3D2x)
Hence:
![T_x(x,y)=88*(u^{-2})*\frac{\partial u}{\partial x}](https://tex.z-dn.net/?f=T_x%28x%2Cy%29%3D88%2A%28u%5E%7B-2%7D%29%2A%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D)
Symplying the expression and replacing the value of u:
![T_x(x,y)=\frac{-176x}{(2+x^2+y^2)^2}](https://tex.z-dn.net/?f=T_x%28x%2Cy%29%3D%5Cfrac%7B-176x%7D%7B%282%2Bx%5E2%2By%5E2%29%5E2%7D)
The partial derivative with respect to y is:
Using the chain rule:
![\frac{\partial u}{\partial y}=2y](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20y%7D%3D2y)
Hence:
![T_y(x,y)=88*(u^{-2})*\frac{\partial u}{\partial y}](https://tex.z-dn.net/?f=T_y%28x%2Cy%29%3D88%2A%28u%5E%7B-2%7D%29%2A%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20y%7D)
Symplying the expression and replacing the value of u:
![T_y(x,y)=\frac{-176y}{(2+x^2+y^2)^2}](https://tex.z-dn.net/?f=T_y%28x%2Cy%29%3D%5Cfrac%7B-176y%7D%7B%282%2Bx%5E2%2By%5E2%29%5E2%7D)
Therefore:
![D_uT(x,y)=(\frac{1}{\sqrt{2} } )*(\frac{-176x}{(2+x^2+y^2)^2} -\frac{176y}{(2+x^2+y^2)^2})](https://tex.z-dn.net/?f=D_uT%28x%2Cy%29%3D%28%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%20%7D%20%29%2A%28%5Cfrac%7B-176x%7D%7B%282%2Bx%5E2%2By%5E2%29%5E2%7D%20-%5Cfrac%7B176y%7D%7B%282%2Bx%5E2%2By%5E2%29%5E2%7D%29)
Evaluating the point (3,1)
![D_uT(3,1)=(\frac{1}{\sqrt{2} } )*(\frac{-176(3)-176(1)}{(2+3^2+1^2)^2})=(\frac{1}{\sqrt{2} })* (-\frac{704}{144})=(\frac{1}{\sqrt{2} }) ( - \frac{44}{9})\approx -3.46](https://tex.z-dn.net/?f=D_uT%283%2C1%29%3D%28%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%20%7D%20%29%2A%28%5Cfrac%7B-176%283%29-176%281%29%7D%7B%282%2B3%5E2%2B1%5E2%29%5E2%7D%29%3D%28%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%20%7D%29%2A%20%28-%5Cfrac%7B704%7D%7B144%7D%29%3D%28%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%20%7D%29%20%28%20-%20%5Cfrac%7B44%7D%7B9%7D%29%5Capprox%20-3.46)
Answer:
223454353.000000000000000000
Step-by-step explanation:
A refrigeration system of 60 kW capacity at an evaporator temperature of – 40°C and a condenser temperature of 20°C is needed in a food storage locker. The refrigerant R134a is sub-cooled by 10°C before entering the expansion valve. The vapour is 0.98 dry as it leaves the evaporator coil. The compression in the compressor is of adiabatic type.
![\qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctextit%7Bperfect%20square%20trinomial%7D%20%5C%5C%5C%5C%20%28a%5Cpm%20b%29%5E2%5Cimplies%20a%5E2%5Cpm%20%5Cstackrel%7B%5Cstackrel%7B%5Ctext%7B%5Csmall%202%7D%5Ccdot%20%5Csqrt%7B%5Ctextit%7B%5Csmall%20a%7D%5E2%7D%5Ccdot%20%5Csqrt%7B%5Ctextit%7B%5Csmall%20b%7D%5E2%7D%7D%7B%5Cdownarrow%20%7D%7D%7B2ab%7D%20%2B%20b%5E2)
the idea behind the completion of the square is simply using a perfect square trinomial, hmmm usually we do that by using our very good friend Mr Zero, 0.
if we look at the 2nd step, we have a group as x² - x, hmmm so we need a third element, which will be squared.
keeping in mind that the middle term of the perfect square trinomial is simply the product of the roots of "a" and "b", so in this case the middle term is "-x", and the 1st term is x², so we can say that
![\stackrel{middle~term}{2(\sqrt{x^2})(\sqrt{b^2})~}~ = ~~\stackrel{middle~term}{-x}\implies 2xb~~ = ~~~~ = ~~-x \\\\\\ b=\cfrac{-x}{2x} \implies b=-\cfrac{1}{2}](https://tex.z-dn.net/?f=%5Cstackrel%7Bmiddle~term%7D%7B2%28%5Csqrt%7Bx%5E2%7D%29%28%5Csqrt%7Bb%5E2%7D%29~%7D~%20%3D%20~~%5Cstackrel%7Bmiddle~term%7D%7B-x%7D%5Cimplies%202xb~~%20%3D%20~~~~%20%3D%20~~-x%20%5C%5C%5C%5C%5C%5C%20b%3D%5Ccfrac%7B-x%7D%7B2x%7D%20%5Cimplies%20b%3D-%5Ccfrac%7B1%7D%7B2%7D)
so that means that our missing third term for a perfect square trinomial is simply 1/2, now we'll go to our good friend Mr Zero, if we add (1/2)², we have to also subtract (1/2)², because all we're really doing is borrowing from Zero, so we'll be including then +(1/2)² and -(1/2)², keeping in mind that 1/4 - 1/4 = 0, so let's do that.
![-3~~ = ~~-2\left[ x^2-x+\left( \cfrac{1}{2} \right)^2 ~~ - ~~\left( \cfrac{1}{2} \right)^2\right]\implies -3=-2\left(x^2-x+\cfrac{1}{4}-\cfrac{1}{4} \right) \\\\\\ -3=-2\left(x^2-x+\cfrac{1}{4} \right)+(-2)-\cfrac{1}{4}\implies -3=-2\left(x^2-x+\cfrac{1}{4} \right)+\cfrac{1}{2} \\\\\\ -3-\cfrac{1}{2}=-2\left(x^2-x+\cfrac{1}{4} \right)\implies -\cfrac{7}{2}=-2\left(x-\cfrac{1}{2} \right)^2\implies \cfrac{7}{4}=\left(x-\cfrac{1}{2} \right)^2](https://tex.z-dn.net/?f=-3~~%20%3D%20~~-2%5Cleft%5B%20x%5E2-x%2B%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%20~~%20-%20~~%5Cleft%28%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%5Cright%5D%5Cimplies%20-3%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D-%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20-3%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%2B%28-2%29-%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20-3%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%2B%5Ccfrac%7B1%7D%7B2%7D%20%5C%5C%5C%5C%5C%5C%20-3-%5Ccfrac%7B1%7D%7B2%7D%3D-2%5Cleft%28x%5E2-x%2B%5Ccfrac%7B1%7D%7B4%7D%20%5Cright%29%5Cimplies%20-%5Ccfrac%7B7%7D%7B2%7D%3D-2%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%5Cimplies%20%5Ccfrac%7B7%7D%7B4%7D%3D%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2)
![~\dotfill\\\\ \pm\sqrt{\cfrac{7}{4}}=x-\cfrac{1}{2}\implies \cfrac{\pm\sqrt{7}}{2}=x-\cfrac{1}{2}\implies \cfrac{\pm\sqrt{7}}{2}+\cfrac{1}{2}=x \implies \cfrac{\pm\sqrt{7}+1}{2}=x](https://tex.z-dn.net/?f=~%5Cdotfill%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B%5Ccfrac%7B7%7D%7B4%7D%7D%3Dx-%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20%5Ccfrac%7B%5Cpm%5Csqrt%7B7%7D%7D%7B2%7D%3Dx-%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20%5Ccfrac%7B%5Cpm%5Csqrt%7B7%7D%7D%7B2%7D%2B%5Ccfrac%7B1%7D%7B2%7D%3Dx%20%5Cimplies%20%5Ccfrac%7B%5Cpm%5Csqrt%7B7%7D%2B1%7D%7B2%7D%3Dx)
Area depends on the product of sides,
so if the sides are shortened by a factor of 2, area will reduce by a factor of 4. (2×2)
new area = 10/4=2.5 cm²
(1) 3x+2y=-1
(2) 6x+4y=0
This system of equation has no solution. Proof
multiply all the term of (1) by two ==>6x+4y=-2
Notice that in (2) you have the same thing but equal to zero.
Same quantity can't equal 0 & -2. So no solution