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In-s [12.5K]
3 years ago
11

Solve The Inequality

Mathematics
2 answers:
sineoko [7]3 years ago
8 0

Answer:

B

Step-by-step explanation:

I just solved it now

you are welcome

Gelneren [198K]3 years ago
5 0

Answer:

x∈[13,+∞]

Step-by-step explanation:

x≥13

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Verify that the function u(x, y, z) = log x^2 + y^2 is a solution of the two dimensional Laplace equation u_xx + u_yy = 0 everyw
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Answer:

The function  u(x,y,z)=log ( x^{2} +y^{2}) is indeed a solution of the two dimensional Laplace equation  u_{xx} +u_{yy} =0.

The wave equation  u_{tt} =u_{xx} is satisfied by the function u(x,t)=cos(4x)cos(4t) but not by the function u(x,t)=f(x-t)+f(x+1).

Step-by-step explanation:

To verify that the function  u(x,y,z)=log ( x^{2} +y^{2}) is a solution of the 2D Laplace equation we calculate the second partial derivative with respect to x and then with respect to t.

u_{xx}=\frac{2}{ln(10)}((x^{2} +y^{2})^{-1} -2x^{2} (x^{2} +y^{2})^{-2})

u_{yy}=\frac{2}{ln(10)}((x^{2} +y^{2})^{-1} -2y^{2} (x^{2} +y^{2})^{-2})

then we introduce it in the equation  u_{xx} +u_{yy} =0

we get that  \frac{2}{ln(10)} (\frac{2}{(x^{2}+y^{2}) } - \frac{2}{(x^{2}+y^{2} ) } )=0

To see if the functions 1) u(x,t)=cos(4x)cos(4t) and 2)    u(x,t)=f(x-t)+f(x+1) solve the wave equation we have to calculate the second partial derivative with respect to x and the with respect to t for each function. Then we see if they are equal.

1)  u_{xx}=-16 cos (4x) cos (4t)

   u_{tt}=-16cos(4x)cos(4t)

we see for the above expressions that  u_{tt} =u_{xx}

2) with this function we will have to use the chain rule

 If we call  s=x-t and  w=x+1  then we have that

 u(x,t)=f(x-t)+f(x+1)=f(s)+f(w)

So  \frac{\partial u}{\partial x}=\frac{df}{ds}\frac{\partial s}{\partial x} +\frac{df}{dw} \frac{\partial w}{\partial x}

because we have  \frac{\partial s}{\partial x} =1 and   \frac{\partial w}{\partial x} =1

then  \frac{\partial u}{\partial x} =f'(s)+f'(w)

⇒ \frac{\partial^{2} u }{\partial x^{2} } =\frac{\partial}{\partial x} (f'(s))+ \frac{\partial}{\partial x} (f'(w))

⇒\frac{\partial^{2} u }{ \partial x^{2} } =\frac{d}{ds} (f'(s))\frac{\partial s}{\partial x} +\frac{d}{ds} (f'(w))\frac{\partial w}{\partial x}

⇒ \frac{\partial^{2} u }{ \partial x^{2} } =f''(s)+f''(w)

Regarding the derivatives with respect to time

\frac{\partial u}{\partial t}=\frac{df}{ds} \frac{\partial s}{\partial t}+\frac{df}{dw} \frac{\partial w}{\partial t}=-\frac{df}{ds} =-f'(s)

then  \frac{\partial^{2} u }{\partial t^{2} } =\frac{\partial}{\partial t} (-f'(s))=-\frac{d}{ds} (f'(s))\frac{\partial s}{\partial t} =f''(s)

we see that  \frac{\partial^{2} u }{ \partial x^{2} } =f''(s)+f''(w) \neq f''(s)=\frac{\partial^{2} u }{\partial t^{2} }

u(x,t)=f(x-t)+f(x+1)  doesn´t satisfy the wave equation.

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The diagram is shown in the picture. The real numbers are divided into rational numbers and irrational numbers. Example of a rational number is any fraction like 1/2, while an irrational number is one that cannot be expressed as a fraction like π. Then, integers are also part of the rational numbers which include positive and negative whole number like -5, and +4. Whole numbers are positive numbers from 0 to whatever number. Lastly, the natural numbers are whole numbers with the exception of zero. So, these are numbers from 1 to infinity.

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