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lesantik [10]
3 years ago
6

Find the derivative of the function at P 0 in the direction of A. ​f(x,y,z) = 3 e^x cos(yz)​, P0 (0, 0, 0), A = - i + 2 j + 3k

Mathematics
1 answer:
Alik [6]3 years ago
3 0

The derivative of f(x,y,z) at a point p_0=(x_0,y_0,z_0) in the direction of a vector \vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k is

\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}

We have

f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k

and

\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}

Then the derivative at p_0 in the direction of \vec a is

3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}

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Answer:

Step-by-step explanation:

-2x + 3y – 4z = 8    ----------------(I)

5x – 3y + 5z = -8    -----------------(II)

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Add equation (I) & (II) and thus y will be eliminated

(I)        -2x + 3y – 4z = 8

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Multiply equation (II) by (-1) and then add with equation (III). Thus y will be eliminated.

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                     2x         -2z   = 16   ---------------(B)

Multiply equation (A) by 2 and then add. Thus z will be eliminated and we will get the value of x

(A) * 2          6x + 2z = 0

(B)                 <u>2x - 2z  = 16</u>   {Add}

                     8x         = 16

Divide both sides by 8

                             x = 16/8

                          x = 2

Plugin x = 2 in equation (A)

3x + z = 0

3*2 + z = 0

 6 + z = 0

       z = -6

Plug in x = 2 and z  = - 6 in equation (I)

-2x +3y - 4z = 8

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               3y = 8 - 20

               3y = -12

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